wrong_submission_id
stringlengths 10
10
| problem_id
stringlengths 6
6
| user_id
stringlengths 10
10
| time_limit
float64 1k
8k
| memory_limit
float64 131k
1.05M
| wrong_status
stringclasses 2
values | wrong_cpu_time
float64 10
40k
| wrong_memory
float64 2.94k
3.37M
| wrong_code_size
int64 1
15.5k
| problem_description
stringlengths 1
4.75k
| wrong_code
stringlengths 1
6.92k
| acc_submission_id
stringlengths 10
10
| acc_status
stringclasses 1
value | acc_cpu_time
float64 10
27.8k
| acc_memory
float64 2.94k
960k
| acc_code_size
int64 19
14.9k
| acc_code
stringlengths 19
14.9k
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s416797330
|
p02578
|
u684849102
| 2,000
| 1,048,576
|
Wrong Answer
| 130
| 32,364
| 165
|
N persons are standing in a row. The height of the i-th person from the front is A_i. We want to have each person stand on a stool of some heights - at least zero - so that the following condition is satisfied for every person: Condition: Nobody in front of the person is taller than the person. Here, the height of a person includes the stool. Find the minimum total height of the stools needed to meet this goal.
|
n = int(input())
A = list(map(int, input().split()))
a=0
result=0
for i in range(n-1):
if A[i] > A[i+1]:
a = A[i] - A[i+1]
result += a
print(result)
|
s119727384
|
Accepted
| 154
| 32,188
| 181
|
n = int(input())
A = list(map(int, input().split()))
a=0
result=0
for i in range(n-1):
if A[i] > A[i+1]:
a = A[i] - A[i+1]
A[i+1] += a
result += a
print(result)
|
s168291316
|
p03470
|
u150985282
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 161
|
An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?
|
N = int(input())
d = list(map(int, [input() for i in range(N)]))
d.sort()
count = 0
for i in range(0, N-1):
if d[i] < d[i+1]:
count += 1
print(count)
|
s415994459
|
Accepted
| 17
| 3,064
| 219
|
N = int(input())
d = list(map(int, [input() for i in range(N)]))
bucket = [0 for i in range(110)]
for i in range(N):
bucket[d[i]] += 1
count = 0
for i in range(110):
if bucket[i] != 0:
count += 1
print(count)
|
s871243969
|
p03408
|
u129978636
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,188
| 291
|
Takahashi has N blue cards and M red cards. A string is written on each card. The string written on the i-th blue card is s_i, and the string written on the i-th red card is t_i. Takahashi will now announce a string, and then check every card. Each time he finds a blue card with the string announced by him, he will earn 1 yen (the currency of Japan); each time he finds a red card with that string, he will lose 1 yen. Here, we only consider the case where the string announced by Takahashi and the string on the card are exactly the same. For example, if he announces `atcoder`, he will not earn money even if there are blue cards with `atcoderr`, `atcode`, `btcoder`, and so on. (On the other hand, he will not lose money even if there are red cards with such strings, either.) At most how much can he earn on balance? Note that the same string may be written on multiple cards.
|
N = int( input())
s = list()
for i in range(N):
S = input()
s.append(S)
M = int( input())
t = list()
for j in range(M):
T = input()
t.append(T)
judge = 0
for l in range(N):
money = s.count(s[i]) - t.count(s[i])
if( judge < money):
judge = money
print('judge')
|
s456691019
|
Accepted
| 17
| 3,060
| 256
|
N = int( input())
s = [ input() for i in range(N)]
M = int(input())
t = [ input() for j in range(M)]
judge = 0
for l in range(N):
money = s.count(s[l]) - t.count(s[l])
if( judge < money):
judge = money
else:
continue
print(judge)
|
s848817675
|
p03854
|
u693048766
| 2,000
| 262,144
|
Wrong Answer
| 59
| 3,188
| 653
|
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
s = input()
curs = [0]
while True:
next_curs = []
break_ = False
for c in curs:
#print(c)
if s[c:c+5] == 'dream':
next_curs.append( c+5 )
if len(s) == c+5:
break_ = True
if s[c:c+7] == 'dreamer':
next_curs.append( c+7 )
if len(s) == c+7:
break_ = True
if s[c:c+5] == 'erase':
next_curs.append( c+5 )
if len(s) == c+5:
break_ = True
if s[c:c+6] == 'eraser':
next_curs.append( c+6 )
if len(s) == c+6:
break_ = True
if next_curs == []:
print('No')
break
if break_:
print('Yes')
break
curs = next_curs
|
s281689260
|
Accepted
| 60
| 3,188
| 653
|
s = input()
curs = [0]
while True:
next_curs = []
break_ = False
for c in curs:
#print(c)
if s[c:c+5] == 'dream':
next_curs.append( c+5 )
if len(s) == c+5:
break_ = True
if s[c:c+7] == 'dreamer':
next_curs.append( c+7 )
if len(s) == c+7:
break_ = True
if s[c:c+5] == 'erase':
next_curs.append( c+5 )
if len(s) == c+5:
break_ = True
if s[c:c+6] == 'eraser':
next_curs.append( c+6 )
if len(s) == c+6:
break_ = True
if next_curs == []:
print('NO')
break
if break_:
print('YES')
break
curs = next_curs
|
s618122535
|
p04012
|
u677393869
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 100
|
Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful.
|
N=input()
for i in N:
if N.count(i)%2==0:
pass
else:
print("NO")
exit()
print("YES")
|
s846259884
|
Accepted
| 25
| 9,004
| 239
|
N = input()
ans_num = 0
set_N = list(set(N))
for i in set_N:
if N.count(i) % 2 == 1:
ans_num = 1
break
else:
continue
if ans_num == 0:
print('Yes')
else :
print('No')
|
s485439972
|
p03455
|
u728774856
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 88
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b = map(int, input().split())
if (a * b ) % 2:
print('odd')
else:
print('even')
|
s298065925
|
Accepted
| 17
| 2,940
| 100
|
a, b = map(int, input().split())
if a % 2 == 0 or b % 2 == 0:
print('Even')
else:
print('Odd')
|
s969945549
|
p03998
|
u703890795
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 571
|
Alice, Bob and Charlie are playing _Card Game for Three_ , as below: * At first, each of the three players has a deck consisting of some number of cards. Each card has a letter `a`, `b` or `c` written on it. The orders of the cards in the decks cannot be rearranged. * The players take turns. Alice goes first. * If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says `a`, Alice takes the next turn.) * If the current player's deck is empty, the game ends and the current player wins the game. You are given the initial decks of the players. More specifically, you are given three strings S_A, S_B and S_C. The i-th (1≦i≦|S_A|) letter in S_A is the letter on the i-th card in Alice's initial deck. S_B and S_C describes Bob's and Charlie's initial decks in the same way. Determine the winner of the game.
|
Sa = input()
Sb = input()
Sc = input()
pl = 0
while(True):
if pl == 0:
s = Sa[0]
if s == "a":
pl = 0
elif s == "b":
pl = 1
else:
pl = 2
Sa = Sa[1:]
elif pl == 1:
s = Sb[0]
if s == "a":
pl = 0
elif s == "b":
pl = 1
else:
pl = 2
Sb = Sb[1:]
else:
s = Sc[0]
if s == "a":
pl = 0
elif s == "b":
pl = 1
else:
pl = 2
Sc = Sc[1:]
if len(Sa)==0:
print("A")
break
elif len(Sb)==0:
print("B")
break
elif len(Sc)==0:
print("C")
break
|
s484993650
|
Accepted
| 17
| 3,064
| 601
|
Sa = input()
Sb = input()
Sc = input()
pl = 0
while(True):
if len(Sa)==0 and pl==0:
print("A")
break
elif len(Sb)==0 and pl==1:
print("B")
break
elif len(Sc)==0 and pl==2:
print("C")
break
if pl == 0:
s = Sa[0]
if s == "a":
pl = 0
elif s == "b":
pl = 1
else:
pl = 2
Sa = Sa[1:]
elif pl == 1:
s = Sb[0]
if s == "a":
pl = 0
elif s == "b":
pl = 1
else:
pl = 2
Sb = Sb[1:]
else:
s = Sc[0]
if s == "a":
pl = 0
elif s == "b":
pl = 1
else:
pl = 2
Sc = Sc[1:]
|
s047332820
|
p00168
|
u844945939
| 1,000
| 131,072
|
Wrong Answer
| 30
| 6,748
| 165
|
一郎君の家の裏山には観音堂があります。この観音堂まではふもとから 30 段の階段があり、一郎君は、毎日のように観音堂まで遊びに行きます。一郎君は階段を１足で３段まで上がることができます。遊んでいるうちに階段の上り方の種類（段の飛ばし方の個数）が非常にたくさんあることに気がつきました。 そこで、一日に 10 種類の上り方をし、すべての上り方を試そうと考えました。しかし数学を熟知しているあなたはそんなことでは一郎君の寿命が尽きてしまうことを知っているはずです。 一郎君の計画が実現不可能であることを一郎君に納得させるために、階段の段数 n を入力とし、一日に 10 種類の上り方をするとして、一郎君がすべての上り方を実行するのに要する年数を出力するプログラムを作成してください。一年は 365 日として計算してください。一日でも必要なら一年とします。365 日なら 1 年であり、366 日なら 2 年となります。
|
a = [1, 1, 2]
for i in range(3, 31):
a.append(sum(a[-3:]))
while True:
n = int(input())
if not n:
break
print(((a[n] + 9) / 10 + 364) / 365)
|
s391713949
|
Accepted
| 30
| 6,720
| 161
|
a = [1]
for i in range(1, 31):
a.append(sum(a[-3:]))
while True:
n = int(input())
if not n:
break
print(((a[n] + 9) // 10 + 364) // 365)
|
s974875879
|
p03593
|
u405660020
| 2,000
| 262,144
|
Wrong Answer
| 24
| 3,680
| 810
|
We have an H-by-W matrix. Let a_{ij} be the element at the i-th row from the top and j-th column from the left. In this matrix, each a_{ij} is a lowercase English letter. Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A. Here, he wants to satisfy the following condition: * Every row and column in A' can be read as a palindrome. Determine whether he can create a matrix satisfying the condition.
|
from collections import Counter
import math
h, w = map(int,input().split())
a = [list(input()) for _ in range(h)]
c=Counter(sum(a,[]))
if h%2==0 and w%2==0:
g1=0
g2=0
g4=(h//2)*(w//2)
elif h%2==1 and w%2==0:
g1=0
g2=w//2
g4=(h//2)*(w//2)
elif h%2==0 and w%2==1:
g1=0
g2=h//2
g4=(h//2)*(w//2)
else:
g1=1
g2=(h//2)+(w//2)
g4=(h//2)*(w//2)
print(c)
g1_cnt=0
g2_cnt=0
g4_cnt=0
for key in c:
if c[key]%4==1:
g1_cnt+=1
g4_cnt+=c[key]//4
elif c[key]%4==2:
g2_cnt+=1
g4_cnt+=c[key]//4
elif c[key]%4==2:
g1_cnt+=1
g2_cnt+=1
g4_cnt+=c[key]//4
else:
g4_cnt+=c[key]//4
# print(g1,g2,g4)
# print(g1_cnt,g2_cnt,g4_cnt)
print('Yes' if g1==g1_cnt and g2==g2_cnt and g4==g4_cnt else 'No')
|
s206634291
|
Accepted
| 23
| 3,680
| 927
|
from collections import Counter
import math
h, w = map(int,input().split())
a = [list(input()) for _ in range(h)]
c=Counter(sum(a,[]))
if h%2==0 and w%2==0:
g1=0
g2=0
g4=(h//2)*(w//2)
elif h%2==1 and w%2==0:
g1=0
g2=w//2
g4=(h//2)*(w//2)
elif h%2==0 and w%2==1:
g1=0
g2=h//2
g4=(h//2)*(w//2)
else:
g1=1
g2=(h//2)+(w//2)
g4=(h//2)*(w//2)
g1_cnt=0
g2_cnt=0
g4_cnt=0
for key in c:
if c[key]%4==1:
g1_cnt+=1
g4_cnt+=c[key]//4
elif c[key]%4==2:
g2_cnt+=1
g4_cnt+=c[key]//4
elif c[key]%4==2:
g1_cnt+=1
g2_cnt+=1
g4_cnt+=c[key]//4
else:
g4_cnt+=c[key]//4
# print(g1,g2,g4)
# print(g1_cnt,g2_cnt,g4_cnt)
flag=False
if g1==g1_cnt:
if g2==g2_cnt and g4==g4_cnt:
flag=True
elif g2_cnt<g4_cnt and g2>g4 and g2*2+g4*4==g2_cnt*2+g4_cnt*4:
flag=True
print('Yes' if flag else 'No')
|
s047455216
|
p03338
|
u581603131
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 79
|
You are given a string S of length N consisting of lowercase English letters. We will cut this string at one position into two strings X and Y. Here, we would like to maximize the number of different letters contained in both X and Y. Find the largest possible number of different letters contained in both X and Y when we cut the string at the optimal position.
|
N = int(input())
S = input()
print(max(len(set[:i]&set[i:])) for i in range(N))
|
s411296864
|
Accepted
| 18
| 3,060
| 85
|
N = int(input())
S = input()
print(max(len(set(S[:i])&set(S[i:])) for i in range(N)))
|
s055604428
|
p03095
|
u595893956
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 86
|
You are given a string S of length N. Among its subsequences, count the ones such that all characters are different, modulo 10^9+7. Two subsequences are considered different if their characters come from different positions in the string, even if they are the same as strings. Here, a subsequence of a string is a concatenation of **one or more** characters from the string without changing the order.
|
s=input()
ret=1
for i in range(97, 97+26):
x=s.count(chr(i))
ret*=x+1
print(ret-1)
|
s090911769
|
Accepted
| 20
| 3,188
| 128
|
n=input()
s=input()
ret=1
for i in range(97, 97+26):
x=s.count(chr(i))
ret*=x+1
ret%=1000000007
print((ret-1)%1000000007)
|
s925192587
|
p02393
|
u619570677
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,452
| 264
|
Write a program which reads three integers, and prints them in ascending order.
|
a,b,c = list(map(int,input().split()))
if a < b:
if a < c:
if b < c:
print(a < b < c)
else:
print("a < c < b")
else:
print("c < a < b")
elif a > b:
if b < c:
if a < c:
print("b < a < c")
else:
print("b < c < a")
else:
print("c < b < a")
|
s462821406
|
Accepted
| 20
| 7,656
| 80
|
num = list(map(int, input().split()))
num.sort()
print(num[0], num[1], num[2])
|
s349681551
|
p02406
|
u870718588
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,460
| 249
|
In programming languages like C/C++, a goto statement provides an unconditional jump from the "goto" to a labeled statement. For example, a statement "goto CHECK_NUM;" is executed, control of the program jumps to CHECK_NUM. Using these constructs, you can implement, for example, loops. Note that use of goto statement is highly discouraged, because it is difficult to trace the control flow of a program which includes goto. Write a program which does precisely the same thing as the following program (this example is wrtten in C++). Let's try to write the program without goto statements. void call(int n){ int i = 1; CHECK_NUM: int x = i; if ( x % 3 == 0 ){ cout << " " << i; goto END_CHECK_NUM; } INCLUDE3: if ( x % 10 == 3 ){ cout << " " << i; goto END_CHECK_NUM; } x /= 10; if ( x ) goto INCLUDE3; END_CHECK_NUM: if ( ++i <= n ) goto CHECK_NUM; cout << endl; }
|
n = int(input())
for i in range(1, n + 1):
x = i
if x % 3 == 0:
print(" ", i, sep="")
else:
while x > 0:
if x % 10 == 3:
print(" ", i, sep="")
break
x //= 10
print()
|
s600654312
|
Accepted
| 30
| 8,216
| 124
|
n = int(input())
for i in range(1, n + 1):
if i % 3 == 0 or "3" in str(i):
print(" ", i, sep="", end="")
print()
|
s971259168
|
p03448
|
u814663076
| 2,000
| 262,144
|
Wrong Answer
| 48
| 3,060
| 254
|
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
import sys
sys.setrecursionlimit(10**6)
def input(): return sys.stdin.readline()
A, B, C, X = [int(input()) for i in range(4)]
cnt = 0
for a in range(A):
for b in range(B):
for c in range(C):
if 500*a + 100*b + 50*c == X:
cnt += 1
print(cnt)
|
s093346295
|
Accepted
| 54
| 3,060
| 260
|
import sys
sys.setrecursionlimit(10**6)
def input(): return sys.stdin.readline()
A, B, C, X = [int(input()) for i in range(4)]
cnt = 0
for a in range(A+1):
for b in range(B+1):
for c in range(C+1):
if 500*a + 100*b + 50*c == X:
cnt += 1
print(cnt)
|
s763739457
|
p03090
|
u674885198
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 4,244
| 396
|
You are given an integer N. Build an undirected graph with N vertices with indices 1 to N that satisfies the following two conditions: * The graph is simple and connected. * There exists an integer S such that, for every vertex, the sum of the indices of the vertices adjacent to that vertex is S. It can be proved that at least one such graph exists under the constraints of this problem.
|
n = int(input())
if n%2==0:
lis=[[i,n+1-i] for i in range(1,int(n/2)+1)]
else:
lis=[[i,n-i] for i in range(1,int((n-1)/2)+1)] + [[n]]
connect_lis=[]
for i in lis:
for j in i:
for k in lis:
if k!=i:
for p in k:
if j < p:
connect_lis.append([j,p])
for i in connect_lis:
print(i[0],'',i[1])
|
s292905651
|
Accepted
| 26
| 4,124
| 416
|
n = int(input())
if n%2==0:
lis=[[i,n+1-i] for i in range(1,int(n/2)+1)]
else:
lis=[[i,n-i] for i in range(1,int((n-1)/2)+1)] + [[n]]
connect_lis=[]
for i in lis:
for j in i:
for k in lis:
if k!=i:
for p in k:
if j < p:
connect_lis.append([j,p])
print(len(connect_lis))
for i in connect_lis:
print(i[0],i[1])
|
s012256307
|
p02390
|
u498511622
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,504
| 82
|
Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively.
|
s=int(input())
h=s/3600
s=s%3600
m=s/60
s=s%60
print(str(h),":",str(m),":",str(s))
|
s550008947
|
Accepted
| 20
| 7,740
| 143
|
import math
s=int(input())
h=math.floor(s/3600)
s=math.floor(s%3600)
m=math.floor(s/60)
s=math.floor(s%60)
print(str(h)+":"+str(m)+":"+str(s))
|
s711063757
|
p03149
|
u794173881
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,064
| 324
|
You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974".
|
n1,n2,n3,n4 = map(int,input().split())
if n1==1 or n2==1 or n3==1 or n4==1:
if n1==7 or n2==7 or n3==7 or n4==7:
if n1==9 or n2==9 or n3==9 or n4==9:
if n1==4 or n2==4 or n3==4 or n4==4:
print("Yes")
else:
print("No")
else:
print("No")
else:
print("No")
else:
print("No")
|
s090814383
|
Accepted
| 17
| 3,064
| 325
|
n1,n2,n3,n4 = map(int,input().split())
if n1==1 or n2==1 or n3==1 or n4==1:
if n1==7 or n2==7 or n3==7 or n4==7:
if n1==9 or n2==9 or n3==9 or n4==9:
if n1==4 or n2==4 or n3==4 or n4==4:
print("YES")
else:
print("NO")
else:
print("NO")
else:
print("NO")
else:
print("NO")
|
s440868245
|
p02408
|
u845643816
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,760
| 224
|
Taro is going to play a card game. However, now he has only n cards, even though there should be 52 cards (he has no Jokers). The 52 cards include 13 ranks of each of the four suits: spade, heart, club and diamond.
|
import itertools
n = int(input())
c = list(itertools.product(['S', 'H', 'C', 'D'], range(1,14)))
for _ in range(n):
s, num = input().split()
c.remove((s, int(num)))
for i in range(4*13 - n):
print(*c, sep = '\n')
|
s764087573
|
Accepted
| 30
| 7,744
| 226
|
import itertools
n = int(input())
c = list(itertools.product(['S', 'H', 'C', 'D'], range(1,14)))
for _ in range(n):
s, num = input().split()
c.remove((s, int(num)))
for i in range(4*13 - n):
print(c[i][0], c[i][1])
|
s993814438
|
p04030
|
u016622494
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 283
|
Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key. To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string: * The `0` key: a letter `0` will be inserted to the right of the string. * The `1` key: a letter `1` will be inserted to the right of the string. * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted. Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?
|
S = input()
str = []
print(len(S))
result = ""
flg = 0
for i in range(len(S)):
if flg == 1:
flg = 0
continue
if S[len(S)-1 - i] == 'b':
flg = 1
else:
str.append(S[len(S) - 1 - i])
str.reverse()
for i in str:
result += i
print(result)
|
s642670821
|
Accepted
| 17
| 2,940
| 161
|
S = input()
str = []
result = ""
flg = 0
for i in range(len(S)):
if S[i] == 'B':
result = result[:-1]
else:
result += S[i]
print(result)
|
s043054890
|
p03720
|
u694946470
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 229
|
There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
|
N, M = map(int,input().split())
a = []
b = []
for i in range(M):
a.append(list(map(int, input().split())))
print (a)
a=[flatten for inner in a for flatten in inner]
for i in range(N):
b.append((a.count(i+1)))
print (b)
|
s353559777
|
Accepted
| 17
| 3,060
| 273
|
N, M = map(int,input().split())
a = []
b = []
for i in range(M):
a.append(list(map(int, input().split())))
a=[flatten for inner in a for flatten in inner]
for i in range(N):
b.append((a.count(i+1)))
c = list(map(str, b))
answer = '\n'.join(c)
print(answer)
|
s074075341
|
p03471
|
u590048048
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 3,572
| 678
|
The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough.
|
n, y = map(int, input().split())
y = y // 1000
print(n, y)
def find_by_for(v, n):
"""
the tuple by example
((0, 0, 0), (10, 5, 1), 100, 10)
"""
queue = []
queue.append(((0, 0, 0), (1, 5, 10), v, n))
while queue:
xs, fv, v, n = queue.pop()
if v == 0 and n == 0:
return "%d %d %d"%(xs[2], xs[1], xs[0])
elif not fv or v <= 0 or n <= 0:
pass
else:
queue.append((xs, fv[:-1], v, n))
z = list(xs)
idx = len(fv) -1
z[idx] +=1
queue.append((tuple(z), fv, v - fv[idx], n -1))
return "-1 -1 -1"
print(find_by_for(y, n))
|
s050010105
|
Accepted
| 19
| 3,064
| 920
|
def find_min(v):
x = v % 10
return (x % 5, x//5, v // 10)
def find_max(v):
return (v, 0, 0)
def conv10to5(xs):
return (xs[0], xs[1] +2, xs[2] - 1)
def undoconv10to5(xs):
return (xs[0], xs[1] -2, xs[2] + 1)
def conv5to1(xs):
return (xs[0]+5, xs[1] -1, xs[2])
def next(xs):
if xs[2] == 0:
if xs[1] == 0:
return None
else:
return conv5to1(xs)
else:
if xs[1] < 7:
return conv10to5(xs)
else:
return conv5to1(undoconv10to5(undoconv10to5(undoconv10to5(xs))))
def find(v, n):
xs = find_min(v)
m = sum(xs)
if m > n or n > v:
return (-1, -1, -1)
while xs:
if sum(xs) == n:
return xs
xs = next(xs)
return (-1, -1, -1)
if __name__ == "__main__":
n, y = map(int, input().split())
y = y // 1000
print("{2} {1} {0}".format(*find(y, n)))
|
s848056225
|
p03503
|
u590048048
| 2,000
| 262,144
|
Wrong Answer
| 240
| 3,064
| 520
|
Joisino is planning to open a shop in a shopping street. Each of the five weekdays is divided into two periods, the morning and the evening. For each of those ten periods, a shop must be either open during the whole period, or closed during the whole period. Naturally, a shop must be open during at least one of those periods. There are already N stores in the street, numbered 1 through N. You are given information of the business hours of those shops, F_{i,j,k}. If F_{i,j,k}=1, Shop i is open during Period k on Day j (this notation is explained below); if F_{i,j,k}=0, Shop i is closed during that period. Here, the days of the week are denoted as follows. Monday: Day 1, Tuesday: Day 2, Wednesday: Day 3, Thursday: Day 4, Friday: Day 5. Also, the morning is denoted as Period 1, and the afternoon is denoted as Period 2. Let c_i be the number of periods during which both Shop i and Joisino's shop are open. Then, the profit of Joisino's shop will be P_{1,c_1}+P_{2,c_2}+...+P_{N,c_N}. Find the maximum possible profit of Joisino's shop when she decides whether her shop is open during each period, making sure that it is open during at least one period.
|
def profit(xs, n, F, P):
return sum([P[i][sum(map(lambda x: x[0]*x[1], zip(xs, F[i])))] for i in range(n)])
def vector(k):
return [1 if k & 1 << i else 0 for i in range(10)]
def foo(n, F, P):
return max([profit(vector(i), n, F, P) for pat in range(1, 1024)])
if __name__ == "__main__":
n = int(input())
F = {}
P = {}
for i in range(n):
F[i] = list(map(int, input().split()))
for i in range(n):
P[i] = list(map(int, input().split()))
print(foo(n, F, P))
|
s999096484
|
Accepted
| 239
| 3,188
| 522
|
def profit(xs, n, F, P):
return sum([P[i][sum(map(lambda x: x[0]*x[1], zip(xs, F[i])))] for i in range(n)])
def vector(k):
return [1 if k & 1 << i else 0 for i in range(10)]
def foo(n, F, P):
return max([profit(vector(pat), n, F, P) for pat in range(1, 1024)])
if __name__ == "__main__":
n = int(input())
F = {}
P = {}
for i in range(n):
F[i] = list(map(int, input().split()))
for i in range(n):
P[i] = list(map(int, input().split()))
print(foo(n, F, P))
|
s019402865
|
p02264
|
u548155360
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,616
| 331
|
_n_ _q_ _name 1 time1_ _name 2 time2_ ... _name n timen_ In the first line the number of processes _n_ and the quantum _q_ are given separated by a single space. In the following _n_ lines, names and times for the _n_ processes are given. _name i_ and _time i_ are separated by a single space.
|
n, q = map(int, input().split())
p = []
time = 0
for i in range(n):
p.append(input().split())
p[i][1] = int(p[i][1])
print("{0} {1}".format(p[0][0], time))
while(len(p) > 0):
if (p[0][1]<= q):
time += p[0][1]
print("{0} {1}".format(p[0][0], time))
p.pop(0)
else:
time += q
p[0][1] -= q
p.append(p[0])
p.pop(0)
|
s834183402
|
Accepted
| 820
| 19,428
| 291
|
n, q = map(int, input().split())
p = []
time = 0
for i in range(n):
p.append(input().split())
p[i][1] = int(p[i][1])
while(len(p) > 0):
if (p[0][1]<= q):
time += p[0][1]
print("{0} {1}".format(p[0][0], time))
p.pop(0)
else:
time += q
p[0][1] -= q
p.append(p[0])
p.pop(0)
|
s720864444
|
p02607
|
u505026996
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 9,160
| 121
|
We have N squares assigned the numbers 1,2,3,\ldots,N. Each square has an integer written on it, and the integer written on Square i is a_i. How many squares i satisfy both of the following conditions? * The assigned number, i, is odd. * The written integer is odd.
|
lst=list(map(int,input().split()))
cnt=0
for i,j in enumerate(lst,1):
if i%2!=0 and j%2!=0:
cnt+=1
print(cnt)
|
s382812097
|
Accepted
| 26
| 9,116
| 202
|
n=int(input())
lst=list(map(int,input().split()))
cnt=0
for i,j in enumerate(lst,1):
if i%2==0:
pass
else:
if j%2==0:
pass
else:
cnt+=1
print(cnt)
|
s764439826
|
p02393
|
u933096856
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,680
| 45
|
Write a program which reads three integers, and prints them in ascending order.
|
print(list(sorted(map(int,input().split()))))
|
s595915061
|
Accepted
| 50
| 7,672
| 46
|
print(*list(sorted(map(int,input().split()))))
|
s702991252
|
p00004
|
u957021485
| 1,000
| 131,072
|
Time Limit Exceeded
| 17,630
| 7,676
| 390
|
Write a program which solve a simultaneous equation: ax + by = c dx + ey = f The program should print x and y for given a, b, c, d, e and f (-1,000 ≤ a, b, c, d, e, f ≤ 1,000). You can suppose that given equation has a unique solution.
|
import itertools
import operator
dataset = []
while True:
try:
dataset.append(input())
except EOFError:
break
for item in dataset:
a, b, c, d, e, f = [int(i) for i in item.split()]
for i in range(-1000, 1001):
for j in range(-1000, 1001):
if a * i + b * j == c and d * i + e * j == f:
print("{:.3f} {:.3f}".format(i, j))
|
s494389333
|
Accepted
| 20
| 7,588
| 268
|
import sys
dataset = sys.stdin.readlines()
for item in dataset:
a, b, c, d, e, f = list(map(int, item.split()))
y = (c * d - f * a)/(b * d - e * a)
x = (c * e - b * f) / (a * e - b * d)
if x == 0:
x = 0
print("{:.3f} {:.3f}".format(x, y))
|
s095948087
|
p03456
|
u413377603
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,068
| 63
|
AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.
|
import math
a,b=map(str,input().split())
d=math.sqrt(int(a+b))
|
s310708390
|
Accepted
| 18
| 2,940
| 103
|
import math
a,b=input().split()
d=int(a+b)
if math.sqrt(d)%1==0:
print("Yes")
else:
print("No")
|
s114630536
|
p03386
|
u951601135
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,060
| 99
|
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
a,b,k=map(int,input().split())
for i in range(a,a+k):
print(i)
for i in range(b-k,b):
print(i)
|
s841698849
|
Accepted
| 22
| 3,060
| 144
|
a,b,k=map(int,input().split())
r=range(a,b+1)
for i in sorted(set(r[:k])|set(r[-k:])):print(i)
|
s085581626
|
p02613
|
u727787724
| 2,000
| 1,048,576
|
Wrong Answer
| 148
| 9,136
| 214
|
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
n=int(input())
ans=[0]*4
for i in range(n):
s=input()
if s=='AC':
ans[0]+=1
elif s=='WA':
ans[1]+=1
elif s=='TLE':
ans[2]+=1
else:
ans[3]+=1
for i in range(4):
print("AC x "+str(ans[i]))
|
s661605824
|
Accepted
| 148
| 9,024
| 276
|
n=int(input())
ans=[0]*4
for i in range(n):
s=input()
if s=='AC':
ans[0]+=1
elif s=='WA':
ans[1]+=1
elif s=='TLE':
ans[2]+=1
else:
ans[3]+=1
print("AC x "+str(ans[0]))
print("WA x "+str(ans[1]))
print("TLE x "+str(ans[2]))
print("RE x "+str(ans[3]))
|
s547810987
|
p03644
|
u102242691
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 82
|
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
n = int(input())
count = 0
while 2 ** count <= 100:
count += 1
print(count)
|
s699693925
|
Accepted
| 17
| 2,940
| 89
|
n = int(input())
count = 0
while 2 ** count <= n:
count += 1
print(2 ** (count-1))
|
s456278933
|
p03129
|
u518064858
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 83
|
Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.
|
n,k=map(int,input().split())
x=n//2
if x<k:
print("NO")
else:
print("YES")
|
s892349247
|
Accepted
| 17
| 2,940
| 87
|
n,k=map(int,input().split())
x=-(-n//2)
if x<k:
print("NO")
else:
print("YES")
|
s144121063
|
p00043
|
u724963150
| 1,000
| 131,072
|
Time Limit Exceeded
| 40,000
| 7,580
| 835
|
1 〜 9 の数字を 14 個組み合わせて完成させるパズルがあります。与えられた 13 個の数字にもうひとつ数字を付け加えて完成させます。 パズルの完成条件は * 同じ数字を２つ組み合わせたものが必ずひとつ必要です。 * 残りの12 個の数字は、３個の数字の組み合わせ４つです。 ３個の数字の組み合わせ方は、同じ数字を３つ組み合わせたものか、または３つの連続する数字を組み合わせたものです。ただし、9 1 2 のような並びは連続する数字とは認められません。 * 同じ数字は4 回まで使えます。 13 個の数字からなる文字列を読み込んで、パズルを完成することができる数字を昇順に全て出力するプログラムを作成してください。なお、1〜9 のどの数字を付け加えてもパズルを完成させることができないときは 0 を出力してください。 例えば与えられた文字列が 3456666777999 の場合 「2」があれば、 234 567 666 77 999 「3」があれば、 33 456 666 777 999 「5」があれば、 345 567 666 77 999 「8」があれば、 345 666 678 77 999 というふうに、2 3 5 8 のいずれかの数字が付け加えられるとパズルは完成します。「6」でも整いますが、5 回目の使用になるので、この例では使えないことに注意してください。
|
def Solve(c,s):
if s:
for i in range(9):
if c[i]>4:return False
elif c[i]>=2:
cc=c[:]
cc[i]-=2
if Solve(cc,False):return True
else:
if c.count(0)+c.count(3)==9:return True
else:
for i in range(9):
if c[i]<0:return False
elif c[i]>=1 and i<7:
cc=c[:]
for j in range(3):
cc[i+j]-=1
if Solve(cc,False):return True
_in=""
while True:
ans=[]
try:_in="9118992346175"
except EOFError:break
for i in range(9):
l=sorted((_in+str(i+1))[::1])
count=[l.count(str(j+1))for j in range(9)]
if Solve(count,True):ans.append(str(i+1))
print(0 if len(ans)==0 else " ".join(ans))
|
s613447063
|
Accepted
| 80
| 7,472
| 972
|
def Solve(c,s):
if s:
if max(c)>4:return False
for i in range(9):
if c[i]>=2:
cc=c[:]
cc[i]-=2
if Solve(cc,False):return True
else:
check=0
for i in range(4):check+=c.count(3*i)
if check==9:return True
else:
for i in range(7):
if c[i]>=1:
cc=c[:]
isneg=False
for j in range(3):
cc[i+j]-=1
if cc[i+j]<0:
isneg=True
break
if isneg==False and Solve(cc,False):return True
_in=""
while True:
ans=[]
try:_in=input()
except EOFError:break
for i in range(9):
l=(_in+str(i+1))[::1]
count=[l.count(str(j+1))for j in range(9)]
if Solve(count,True):ans.append(str(i+1))
print(0 if len(ans)==0 else " ".join(ans))
|
s616405815
|
p00015
|
u650459696
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,540
| 143
|
A country has a budget of more than 81 trillion yen. We want to process such data, but conventional integer type which uses signed 32 bit can represent up to 2,147,483,647. Your task is to write a program which reads two integers (more than or equal to zero), and prints a sum of these integers. If given integers or the sum have more than 80 digits, print "overflow".
|
l = int(input())
for i in range(l):
a = int(input()) + int(input())
if(a >= 1E79):
print('overflow')
else:
print(a)
|
s171687918
|
Accepted
| 50
| 7,572
| 150
|
l = int(input())
for i in range(l):
a = int(input()) + int(input())
if(len(str(a)) > 80):
print('overflow')
else:
print(a)
|
s440951212
|
p03713
|
u761989513
| 2,000
| 262,144
|
Wrong Answer
| 255
| 3,064
| 316
|
There is a bar of chocolate with a height of H blocks and a width of W blocks. Snuke is dividing this bar into exactly three pieces. He can only cut the bar along borders of blocks, and the shape of each piece must be a rectangle. Snuke is trying to divide the bar as evenly as possible. More specifically, he is trying to minimize S_{max} \- S_{min}, where S_{max} is the area (the number of blocks contained) of the largest piece, and S_{min} is the area of the smallest piece. Find the minimum possible value of S_{max} - S_{min}.
|
h, w = map(int, input().split())
ans = float("inf")
for i in range(h):
choco = [i * w, (h - i) * (w // 2), (h - i) * (w - w // 2)]
ans = min(ans, max(choco) - min(choco))
for i in range(w):
choco = [i * h, (w - i) * (h // 2), (w - i) * (h - h // 2)]
ans = min(ans, max(choco) - min(choco))
print(ans)
|
s738421579
|
Accepted
| 517
| 3,064
| 582
|
h, w = map(int, input().split())
ans = float("inf")
for i in range(h):
choco = [i * w, (h - i) * (w // 2), (h - i) * (w - w // 2)]
ans = min(ans, max(choco) - min(choco))
for i in range(w):
choco = [i * h, (w - i) * (h // 2), (w - i) * (h - h // 2)]
ans = min(ans, max(choco) - min(choco))
for i in range(h):
choco = [i * w, w * ((h - i) // 2), w * ((h - i) - (h - i) // 2)]
ans = min(ans, max(choco) - min(choco))
for i in range(w):
choco = [i * h, h * ((w - i) // 2), h * ((w - i) - (w - i) // 2)]
ans = min(ans, max(choco) - min(choco))
print(ans)
|
s715535817
|
p02612
|
u942697937
| 2,000
| 1,048,576
|
Wrong Answer
| 29
| 9,136
| 34
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N = int(input())
print(N % 1000)
|
s717702269
|
Accepted
| 28
| 9,160
| 82
|
N = int(input())
if N % 1000 == 0:
print(0)
else:
print(1000 - N % 1000)
|
s975502520
|
p00065
|
u811733736
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,800
| 920
|
取引先の顧客番号と取引日を月ごとに記録したデータがあります。今月のデータと先月のデータを読み込んで、先月から２ヶ月連続で取引のある会社の顧客番号と取引のあった回数を出力するプログラムを作成してください。ただし、月々の取引先数は 1,000 社以内です。
|
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0065
"""
import sys
from collections import Counter
def analyze_data(this_month, last_month):
result = []
tm = Counter(this_month)
lm = Counter(last_month)
for ele in lm:
if ele in tm:
c = lm[ele] + tm[ele]
result.append([ele, c])
return result
def main(args):
this_month = []
last_month = []
month = this_month
for line in sys.stdin:
if len(line) == 1: # ??\??????????????????
month = last_month
else:
id, date = line.strip().split(',')
month.append(id)
#this_month = [123, 56, 34]
#last_month = [123, 56, 123]
result = analyze_data(this_month, last_month)
result.sort()
for d in result:
print('{} {}'.format(d[0], d[1]))
if __name__ == '__main__':
main(sys.argv[1:])
|
s730385182
|
Accepted
| 30
| 7,896
| 964
|
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0065
"""
import sys
from collections import Counter
def analyze_data(this_month, last_month):
result = []
tm = Counter(this_month)
lm = Counter(last_month)
for ele in lm:
if ele in tm:
c = lm[ele] + tm[ele]
result.append([ele, c])
result.sort()
return result
def main(args):
this_month = []
last_month = []
month = this_month
for line in sys.stdin:
if len(line) == 1: # ??\??????????????????
month = last_month
else:
id, date = line.strip().split(',')
month.append(int(id))
# this_month = [1, 123, 56, 34, 23, 1, 23]
result = analyze_data(this_month, last_month)
for d in result:
print('{} {}'.format(d[0], d[1]))
if __name__ == '__main__':
main(sys.argv[1:])
|
s385742857
|
p03006
|
u891847179
| 2,000
| 1,048,576
|
Wrong Answer
| 123
| 27,324
| 1,586
|
There are N balls in a two-dimensional plane. The i-th ball is at coordinates (x_i, y_i). We will collect all of these balls, by choosing two integers p and q such that p \neq 0 or q \neq 0 and then repeating the following operation: * Choose a ball remaining in the plane and collect it. Let (a, b) be the coordinates of this ball. If we collected a ball at coordinates (a - p, b - q) in the previous operation, the cost of this operation is 0. Otherwise, including when this is the first time to do this operation, the cost of this operation is 1. Find the minimum total cost required to collect all the balls when we optimally choose p and q.
|
# # Make IO faster
# import sys
# input = sys.stdin.readline
# X = input()
# N = int(input())
# X, Y = map(int, input().split())
for N lines
# XY = [list(map(int, input().split())) for _ in range(N)]
# from IPython import embed; embed(); exit();
import sys, re
from collections import deque, defaultdict, Counter
from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians
from itertools import accumulate, permutations, combinations, product
from operator import itemgetter, mul
from copy import deepcopy
from string import ascii_lowercase, ascii_uppercase, digits
from bisect import bisect, bisect_left
from fractions import gcd
from heapq import heappush, heappop
from functools import reduce
import numpy as np
def input(): return sys.stdin.readline().strip()
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))
def ZIP(n): return zip(*(MAP() for _ in range(n)))
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
mod = 10 ** 9 + 7
N = INT()
coords = []
for i in range(N):
x, y = MAP()
coords.append((x, y))
dic = defaultdict(int)
pq_pairs = []
for i in range(N):
for j in range(N):
if i >= j: continue
p = coords[i][0] - coords[j][0]
q = coords[i][1] - coords[j][1]
if p < 0:
p = -p
q = -q
dic[(p, q)] += 1
ans = 0
for k, v in dic.items():
if v > ans:
ans = v
print(N - ans)
|
s841929504
|
Accepted
| 35
| 9,648
| 617
|
ma = lambda :map(int,input().split())
lma = lambda :list(map(int,input().split()))
tma = lambda :tuple(map(int,input().split()))
ni = lambda:int(input())
yn = lambda fl:print("Yes") if fl else print("No")
import collections
import math
import itertools
import heapq as hq
n = ni()
if n==1:
print(1)
exit()
xy = []
for i in range(n):
xy.append(tma())
co = collections.Counter()
for i in range(n):
for j in range(n):
if i==j:
continue
x1,y1 = xy[i]
x2,y2 = xy[j]
co[(x2-x1,y2-y1)] +=1
ans = 10**9
for num in co.values():
ans = min(ans,n-num)
print(ans)
|
s460946295
|
p03455
|
u649558044
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 110
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
from math import sqrt
n = int(input().replace(' ',''))
m = int(sqrt(n)) ** 2
print('Yes' if n == m else 'No')
|
s200923607
|
Accepted
| 17
| 2,940
| 68
|
print('Odd' if eval(input().replace(' ', '*')) % 2 == 1 else 'Even')
|
s292113724
|
p03525
|
u905582793
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,064
| 479
|
In CODE FESTIVAL XXXX, there are N+1 participants from all over the world, including Takahashi. Takahashi checked and found that the _time gap_ (defined below) between the local times in his city and the i-th person's city was D_i hours. The time gap between two cities is defined as follows. For two cities A and B, if the local time in city B is d o'clock at the moment when the local time in city A is 0 o'clock, then the time gap between these two cities is defined to be min(d,24-d) hours. Here, we are using 24-hour notation. That is, the local time in the i-th person's city is either d o'clock or 24-d o'clock at the moment when the local time in Takahashi's city is 0 o'clock, for example. Then, for each pair of two people chosen from the N+1 people, he wrote out the time gap between their cities. Let the smallest time gap among them be s hours. Find the maximum possible value of s.
|
n=int(input())
d=list(map(int,input().split()))
d.sort()
time=[0]*24
time[0]=1
for i in range(n):
if d[i]==0:
print(0)
exit()
if i%2:
if time[d[i]]==1:
print(0)
exit()
time[d[i]]=1
else:
if time[24-d[i]]==1:
print(0)
exit()
time[24-d[i]]=1
flg=0
ans=0
cnt=0
for i in range(25):
i=i%24
if time[i]==0 and not flg:
flg=1
cnt=1
elif time[i]==0:
cnt+=1
else:
ans=max(ans,cnt)
cnt=0
print(ans)
|
s717561995
|
Accepted
| 17
| 3,064
| 314
|
n=int(input())
d=list(map(int,input().split()))
d.sort()
time=[0]*24
time[0]=1
for i in range(n):
if i%2:
time[d[i]]+=1
else:
time[-d[i]]+=1
if max(time)>1:
print(0)
exit()
ans=[]
cnt=1
for i in range(1,25):
i=i%24
if time[i]==0:
cnt+=1
else:
ans.append(cnt)
cnt=1
print(min(ans))
|
s204344691
|
p02277
|
u831244171
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,736
| 827
|
Let's arrange a deck of cards. Your task is to sort totally n cards. A card consists of a part of a suit (S, H, C or D) and an number. Write a program which sorts such cards based on the following pseudocode: Partition(A, p, r) 1 x = A[r] 2 i = p-1 3 for j = p to r-1 4 do if A[j] <= x 5 then i = i+1 6 exchange A[i] and A[j] 7 exchange A[i+1] and A[r] 8 return i+1 Quicksort(A, p, r) 1 if p < r 2 then q = Partition(A, p, r) 3 run Quicksort(A, p, q-1) 4 run Quicksort(A, q+1, r) Here, A is an array which represents a deck of cards and comparison operations are performed based on the numbers. Your program should also report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).
|
def partition(a,b,c,p,r):
x = a[r]
i = p - 1
for j in range(p,r):
if x >= a[j]:
i += 1
a[i],a[j] = a[j],a[i]
b[i],b[j] = b[j],b[i]
c[i],c[j] = c[j],c[i]
a[i+1],a[r] = a[r],a[i+1]
b[i+1],b[r] = b[r],b[i+1]
c[i+1],c[r] = c[r],c[i+1]
return i+1
def quickSort(a,b,c,p,r):
if p < r:
q = partition(a,b,c,p,r-1)
quickSort(a,b,c,p,q-1)
quickSort(a,b,c,q+1,r)
def checkStable(a,b):
for i in range(1,len(a)):
if a[i-1] == a[i]:
if b[i-1] > b[i]:
return "Not Stable"
return "Stable"
n = int(input())
mark = [""]*n
num = [0]*n
count = [i for i in range(n)]
for i in range(n):
mark[i],num[i] = input().split()
num = list(map(int,num))
quickSort(num,mark,count,0,n)
print(checkStable(num,count))
for i in range(n):
print(mark[i],num[i])
|
s912311296
|
Accepted
| 1,130
| 22,668
| 827
|
def partition(a,b,c,p,r):
x = a[r]
i = p - 1
for j in range(p,r):
if x >= a[j]:
i += 1
a[i],a[j] = a[j],a[i]
b[i],b[j] = b[j],b[i]
c[i],c[j] = c[j],c[i]
a[i+1],a[r] = a[r],a[i+1]
b[i+1],b[r] = b[r],b[i+1]
c[i+1],c[r] = c[r],c[i+1]
return i+1
def quickSort(a,b,c,p,r):
if p < r:
q = partition(a,b,c,p,r)
quickSort(a,b,c,p,q-1)
quickSort(a,b,c,q+1,r)
def checkStable(a,b):
for i in range(1,len(a)):
if a[i-1] == a[i]:
if b[i-1] > b[i]:
return "Not stable"
return "Stable"
n = int(input())
mark = [""]*n
num = [0]*n
count = [i for i in range(n)]
for i in range(n):
mark[i],num[i] = input().split()
num = list(map(int,num))
quickSort(num,mark,count,0,n-1)
print(checkStable(num,count))
for i in range(n):
print(mark[i],num[i])
|
s938619320
|
p03998
|
u330310077
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 239
|
Alice, Bob and Charlie are playing _Card Game for Three_ , as below: * At first, each of the three players has a deck consisting of some number of cards. Each card has a letter `a`, `b` or `c` written on it. The orders of the cards in the decks cannot be rearranged. * The players take turns. Alice goes first. * If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says `a`, Alice takes the next turn.) * If the current player's deck is empty, the game ends and the current player wins the game. You are given the initial decks of the players. More specifically, you are given three strings S_A, S_B and S_C. The i-th (1≦i≦|S_A|) letter in S_A is the letter on the i-th card in Alice's initial deck. S_B and S_C describes Bob's and Charlie's initial decks in the same way. Determine the winner of the game.
|
dk = {}
dk["a"] = input()
dk["b"] = input()
dk["c"] = input()
n = "a"
for i in range(0,100):
if len(dk[n]) == 0:
print("Break! finalist is {}".format(n))
break
nn = dk[n][0]
dk[n] = dk[n].lstrip(nn)
n = nn
|
s569632758
|
Accepted
| 17
| 2,940
| 281
|
def game(dk, first):
n = first
while True:
if len(dk[n]) == 0:
return n.upper()
break
nn = dk[n][0]
dk[n] = dk[n][1:]
n = nn
pass
dk = {}
dk["a"] = input()
dk["b"] = input()
dk["c"] = input()
print(game(dk,"a"))
|
s959422081
|
p02409
|
u639421643
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,656
| 547
|
You manage 4 buildings, each of which has 3 floors, each of which consists of 10 rooms. Write a program which reads a sequence of tenant/leaver notices, and reports the number of tenants for each room. For each notice, you are given four integers b, f, r and v which represent that v persons entered to room r of fth floor at building b. If v is negative, it means that −v persons left. Assume that initially no person lives in the building.
|
place = []
bn,fn,rn=[4,3,10]
for i in range(bn):
place.append([])
for j in range(fn):
place[i].append([])
for k in range(rn):
place[i][j].append(0)
count = int(input())
for i in range(count):
b,f,r,v = map(int, input().split())
place[b-1][f-1][r-1] += v
for i in range(bn):
for j in range(fn):
line = ""
for k in range(rn):
line += " "
line += str(place[i][j][k])
print(line)
mark = ""
for i in range(20):
mark += "#"
print(mark)
|
s654372249
|
Accepted
| 20
| 7,716
| 581
|
place = []
bn,fn,rn=[4,3,10]
for i in range(bn):
place.append([])
for j in range(fn):
place[i].append([])
for k in range(rn):
place[i][j].append(0)
count = int(input())
for i in range(count):
b,f,r,v = map(int, input().split())
place[b-1][f-1][r-1] += v
for i in range(bn):
for j in range(fn):
line = ""
for k in range(rn):
line += " "
line += str(place[i][j][k])
print(line)
if(i == bn -1):
break
mark = ""
for i in range(20):
mark += "#"
print(mark)
|
s050515405
|
p03302
|
u787562674
| 2,000
| 1,048,576
|
Wrong Answer
| 21
| 3,316
| 114
|
You are given two integers a and b. Determine if a+b=15 or a\times b=15 or neither holds. Note that a+b=15 and a\times b=15 do not hold at the same time.
|
a, b = map(int, input().split())
if a*b == 15:
print("*")
elif a*b == 15:
print("+")
else:
print("x")
|
s769102046
|
Accepted
| 18
| 2,940
| 114
|
a, b = map(int, input().split())
if a*b == 15:
print("*")
elif a+b == 15:
print("+")
else:
print("x")
|
s117886838
|
p03643
|
u724742135
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 64
|
This contest, _AtCoder Beginner Contest_ , is abbreviated as _ABC_. When we refer to a specific round of ABC, a three-digit number is appended after ABC. For example, ABC680 is the 680th round of ABC. What is the abbreviation for the N-th round of ABC? Write a program to output the answer.
|
from sys import stdin
n = stdin.readline().rstrip()
print('A'+n)
|
s062019291
|
Accepted
| 18
| 2,940
| 66
|
from sys import stdin
n = stdin.readline().rstrip()
print('ABC'+n)
|
s343290034
|
p03494
|
u038408819
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,104
| 3,060
| 145
|
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
N = int(input())
a = list(map(int, input().split()))
ans = 0
while all(A % 2 == 0 for A in a):
A = [i / 2 for i in a]
ans += 1
print(ans)
|
s666590318
|
Accepted
| 19
| 3,060
| 146
|
n = int(input())
a = list(map(int, input().split()))
ans = 0
while all(A % 2 == 0 for A in a):
a = [i / 2 for i in a]
ans += 1
print(ans)
|
s383524169
|
p03964
|
u312078744
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 3,060
| 315
|
AtCoDeer the deer is seeing a quick report of election results on TV. Two candidates are standing for the election: Takahashi and Aoki. The report shows the ratio of the current numbers of votes the two candidates have obtained, but not the actual numbers of votes. AtCoDeer has checked the report N times, and when he checked it for the i-th (1≦i≦N) time, the ratio was T_i:A_i. It is known that each candidate had at least one vote when he checked the report for the first time. Find the minimum possible total number of votes obtained by the two candidates when he checked the report for the N-th time. It can be assumed that the number of votes obtained by each candidate never decreases.
|
n = int(input())
x, y = map(int, input().split())
t, a = x, y
for _ in range(n - 1):
tt, aa = map(int, input().split())
c = 1
if (t >= a):
while (tt * c < t):
c += 1
else:
while (aa * c < a):
c += 1
t = tt * c
a = aa * c
ans = t + a
print(ans)
|
s037163575
|
Accepted
| 42
| 10,480
| 605
|
import math
from fractions import Fraction
from decimal import *
# ex) Fraction(2,6) > 1/3 > 0.33333
n = int(input())
x, y = map(int, input().split())
t, a = x, y
for _ in range(n - 1):
tt, aa = map(int, input().split())
# c = 1
c = max(math.ceil(Fraction(t, tt)), math.ceil(Fraction(a, aa)))
c = max(1, c)
t = tt * c
a = aa * c
ans = t + a
print(ans)
|
s186098341
|
p03251
|
u157850041
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,064
| 332
|
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out.
|
a = input().split()
n = int(a[0])
m = int(a[1])
x = int(a[2])
y = int(a[3])
a = input().split()
a = [int(s) for s in a]
x_max = max(a)
a = input().split()
a = [int(s) for s in a]
y_min = min(a)
print(y <= x_max , y_min <= x , x_max >= y_min)
if y <= x_max or y_min <= x or x_max >= y_min:
print("War")
else:
print("No War")
|
s329666803
|
Accepted
| 17
| 3,064
| 284
|
a = input().split()
n = int(a[0])
m = int(a[1])
x = int(a[2])
y = int(a[3])
a = input().split()
a = [int(s) for s in a]
x_max = max(a)
a = input().split()
a = [int(s) for s in a]
y_min = min(a)
if x_max >= y or y_min <= x or x_max >= y_min:
print("War")
else:
print("No War")
|
s770940876
|
p02393
|
u978086225
| 1,000
| 131,072
|
Wrong Answer
| 30
| 6,716
| 106
|
Write a program which reads three integers, and prints them in ascending order.
|
nums = input().split()
a = int(nums[0])
b = int(nums[1])
c = int(nums[2])
if a < b < c:
print(a, b, c)
|
s155623782
|
Accepted
| 30
| 6,724
| 281
|
nums = input().split()
a = int(nums[0])
b = int(nums[1])
c = int(nums[2])
if a <= b <= c:
print(a, b, c)
elif a <= c <= b:
print(a, c, b)
elif b <= a <= c:
print(b, a, c)
elif b <= c <= a:
print(b, c, a)
elif c <= a <= b:
print(c, a, b)
else:
print(c, b, a)
|
s985344002
|
p02418
|
u650790815
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,292
| 75
|
Write a program which finds a pattern $p$ in a ring shaped text $s$.
|
s = input()
p = input()
if s in p*2:
print('yes')
else:
print('no')
|
s720646633
|
Accepted
| 20
| 7,392
| 56
|
s,p = input()*2,input()
print('Yes' if p in s else 'No')
|
s379600124
|
p03150
|
u480138356
| 2,000
| 1,048,576
|
Wrong Answer
| 21
| 3,316
| 345
|
A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once. Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string.
|
import sys
input = sys.stdin.readline
def ok(s):
for i in range(len(s)):
for j in range(i, len(s)):
print(s[:i] + s[j:])
if s[:i] + s[j:] == "keyence":
return True
return False
def main():
s = input().strip()
print("YES" if ok(s) else "NO")
if __name__ == "__main__":
main()
|
s273694253
|
Accepted
| 18
| 3,060
| 540
|
import sys
input = sys.stdin.readline
def ok(s):
for i in range(len(s)+1):
for j in range(i, len(s)+1):
# for k in range(len(s)):
# if i <= k and k < j:
# print("_", end="")
# else:
# print(s[k], end="")
if s[:i] + s[j:] == "keyence":
return True
return False
def main():
s = input().strip()
print("YES" if ok(s) else "NO")
if __name__ == "__main__":
main()
|
s901687962
|
p02831
|
u723444827
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 325
|
Takahashi is organizing a party. At the party, each guest will receive one or more snack pieces. Takahashi predicts that the number of guests at this party will be A or B. Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted. We assume that a piece cannot be divided and distributed to multiple guests.
|
def gcd(a,b):
if (a % b) == 0:
return b
else:
return gcd(b, a%b)
def lcm(a, b):
y = gcd(a,b)
print(y)
if (y == 0):
return a*b
else:
return a*b/y
A, B = map(int,input().split())
if A < B:
t = A
A = B
B = t
x = int(lcm(A,B))
print(x)
|
s112866236
|
Accepted
| 17
| 3,060
| 312
|
def gcd(a,b):
if (a % b) == 0:
return b
else:
return gcd(b, a%b)
def lcm(a, b):
y = gcd(a,b)
if (y == 0):
return a*b
else:
return a*b/y
A, B = map(int,input().split())
if A < B:
t = A
A = B
B = t
x = int(lcm(A,B))
print(x)
|
s587563227
|
p04043
|
u759412327
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 43
|
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
print(["YES'","NO"][input().count("7")==2])
|
s753543609
|
Accepted
| 27
| 9,056
| 80
|
if sorted(map(int,input().split()))==[5,5,7]:
print("YES")
else:
print("NO")
|
s917535928
|
p03861
|
u976162616
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 166
|
You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x?
|
if __name__ == "__main__":
A,B,C = map(int, input().split())
result = B // C
result -= A // C
if (A % C == 0):
result -= 1
print (result)
|
s385969280
|
Accepted
| 17
| 2,940
| 166
|
if __name__ == "__main__":
A,B,C = map(int, input().split())
result = B // C
result -= A // C
if (A % C == 0):
result += 1
print (result)
|
s005315860
|
p03993
|
u215315599
| 2,000
| 262,144
|
Wrong Answer
| 83
| 14,008
| 145
|
There are N rabbits, numbered 1 through N. The i-th (1≤i≤N) rabbit likes rabbit a_i. Note that no rabbit can like itself, that is, a_i≠i. For a pair of rabbits i and j (i＜j), we call the pair (i，j) a _friendly pair_ if the following condition is met. * Rabbit i likes rabbit j and rabbit j likes rabbit i. Calculate the number of the friendly pairs.
|
N = int(input())
A = list(map(lambda x:int(x)-1,input().split()))
print(A)
ans = 0
for i in range(N):
if A[A[i]] == i: ans += 1
print(ans//2)
|
s955326835
|
Accepted
| 76
| 14,008
| 137
|
N = int(input())
A = list(map(lambda x:int(x)-1,input().split()))
ans = 0
for i in range(N):
if A[A[i]] == i: ans += 1
print(ans//2)
|
s775235681
|
p03504
|
u988402778
| 2,000
| 262,144
|
Wrong Answer
| 775
| 125,180
| 5,374
|
Joisino is planning to record N TV programs with recorders. The TV can receive C channels numbered 1 through C. The i-th program that she wants to record will be broadcast from time s_i to time t_i (including time s_i but not t_i) on Channel c_i. Here, there will never be more than one program that are broadcast on the same channel at the same time. When the recorder is recording a channel from time S to time T (including time S but not T), it cannot record other channels from time S-0.5 to time T (including time S-0.5 but not T). Find the minimum number of recorders required to record the channels so that all the N programs are completely recorded.
|
# using main() makes code faster from the point of view of "access to variables in global name-space"
# for i, a in enumerate(iterable)
# divmod(x, y) returns the tuple (x//y, x%y)
# manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
import sys
sys.setrecursionlimit(10**7)
from itertools import accumulate, combinations, permutations # https://docs.python.org/ja/3/library/itertools.html
from math import factorial
def combinations_count(n, r):
# faster than the following code
# return factorial(n) // (factorial(n - r) * factorial(r))
if n - r < r: r = n - r
if r == 0: return 1
if r == 1: return n
numerator = [n - r + k + 1 for k in range(r)]
denominator = [k + 1 for k in range(r)]
for p in range(2,r+1):
pivot = denominator[p - 1]
if pivot > 1:
offset = (n - r) % p
for k in range(p-1,r,p):
numerator[k - offset] /= pivot
denominator[k] /= pivot
result = 1
for k in range(r):
if numerator[k] > 1:
result *= int(numerator[k])
return result
def combination_with_repetition_count(n, r):
return combinations_count(n + r - 1, r)
from collections import deque, Counter # https://docs.python.org/ja/3/library/collections.html#collections.deque
from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
from operator import itemgetter
# ex1: List.sort(key=itemgetter(1))
# ex2: sorted(tuples, key=itemgetter(1,2))
from functools import reduce
from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
def gcds(numbers):
return reduce(gcd, numbers)
def lcm(x, y):
return (x * y) // gcd(x, y)
def lcms(numbers):
return reduce(lcm, numbers, 1)
# first create factorial_list
# fac_list = mod_factorial_list(n)
INF = 10 ** 18
MOD = 10 ** 6 + 3
modpow = lambda a, n, p = MOD: pow(a, n, p) # Recursive function in python is slow!
def modinv(a, p = MOD):
# evaluate reciprocal using Fermat's little theorem:
# a**(p-1) is identical to 1 (mod p) when a and p is coprime
return modpow(a, p-2, p)
def modinv_list(n, p = MOD):
if n <= 1:
return [0,1][:n+1]
else:
inv_t = [0,1]
for i in range(2, n+1):
inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
return inv_t
def modfactorial_list(n, p = MOD):
if n == 0:
return [1]
else:
l = [0] * (n+1)
tmp = 1
for i in range(1, n+1):
tmp = tmp * i % p
l[i] = tmp
return l
def modcomb(n, k, fac_list = [], p = MOD):
# fac_list = modfactorial_list(100)
# print(modcomb(100, 5, modfactorial_list(100)))
from math import factorial
if n < 0 or k < 0 or n < k: return 0
if n == 0 or k == 0: return 1
if len(fac_list) <= n:
a = factorial(n) % p
b = factorial(k) % p
c = factorial(n-k) % p
else:
a = fac_list[n]
b = fac_list[k]
c = fac_list[n-k]
return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
def modadd(a, b, p = MOD):
return (a + b) % MOD
def modsub(a, b, p = MOD):
return (a - b) % p
def modmul(a, b, p = MOD):
return ((a % p) * (b % p)) % p
def moddiv(a, b, p = MOD):
return modmul(a, modpow(b, p-2, p))
# initialize variables and set inputs
#initialize variables
# to initialize list, use [0] * n
# to initialize two dimentional array, use [[0] * N for _ in range(N)]
# open(0).read() is a convenient method:
# ex) n, m, *x = map(int, open(0).read().split())
# min(x[::2]) - max(x[1::2])
# ex2) *x, = map(int, open(0).read().split())
# don't forget to add comma after *x if only one variable is used
# calculate and output
# ex1) print(*l) => when l = [2, 5, 6], printed 2 5 6
# sys.stdin = open('sample.txt') # for test
# functions used
r = lambda: sys.stdin.readline().strip() #single: int(r()), line: map(int, r().split())
R = lambda: list(map(int, r().split())) # line: R(), lines: [R() for _ in range(n)]
Rmap = lambda: map(int, r().split())
N, C = R()
STC = [R() for _ in range(N)]
tbl = [[0]*(2*(10**5)+5) for _ in range(C)]
# watch out: join tv programmes which are continuously broadcast in one channel
for s, t, c in STC:
c -= 1
if tbl[c][2*s] != 0:
tbl[c][2*s] += 1
tbl[c][2*t] -= 1
elif tbl[c][2*t-1] != 0:
tbl[c][2*t-1] -= 1
tbl[c][2*s-1] += 1
else:
tbl[c][2*s-1] = 1
tbl[c][2*t] = -1
# for s, t, c in STC:
# c -= 1
# if tbl[c][s] == -1:
# tbl[c][s] = 0
# tbl[c][t] = -1
# elif tbl[c][t-1] == 1:
# tbl[c][s-1] = 1
# tbl[c][t-1] = 0
# else:
# tbl[c][s-1] = 1
# tbl[c][t] = -1
tbl2 = [list(accumulate(a)) for a in tbl]
ans = max([sum(x) for x in zip(*tbl2)])
print(ans)
# res = 0
# cc = 0
# for j in range(C):
# if tbl2[j][i]:
# cc += 1
# res = max(res, cc)
# print(res)
# if __name__ == '__main__':
# main()
|
s054934209
|
Accepted
| 1,638
| 123,388
| 1,016
|
import sys
from itertools import accumulate
# functions used
r = lambda: sys.stdin.readline().strip() #single: int(r()), line: map(int, r().split())
R = lambda: list(map(int, r().split())) # line: R(), lines: [R() for _ in range(n)]
Rmap = lambda: map(int, r().split())
N, C = R()
STC = [R() for _ in range(N)]
# 0.5sec as one unit
tbl = [[0]*(2*(10**5)+5) for _ in range(C)]
for s, t, c in STC:
c -= 1
# if we can record the programms continuously (ver1)
if tbl[c][2*s] != 0:
tbl[c][2*s] += 1
tbl[c][2*t] -= 1
# if we can record the programms continuously (ver2)
elif tbl[c][2*t-1] != 0:
tbl[c][2*t-1] -= 1
tbl[c][2*s-1] += 1
# else:
else:
tbl[c][2*s-1] = 1
tbl[c][2*t] = -1
tbl2 = [list(accumulate(a)) for a in tbl]
# ans = max([sum(x) for x in zip(*tbl2)])
# print(ans)
res = 0
for i in range(200002):
cc = 0
for j in range(C):
if tbl2[j][i] == 1:
cc += 1
res = max(res, cc)
print(res)
|
s798665800
|
p02390
|
u090921599
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,456
| 1
|
Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively.
|
s470924865
|
Accepted
| 20
| 5,584
| 94
|
S = int(input())
h = S // 3600
m = S % 3600 // 60
s = S % 60
print(h, ':',m, ':', s, sep='')
|
|
s979766410
|
p03997
|
u391059484
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 61
|
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a, b, h = [int(input()) for i in range(3)]
print((a + b)*h/2)
|
s476192490
|
Accepted
| 17
| 2,940
| 66
|
a, b, h = [int(input()) for i in range(3)]
print(int((a + b)*h/2))
|
s369624092
|
p03141
|
u404676457
| 2,000
| 1,048,576
|
Wrong Answer
| 609
| 48,736
| 361
|
There are N dishes of cuisine placed in front of Takahashi and Aoki. For convenience, we call these dishes Dish 1, Dish 2, ..., Dish N. When Takahashi eats Dish i, he earns A_i points of _happiness_ ; when Aoki eats Dish i, she earns B_i points of happiness. Starting from Takahashi, they alternately choose one dish and eat it, until there is no more dish to eat. Here, both of them choose dishes so that the following value is maximized: "the sum of the happiness he/she will earn in the end" minus "the sum of the happiness the other person will earn in the end". Find the value: "the sum of the happiness Takahashi earns in the end" minus "the sum of the happiness Aoki earns in the end".
|
n = int(input())
ab = [list(map(int, input().split())) for _ in range(n)]
div = [[abs(ab[i][1] - ab[i][0]), ab[i][0], ab[i][1]] for i in range(n)]
div = sorted(div, key=lambda x: x[0], reverse=True)
print(div)
counta = 0
countb = 0
for i in range(n):
if i % 2 == 0:
counta += div[i][1]
else:
countb += div[i][2]
print(counta - countb)
|
s198239765
|
Accepted
| 556
| 40,136
| 344
|
n = int(input())
ab = [list(map(int, input().split())) for i in range(n)]
div = [[ab[i][1] + ab[i][0], ab[i][0], ab[i][1]] for i in range(n)]
div = sorted(div, key=lambda x: x[0], reverse=True)
counta = 0
countb = 0
for i in range(n):
if i % 2 == 0:
counta += div[i][1]
else:
countb += div[i][2]
print(counta - countb)
|
s860133947
|
p04030
|
u976225138
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 134
|
Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key. To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string: * The `0` key: a letter `0` will be inserted to the right of the string. * The `1` key: a letter `1` will be inserted to the right of the string. * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted. Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?
|
ans = ""
for s in input():
if s == "B":
if ans:
ans = ans[:-2]
else:
ans += s
else:
print(ans)
|
s593658998
|
Accepted
| 17
| 3,060
| 131
|
ans = ""
for s in input():
if s == "B" and ans:
ans = ans[:-1]
elif s != "B":
ans += s
else:
print(ans)
|
s970588785
|
p03836
|
u057109575
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 486
|
Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him.
|
sx, sy, tx, ty = map(int, input().split())
def func(a, b, x, y):
ans = ''
if x > a:
ans += 'R' * (x - a)
if x < a:
ans += 'L' * (a - x)
if y > b:
ans += 'U' * (y - b)
if y < b:
ans += 'D' * (b - y)
return ans
print('U' + func(sx + 1, sy, tx + 1, ty) + 'D' \
+ 'R' + func(tx, ty + 1, sx, sy + 1) + 'L' \
+ 'L' + func(sx - 1, sy, tx - 1, ty) + 'R' \
+ 'D' + func(tx, ty - 1, sx, sy - 1) + 'U')
|
s334642211
|
Accepted
| 17
| 3,060
| 269
|
sx, sy, tx, ty = map(int, input().split())
print('U' * (ty - sy) + 'R' * (tx - sx) \
+ 'D' * (ty - sy) + 'L' * (tx - sx) \
+ 'L' + 'U' * (ty - sy + 1) + 'R' * (tx - sx + 1) + 'D' \
+ 'R' + 'D' * (ty - sy + 1) + 'L' * (tx - sx + 1) + 'U' )
|
s121397555
|
p03379
|
u988402778
| 2,000
| 262,144
|
Wrong Answer
| 316
| 25,220
| 234
|
When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l. You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i. Find B_i for each i = 1, 2, ..., N.
|
n = int(input())
x = [int(i) for i in input().split()]
sortx = sorted(x)
mid_left = sortx[(n//2)-1]
mid_right = sortx[(n//2)]
for i in range(n):
if sortx[i] <= mid_left:
print(mid_right)
else:
print(mid_left)
|
s951041911
|
Accepted
| 305
| 25,220
| 230
|
n = int(input())
x = [int(i) for i in input().split()]
sortx = sorted(x)
mid_left = sortx[(n//2)-1]
mid_right = sortx[(n//2)]
for i in range(n):
if x[i] <= mid_left:
print(mid_right)
else:
print(mid_left)
|
s377078893
|
p03377
|
u565464228
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 95
|
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
a, b, x = map(int, input().split())
if a <= x <= a + b:
print("Yes")
else:
print("No")
|
s806566434
|
Accepted
| 17
| 2,940
| 119
|
a, b, x = map(int, input().split(" "))
# x>=a and x<=a+b
if x>=a and x<=a+b:
print("YES")
else:
print("NO")
|
s016733910
|
p03545
|
u714533789
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 164
|
Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.
|
from itertools import product
s = input()
for i in product('+-', repeat=3):
t = ''.join([a+b for a, b in zip(list(i), s)])
if eval(t) == 7:
print(t+'=7');exit()
|
s790452716
|
Accepted
| 17
| 3,060
| 174
|
from itertools import product
s = input()
for ops in product('+-', repeat=3):
t = ''.join([a+b for a, b in zip(s, ops)])
t += s[-1]
if eval(t) == 7:
print(t+'=7');exit()
|
s938749175
|
p03992
|
u016843859
| 2,000
| 262,144
|
Wrong Answer
| 29
| 9,124
| 113
|
This contest is `CODE FESTIVAL`. However, Mr. Takahashi always writes it `CODEFESTIVAL`, omitting the single space between `CODE` and `FESTIVAL`. So he has decided to make a program that puts the single space he omitted. You are given a string s with 12 letters. Output the string putting a single space between the first 4 letters and last 8 letters in the string s.
|
s=input()
for i in range(4):
print(s[i],end="")
print(" ",end="")
for i in range(8):
print(s[i+4],end="")
|
s635264833
|
Accepted
| 20
| 9,052
| 123
|
s=input()
for i in range(4):
print(s[i],end="")
print(" ",end="")
for i in range(8):
print(s[i+4],end="")
print("")
|
s203535397
|
p03524
|
u871841829
| 2,000
| 262,144
|
Wrong Answer
| 26
| 3,564
| 292
|
Snuke has a string S consisting of three kinds of letters: `a`, `b` and `c`. He has a phobia for palindromes, and wants to permute the characters in S so that S will not contain a palindrome of length 2 or more as a substring. Determine whether this is possible.
|
from collections import Counter
S = input()
d = Counter(S)
na = 0 if "a" not in d.keys() else d["a"]
nb = 0 if "b" not in d.keys() else d["b"]
nc = 0 if "c" not in d.keys() else d["c"]
if abs(na - nb) <= 1 and abs(nc - na) <= 1 and abs(nb - nc) <= 1:
print("Yes")
else:
print("No")
|
s193012830
|
Accepted
| 26
| 3,564
| 291
|
from collections import Counter
S = input()
d = Counter(S)
na = 0 if "a" not in d.keys() else d["a"]
nb = 0 if "b" not in d.keys() else d["b"]
nc = 0 if "c" not in d.keys() else d["c"]
if abs(na - nb) <= 1 and abs(nc - na) <= 1 and abs(nb - nc) <= 1:
print("YES")
else:
print("NO")
|
s872617581
|
p02418
|
u656153606
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,352
| 101
|
Write a program which finds a pattern $p$ in a ring shaped text $s$.
|
s = list(input())
p = input()
s.extend(s)
if p in s:
print("Yes")
else:
print("No")
print(s)
|
s363028577
|
Accepted
| 70
| 7,452
| 80
|
s = input()
p = input()
s += s
if p in s:
print("Yes")
else:
print("No")
|
s710039246
|
p03555
|
u746419473
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 97
|
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
|
h = input()
l = input()
print("Yes" if h[0] == l[2] and l[0] == h[2] and h[1] == l[1] else "No")
|
s162738624
|
Accepted
| 17
| 2,940
| 78
|
s = input()
s += input()
print("YES" if s == "".join(reversed(s)) else "NO")
|
s023901645
|
p00006
|
u777299405
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,308
| 24
|
Write a program which reverses a given string str.
|
print(reversed(input()))
|
s914070854
|
Accepted
| 20
| 7,412
| 20
|
print(input()[::-1])
|
s849413610
|
p00506
|
u150984829
| 8,000
| 131,072
|
Wrong Answer
| 20
| 5,644
| 197
|
入力ファイルの１行目に正整数 n が書いてあり， ２行目には半角空白文字１つを区切りとして， n 個の正整数が書いてある． n は 2 または 3 であり， ２行目に書かれているどの整数も値は 108 以下である． これら２個または３個の数の公約数をすべて求め， 小さい方から順に1行に1個ずつ出力せよ． 自明な公約数（「１」）も出力すること． 出力ファイルにおいては， 出力の最後行にも改行コードを入れること．
|
input()
n=sorted(list(map(int,input().split())))
m=n[0]
a=[]
for x in range(1,int(m**.5)+1):
if m%x==0:a+=[x,m/x]
for c in sorted(a):
for k in n[1:]:
if k%c:break
else:print(c)
|
s905135484
|
Accepted
| 20
| 5,636
| 207
|
input()
n=sorted(list(map(int,input().split())))
m=n[0]
a=set()
for x in range(1,int(m**.5)+1):
if m%x==0:a|={x,m//x}
for c in sorted(list(a)):
for k in n[1:]:
if k%c:break
else:print(c)
|
s104901018
|
p02694
|
u219494936
| 2,000
| 1,048,576
|
Wrong Answer
| 33
| 9,148
| 117
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
X = int(input())
K = 100
i = 1
while 1:
K *= 1.01
K = int(K)
if K > X:
break
i += 1
print(i)
|
s264604303
|
Accepted
| 31
| 8,952
| 116
|
X = int(input())
K = 100
i = 1
while 1:
K *= 101
K //= 100
if K >= X:
break
i += 1
print(i)
|
s962142491
|
p02255
|
u534156032
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,716
| 292
|
Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step.
|
def insersionSort(A, N):
print(" ".join(map(str,A)))
for i in range(1,N-1):
v = A[i]
j = i - 1
while j >= 0 and A[j] > v:
A[j+1] = A[j]
j -= 1
A[j+1] =v
print(" ".join(map(str,A)))
n = int(input())
a = [int(i) for i in input().split()]
insersionSort(a, n)
|
s286230665
|
Accepted
| 50
| 7,720
| 217
|
n = int(input())
a = [int(i) for i in input().split()]
for i in range(n):
v = a[i]
j = i - 1
while j >= 0 and a[j] > v:
a[j+1] = a[j]
j -= 1
a[j+1] = v
print(" ".join(map(str, a)))
|
s813591722
|
p03494
|
u363836311
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,060
| 171
|
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
n=int(input())
a=list(map(int, input().split()))
t=0
g=0
for i in range(n):
while a[i]%2==0:
t+=1
a[i]=a[i]//2
if g==0:
g=t
else:
g=min(g,t)
print(g)
|
s221236445
|
Accepted
| 19
| 3,060
| 197
|
n=int(input())
a=list(map(int, input().split()))
t=0
g=0
while g==0:
for i in range(n):
if a[i]%2==0:
a[i]=a[i]//2
t+=1
else:
g=1
print(t//n)
|
s047771218
|
p03673
|
u887207211
| 2,000
| 262,144
|
Wrong Answer
| 2,105
| 21,748
| 133
|
You are given an integer sequence of length n, a_1, ..., a_n. Let us consider performing the following n operations on an empty sequence b. The i-th operation is as follows: 1. Append a_i to the end of b. 2. Reverse the order of the elements in b. Find the sequence b obtained after these n operations.
|
N = int(input())
A = input().split()
B = ''
for i in range(N):
if(i%2 == 0):
B = A[i] + B
else:
B += A[i]
print(B[::-1])
|
s154065349
|
Accepted
| 108
| 23,972
| 112
|
N = int(input())
A = input().split()
if(N%2 == 0):
print(*(A[::-2]+A[::2]))
else:
print(*(A[::-2]+A[1::2]))
|
s240827223
|
p03251
|
u254086528
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,064
| 271
|
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out.
|
n,m,cx,cy = map(int,input().split())
x = list(int(i) for i in input().split())
y = list(int(i) for i in input().split())
x.sort()
y.sort()
z = range(cx+1,cy+1)
ans = "War"
for v in z:
if (v > x[n-1]) and (v <= y[0]):
ans = "No war"
break
print(ans)
|
s863040259
|
Accepted
| 17
| 3,064
| 271
|
n,m,cx,cy = map(int,input().split())
x = list(int(i) for i in input().split())
y = list(int(i) for i in input().split())
x.sort()
y.sort()
z = range(cx+1,cy+1)
ans = "War"
for v in z:
if (v > x[n-1]) and (v <= y[0]):
ans = "No War"
break
print(ans)
|
s714331166
|
p03645
|
u576917603
| 2,000
| 262,144
|
Wrong Answer
| 756
| 49,172
| 251
|
In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands. For convenience, we will call them Island 1, Island 2, ..., Island N. There are M kinds of regular boat services between these islands. Each service connects two islands. The i-th service connects Island a_i and Island b_i. Cat Snuke is on Island 1 now, and wants to go to Island N. However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services. Help him.
|
n,m=map(int,input().split())
a=[[int(i) for i in input().split()]for i in range(m)]
print(a)
s=set()
for i in a:
if i[1]==n:
s.add(i[0])
for i in a:
if i[0]==1 and i[1] in s:
print('POSSIBLE')
exit()
print('IMPOSSIBLE')
|
s499004041
|
Accepted
| 668
| 40,856
| 394
|
n,m=map(int,input().split())
a=[[int(i) for i in input().split()]for i in range(m)]
b=[None]*n
for i in a:
if i[1]==n:
if b[i[0]-1]!=None:
print('POSSIBLE')
exit()
else:
b[i[0]-1]=True
if i[0]==1:
if b[i[1]-1]!=None:
print('POSSIBLE')
exit()
else:
b[i[1]-1]=True
print('IMPOSSIBLE')
|
s690393386
|
p03643
|
u865413330
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 302
|
This contest, _AtCoder Beginner Contest_ , is abbreviated as _ABC_. When we refer to a specific round of ABC, a three-digit number is appended after ABC. For example, ABC680 is the 680th round of ABC. What is the abbreviation for the N-th round of ABC? Write a program to output the answer.
|
n = int(input())
count = 0
currCount = 0
ans = 0
for i in range(n+1):
currCount = 0
currNum = i
while (i % 2) == 0:
if i == 0:
break
i = int(i / 2)
currCount += 1
if count < currCount:
count = currCount
ans = currNum
print(ans, count)
|
s943885956
|
Accepted
| 20
| 2,940
| 28
|
n = input()
print("ABC" + n)
|
s493458084
|
p03711
|
u432805419
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 202
|
Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group.
|
a = list(map(int,input().split()))
b = [1,3,5,7,8,10,12]
c = [4,6,9,11]
if a[0] == 2 or a[1] == 2:
print("No")
elif a[0] in b == a[1] in b or a[0] in c == a[1] in c:
print("Yes")
else:
print("No")
|
s724705489
|
Accepted
| 17
| 2,940
| 124
|
a = list(map(int,input().split()))
b = [0,1,3,1,2,1,2,1,1,2,1,2,1]
if b[a[0]] == b[a[1]]:
print("Yes")
else:
print("No")
|
s140472577
|
p03759
|
u762540523
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 58
|
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b. Determine whether the arrangement of the poles is beautiful.
|
a,b,c=map(int,input().split());print("YNeos"[b-a!=c-b::2])
|
s069559888
|
Accepted
| 17
| 2,940
| 58
|
a,b,c=map(int,input().split());print("YNEOS"[b-a!=c-b::2])
|
s267040045
|
p00004
|
u150984829
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,604
| 111
|
Write a program which solve a simultaneous equation: ax + by = c dx + ey = f The program should print x and y for given a, b, c, d, e and f (-1,000 ≤ a, b, c, d, e, f ≤ 1,000). You can suppose that given equation has a unique solution.
|
import sys
for e in sys.stdin:
a,b,c,d,e,f=map(int,e.split())
print((c*d-a*f)/(b*d-a*e),(c*e-b*f)/(a*e-b*d))
|
s750030173
|
Accepted
| 20
| 5,632
| 121
|
import sys
for e in sys.stdin:
a,b,c,d,e,f=map(int,e.split())
y=(c*d-a*f)/(b*d-a*e)
print(f'{(c-b*y)/a:.3f} {y:.3f}')
|
s675568262
|
p02697
|
u517152997
| 2,000
| 1,048,576
|
Wrong Answer
| 158
| 27,284
| 420
|
You are going to hold a competition of one-to-one game called AtCoder Janken. _(Janken is the Japanese name for Rock-paper-scissors.)_ N players will participate in this competition, and they are given distinct integers from 1 through N. The arena has M playing fields for two players. You need to assign each playing field two distinct integers between 1 and N (inclusive). You cannot assign the same integer to multiple playing fields. The competition consists of N rounds, each of which proceeds as follows: * For each player, if there is a playing field that is assigned the player's integer, the player goes to that field and fight the other player who comes there. * Then, each player adds 1 to its integer. If it becomes N+1, change it to 1. You want to ensure that no player fights the same opponent more than once during the N rounds. Print an assignment of integers to the playing fields satisfying this condition. It can be proved that such an assignment always exists under the constraints given.
|
#
import sys
import math
import numpy as np
import itertools
n,m = (int(i) for i in input().split())
if n % 2 == 0:
if m / 4 <= 1:
for i in range(0,m):
print(i+1, n//2-i)
else:
for i in range(0,n//4+1):
print(i+1,n//2-i)
for i in range(n//2+1,m):
print(i+n//2+1, n-i+n//2)
else:
for i in range(0,m):
print(i+1,n-1-i)
|
s651016806
|
Accepted
| 154
| 27,260
| 383
|
#
import sys
import math
import numpy as np
import itertools
n,m = (int(i) for i in input().split())
if m % 2 == 0:
for i in range(0,m//2):
print(i+1,m+1-i)
for i in range(m//2,m):
print(m//2+2+i,m*2+m//2+1-i)
else:
for i in range(0,m//2):
print(i+1,m-i)
for i in range(m//2+1,m+1):
print(m//2+1+i,m*2+m//2+2-i)
|
s449689500
|
p02408
|
u450020188
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,708
| 358
|
Taro is going to play a card game. However, now he has only n cards, even though there should be 52 cards (he has no Jokers). The 52 cards include 13 ranks of each of the four suits: spade, heart, club and diamond.
|
n = int(input())
cards = [[ s+' '+str(n) for n in range(1,14)] for s in ['S','H','C','D']]
for _ in range(n):
suit,num = input().split()
if suit=='S': cards[0][int(num)-1] = 0
elif suit=='H': cards[1][int(num)-1] = 0
elif suit=='C': cards[2][int(num)-1] = 0
elif suit=='D': cards[3][int(num)-1] = 0
for s in cards:
if n!=0: print(n)
|
s623662261
|
Accepted
| 30
| 7,740
| 378
|
n = int(input())
cards = [[ s+' '+str(n) for n in range(1,14)] for s in ['S','H','C','D']]
for _ in range(n):
suit,num = input().split()
if suit=='S': cards[0][int(num)-1] = 0
elif suit=='H': cards[1][int(num)-1] = 0
elif suit=='C': cards[2][int(num)-1] = 0
elif suit=='D': cards[3][int(num)-1] = 0
for s in cards:
for n in s:
if n!=0: print(n)
|
s790421968
|
p02578
|
u004823354
| 2,000
| 1,048,576
|
Wrong Answer
| 113
| 32,200
| 164
|
N persons are standing in a row. The height of the i-th person from the front is A_i. We want to have each person stand on a stool of some heights - at least zero - so that the following condition is satisfied for every person: Condition: Nobody in front of the person is taller than the person. Here, the height of a person includes the stool. Find the minimum total height of the stools needed to meet this goal.
|
n = int(input())
a = list(map(int,input().split()))
max = 0
sum = 0
for i in range(n):
if a[i] > max:
max = a[i]
else:
sum += a[i] - max
print(sum)
|
s563696995
|
Accepted
| 114
| 32,032
| 167
|
n = int(input())
a = list(map(int,input().split()))
max = 0
sum = 0
for i in range(n):
if a[i] > max:
max = a[i]
else:
sum += (max - a[i])
print(sum)
|
s507259100
|
p03214
|
u163320134
| 2,525
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 162
|
Niwango-kun is an employee of Dwango Co., Ltd. One day, he is asked to generate a thumbnail from a video a user submitted. To generate a thumbnail, he needs to select a frame of the video according to the following procedure: * Get an integer N and N integers a_0, a_1, ..., a_{N-1} as inputs. N denotes the number of the frames of the video, and each a_i denotes the representation of the i-th frame of the video. * Select t-th frame whose representation a_t is nearest to the average of all frame representations. * If there are multiple such frames, select the frame with the smallest index. Find the index t of the frame he should select to generate a thumbnail.
|
n=int(input())
arr=list(map(int,input().split()))
avg=sum(arr)/n
pos=0
diff=10**9
for i in range(n):
if abs(arr[i]-avg)<diff:
diff=abs(arr[i]-avg)
pos=i
|
s613327036
|
Accepted
| 17
| 3,060
| 173
|
n=int(input())
arr=list(map(int,input().split()))
avg=sum(arr)/n
pos=0
diff=10**9
for i in range(n):
if abs(arr[i]-avg)<diff:
diff=abs(arr[i]-avg)
pos=i
print(pos)
|
s471008299
|
p03455
|
u760527120
| 2,000
| 262,144
|
Wrong Answer
| 21
| 3,316
| 88
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b = map(int, input().split())
if a * b % 2 == 0:
print('Odd')
else:
print('Even')
|
s714561013
|
Accepted
| 17
| 2,940
| 88
|
a, b = map(int, input().split())
if a * b % 2 == 0:
print('Even')
else:
print('Odd')
|
s812693896
|
p03549
|
u052332717
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 154
|
Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`. When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases. Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds. Now, he goes through the following process: * Submit the code. * Wait until the code finishes execution on all the cases. * If the code fails to correctly solve some of the M cases, submit it again. * Repeat until the code correctly solve all the cases in one submission. Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).
|
n,m = map(int,input().split())
success = 0.5**m
failure = 1 - success
operation_time = 1900*m + 100*(n-m)
print(success*operation_time//((1-failure)**2))
|
s535519891
|
Accepted
| 18
| 2,940
| 161
|
n,m = map(int,input().split())
success = 0.5**m
failure = 1 - success
operation_time = 1900*m + 100*(n-m)
print(int((success*operation_time//((1-failure)**2))))
|
s348052314
|
p03853
|
u481333386
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 163
|
There is an image with a height of H pixels and a width of W pixels. Each of the pixels is represented by either `.` or `*`. The character representing the pixel at the i-th row from the top and the j-th column from the left, is denoted by C_{i,j}. Extend this image vertically so that its height is doubled. That is, print a image with a height of 2H pixels and a width of W pixels where the pixel at the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division is rounded down).
|
height, width = [int(e) for e in input().split()]
lst = []
for i in range(height):
pict = input()
lst.append(pict)
lst.append(pict)
'\n'.join(lst)
|
s043746513
|
Accepted
| 18
| 3,060
| 170
|
height, width = [int(e) for e in input().split()]
lst = []
for i in range(height):
pict = input()
lst.append(pict)
lst.append(pict)
print('\n'.join(lst))
|
s445265123
|
p03378
|
u798586213
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 307
|
There are N + 1 squares arranged in a row, numbered 0, 1, ..., N from left to right. Initially, you are in Square X. You can freely travel between adjacent squares. Your goal is to reach Square 0 or Square N. However, for each i = 1, 2, ..., M, there is a toll gate in Square A_i, and traveling to Square A_i incurs a cost of 1. It is guaranteed that there is no toll gate in Square 0, Square X and Square N. Find the minimum cost incurred before reaching the goal.
|
N,M,X = (int(i) for i in input().split())
A = list(map(int,input().split()))
print(A)
small = 0
big = 0
for j in A:
if j < X:
continue
else:
small=small+1
for j in A:
if j > X:
continue
else:
big=big+1
if big>=small:
print(small)
else:
print(big)
|
s693118024
|
Accepted
| 17
| 3,060
| 299
|
N,M,X = (int(i) for i in input().split())
A = list(map(int,input().split()))
small = 0
big = 0
for j in A:
if j < X:
continue
else:
small=small+1
for j in A:
if j > X:
continue
else:
big=big+1
if big>=small:
print(small)
else:
print(big)
|
s124614993
|
p03578
|
u544280305
| 2,000
| 262,144
|
Wrong Answer
| 533
| 71,956
| 716
|
Rng is preparing a problem set for a qualification round of CODEFESTIVAL. He has N candidates of problems. The difficulty of the i-th candidate is D_i. There must be M problems in the problem set, and the difficulty of the i-th problem must be T_i. Here, one candidate of a problem cannot be used as multiple problems. Determine whether Rng can complete the problem set without creating new candidates of problems.
|
import sys
from collections import Counter
candnum=int(input())
candiff=list(map(int,input().split()))
pronum=int(input())
candiff=Counter(candiff)
prodiff=list(map(int,input().split()))
prodiff=Counter(candiff)
if pronum>candnum:
print("NO")
sys.exit()
#print(candnum)
print(candiff)
#print(pronum)
print(prodiff)
'''
while candnum>=1 and pronum>=1:
if candiff[candnum-1]<prodiff[pronum-1]:
print("NO")
sys.exit()
candnum-=1
pronum-=1
print("YES")
sys.exit()
'''
#print(candiff)
#print(prodiff)
for k,v in prodiff.items():
if k in candiff:
if candiff[k]<v:
print("NO")
sys.exit()
else:
print("NO")
sys.exit()
print("YES")
|
s741438228
|
Accepted
| 285
| 55,648
| 445
|
import sys
from collections import Counter
candnum=int(input())
candiff=list(map(int,input().split()))
pronum=int(input())
prodiff=list(map(int,input().split()))
candiff=Counter(candiff)
prodiff=Counter(prodiff)
if pronum>candnum:
print("NO")
sys.exit()
for k,v in prodiff.items():
if k in candiff:
if candiff[k]<v:
print("NO")
sys.exit()
else:
print("NO")
sys.exit()
print("YES")
|
s977208080
|
p02612
|
u882853528
| 2,000
| 1,048,576
|
Wrong Answer
| 25
| 9,140
| 32
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N = int(input())
print(N % 1000)
|
s871974639
|
Accepted
| 28
| 9,176
| 81
|
N = int(input())
if N%1000 == 0:
print(0)
else:
print(1000 - (N % 1000))
|
s978108145
|
p03997
|
u702786238
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 68
|
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a = int(input())
b = int(input())
h = int(input())
print((a+b)*h/2)
|
s429020044
|
Accepted
| 17
| 2,940
| 70
|
a = int(input())
b = int(input())
h = int(input())
print((a+b)*h//2)
|
s053428820
|
p03597
|
u284854859
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 43
|
We have an N \times N square grid. We will paint each square in the grid either black or white. If we paint exactly A squares white, how many squares will be painted black?
|
n=int(input())
a=int(input())
print(n*2-a)
|
s115420882
|
Accepted
| 17
| 2,940
| 45
|
n=int(input())
a=int(input())
print(n**2-a)
|
s316767313
|
p03469
|
u419963262
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 28
|
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
|
S=input()
S[3]=="8"
print(S)
|
s501233216
|
Accepted
| 17
| 2,940
| 25
|
print("2018"+input()[4:])
|
s833080534
|
p03578
|
u379692329
| 2,000
| 262,144
|
Wrong Answer
| 2,105
| 35,420
| 343
|
Rng is preparing a problem set for a qualification round of CODEFESTIVAL. He has N candidates of problems. The difficulty of the i-th candidate is D_i. There must be M problems in the problem set, and the difficulty of the i-th problem must be T_i. Here, one candidate of a problem cannot be used as multiple problems. Determine whether Rng can complete the problem set without creating new candidates of problems.
|
N = int(input())
D = [int(_) for _ in input().split()]
M = int(input())
T = [int(_) for _ in input().split()]
D = sorted(D)
T = sorted(T)
Dstart = 0
flag = True
for i in T:
for j in D[Dstart:]:
Dstart += 1
if i == j:
break
if Dstart == N:
flag = False
break
print("YES" if flag else "NO")
|
s748229370
|
Accepted
| 357
| 35,324
| 316
|
N = int(input())
D = [int(_) for _ in input().split()]
M = int(input())
T = [int(_) for _ in input().split()]
D = sorted(D)
T = sorted(T)
Tindex = 0
flag = False
for i in D:
if i == T[Tindex]:
Tindex += 1
if Tindex == M:
flag = True
break
print("YES" if flag else "NO")
|
s174510202
|
p03494
|
u531674264
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 382
|
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
n = input()
a = list(map(int, input().split()))
b = 0
def hantei(a):
b = list(map(lambda x: x % 2, a))
c = 0
for i in range(len(b)):
if a[i] == "0":
c += 0
else:
c += 1
if b == 0:
d = 0
else:
d = 1
return d
while True:
c = hantei(a)
if c == 0:
b += 1
else:
break
print(b)
|
s817130229
|
Accepted
| 21
| 3,064
| 406
|
n = input()
a = list(map(int, input().split()))
c = 0
def han(a):
b = list(map(lambda x: x % 2, a))
c = 0
for i in b:
if i == 0:
c += 1
else:
c -= 1
if c == len(b):return 1
else: return 0
while True:
if han(a) == 1:
a = list(map(lambda x: x / 2, a))
c += 1
if han(a) == 0:
break
c = 0
print(c)
|
s713294081
|
p03695
|
u697690147
| 2,000
| 262,144
|
Wrong Answer
| 28
| 9,136
| 226
|
In AtCoder, a person who has participated in a contest receives a _color_ , which corresponds to the person's rating as follows: * Rating 1-399 : gray * Rating 400-799 : brown * Rating 800-1199 : green * Rating 1200-1599 : cyan * Rating 1600-1999 : blue * Rating 2000-2399 : yellow * Rating 2400-2799 : orange * Rating 2800-3199 : red Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not. Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i. Find the minimum and maximum possible numbers of different colors of the users.
|
N = int(input())
a = list(map(int, input().split()))
c = [0] * 8
r = 0
for i in a:
col = i // 400
if col <= 7:
c[col] = 1
else:
r += 1
c = sum(c)
print(min(8, r) if r>c else c)
print(min(8,r+c))
|
s445314413
|
Accepted
| 25
| 9,128
| 199
|
N = int(input())
a = list(map(int, input().split()))
c = [0] * 8
r = 0
for i in a:
col = i // 400
if col <= 7:
c[col] = 1
else:
r += 1
c = sum(c)
print(max(1, c), r+c)
|
s693145670
|
p03338
|
u345710188
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 189
|
You are given a string S of length N consisting of lowercase English letters. We will cut this string at one position into two strings X and Y. Here, we would like to maximize the number of different letters contained in both X and Y. Find the largest possible number of different letters contained in both X and Y when we cut the string at the optimal position.
|
N = int(input())
S = input()
a = set(S[:1])
b = set(S[1:])
M = a & b
for i in range(2,N):
a = set(S[:i])
b = set(S[i:])
if (a & b) > M:
M = (a & b)
else:
print(M)
break
|
s146676510
|
Accepted
| 17
| 3,060
| 179
|
N = int(input())
S = input()
a = set(S[:1])
b = set(S[1:])
M = len(a & b)
for i in range(2,N):
a = set(S[:i])
b = set(S[i:])
if len(a & b) > M:
M = len(a & b)
print(M)
|
s250226615
|
p00001
|
u350804311
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,684
| 125
|
There is a data which provides heights (in meter) of mountains. The data is only for ten mountains. Write a program which prints heights of the top three mountains in descending order.
|
import sys
a = []
for line in sys.stdin:
a.append(int(line))
a.sort()
print(str(a[0]))
print(str(a[1]))
print(str(a[2]))
|
s717374360
|
Accepted
| 20
| 7,680
| 122
|
import sys
a = []
for line in sys.stdin:
a.append(int(line))
a.sort()
a.reverse()
print(a[0])
print(a[1])
print(a[2])
|
s351287663
|
p03474
|
u242518667
| 2,000
| 262,144
|
Wrong Answer
| 22
| 3,316
| 105
|
The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.
|
a,b=map(int,input().split())
s=input().split('-')
print('YES' if len(s[0])==a and len(s[1])==b else 'NO')
|
s591455705
|
Accepted
| 17
| 2,940
| 105
|
a,b=map(int,input().split())
s=input().split('-')
print('Yes' if len(s[0])==a and len(s[1])==b else 'No')
|
s462523541
|
p00101
|
u748033250
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,576
| 111
|
An English booklet has been created for publicizing Aizu to the world. When you read it carefully, you found a misnomer (an error in writing) on the last name of Masayuki Hoshina, the lord of the Aizu domain. The booklet says "Hoshino" not "Hoshina". Your task is to write a program which replace all the words "Hoshino" with "Hoshina". You can assume that the number of characters in a text is less than or equal to 1000.
|
num = int(input())
for i in range(num):
box = input()
box.replace("Hoshino", "Hoshina")
print(box)
|
s117062116
|
Accepted
| 20
| 7,524
| 108
|
num = int(input())
for i in range(num):
box = input()
print(box.replace("Hoshino", "Hoshina"))
|
s653020566
|
p03434
|
u406767170
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 181
|
We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.
|
n = int(input())
card = list(map(int,input().split()))
card.sort()
for i in range(n//2):
ans = card[2*i]-card[2*i+1]
if n%2==1:
ans += card[-1]
print(ans)
|
s449106143
|
Accepted
| 18
| 3,060
| 196
|
n = int(input())
card = list(map(int,input().split()))
card.sort(reverse=True)
asum = 0
bsum = 0
for i in range(n):
if i%2==0:
asum += card[i]
else:
bsum += card[i]
print(asum-bsum)
|
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