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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s348254811
|
p03047
|
u143903328
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 45
|
Snuke has N integers: 1,2,\ldots,N. He will choose K of them and give those to Takahashi. How many ways are there to choose K consecutive integers?
|
a,b = map(int,input().split())
print(b-a + 1)
|
s632220804
|
Accepted
| 17
| 2,940
| 45
|
a,b = map(int,input().split())
print(a-b + 1)
|
s541211025
|
p02406
|
u659034691
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,564
| 290
|
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; }
|
#15dpro
n=int(input())
#s=
#print(n)
for i in range(1,n-1):
# print(i)
if i%3==0:
print(' '+str(i),end='')
else:
x=i
# print(x)
while x>0 and x%10!=3:
# print(x)
x//=10
if x%10==3:
print(' '+str(i),end='')
|
s562410731
|
Accepted
| 20
| 7,672
| 278
|
#15dpro
n=int(input())
s=''
#print(n)
for i in range(1,n+1):
# print(i)
if i%3==0:
s+=' '+str(i)
else:
x=i
# print(x)
while x>0 and x%10!=3:
# print(x)
x//=10
if x%10==3:
s+=' '+str(i)
print(s)
|
s521787592
|
p03943
|
u920387587
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 182
|
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
a, b, c = map(int, input().split())
count=[a,b,c]
d =max(count)
v=0
count.remove(d)
for i in range(len(count)):
v += count[i]
if v ==d :
print('YES')
else:
print('NO')
|
s250793551
|
Accepted
| 17
| 3,060
| 182
|
a, b, c = map(int, input().split())
count=[a,b,c]
d =max(count)
v=0
count.remove(d)
for i in range(len(count)):
v += count[i]
if v ==d :
print('Yes')
else:
print('No')
|
s538383548
|
p03769
|
u864197622
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 217
|
We will call a string x _good_ if it satisfies the following condition: * Condition: x can be represented as a concatenation of two copies of another string y of length at least 1. For example, `aa` and `bubobubo` are good; an empty string, `a`, `abcabcabc` and `abba` are not good. Eagle and Owl created a puzzle on good strings. Find one string s that satisfies the following conditions. It can be proved that such a string always exists under the constraints in this problem. * 1 ≤ |s| ≤ 200 * Each character of s is one of the 100 characters represented by the integers 1 through 100. * Among the 2^{|s|} subsequences of s, exactly N are good strings.
|
def calc(n, k):
if n==1:
return str(k) + " " + str(k)
if n%2==0:
return calc(n-1, k+1) + " " + str(k) + " " + str(k)
return str(k) + " " + calc((n-1)//2, k+1) + " " + str(k)
print(calc(int(input()), 1))
|
s605882678
|
Accepted
| 18
| 3,064
| 844
|
fact = [1]
for i in range(1, 101):
fact.append(fact[-1]*i)
def C(a, b):
return fact[a]//(fact[b]*fact[a-b])
def sC(a, b):
ret = 0
for i in range(min(b, a-b) + 1):
ret += C(b, i) * C(a - b, i)
return ret
def calc(n, k):
if n<=0:
return []
i = 0
while (1<<(i+1)) -1 <= n:
i += 1
ret = [str(k)] * (i+1)
rem = n-((1<<i) - 1)
kk = k + 1
i += 1
j = i // 2
while j > 0:
while True:
tmp = sC(i, j)
if tmp <= rem:
ret.insert(j, str(kk))
ret.append(str(kk))
kk += 1
rem -= tmp
else:
break
j -= 1
return ret + calc(rem, kk)
N = int(input())
X = calc(N, 1)
print(len(X))
print(" ".join(X))
|
s337417093
|
p03729
|
u150641538
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 98
|
You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`.
|
a,b,c = input().split()
if a[-1] == b[0] and b[-1] == c[0]:
print("Yes")
else:
print("No")
|
s264041995
|
Accepted
| 17
| 2,940
| 98
|
a,b,c = input().split()
if a[-1] == b[0] and b[-1] == c[0]:
print("YES")
else:
print("NO")
|
s372592767
|
p03943
|
u317711717
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 108
|
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
import math
a,b,c=map(int,input().split())
l=[a,b,c]
sorted(l)
print("Yes" if l[0]+l[1]==l[2] else "No")
|
s194728609
|
Accepted
| 17
| 2,940
| 107
|
import math
a,b,c=map(int,input().split())
l=[a,b,c]
l.sort()
print("Yes" if l[0]+l[1]==l[2] else "No")
|
s199639613
|
p03780
|
u593590006
| 2,000
| 262,144
|
Wrong Answer
| 20
| 3,572
| 207
|
AtCoDeer the deer has N cards with positive integers written on them. The number on the i-th card (1≤i≤N) is a_i. Because he loves big numbers, he calls a subset of the cards _good_ when the sum of the numbers written on the cards in the subset, is K or greater. Then, for each card i, he judges whether it is _unnecessary_ or not, as follows: * If, for any good subset of the cards containing card i, the set that can be obtained by eliminating card i from the subset is also good, card i is unnecessary. * Otherwise, card i is NOT unnecessary. Find the number of the unnecessary cards. Here, he judges each card independently, and he does not throw away cards that turn out to be unnecessary.
|
n,k=map(int,input().split())
l=[int(i) for i in input().split()]
#sum of sub >=k
l.sort()
s=sum(l)
if s<k:
print(n)
else:
cnt=0
for i in l:
if s-i<k:
cnt+=1
print(cnt)
|
s576426742
|
Accepted
| 21
| 3,572
| 691
|
N, K = map(int, input().split())
a = list(map(int, input().split()))
a = sorted(a)
ans = N
psum = 0
for i in range(N-1, -1, -1) :
if psum+a[i] < K :
psum += a[i]
else :
ans = min(ans, i)
print(ans)
'''
notes:
1-->we include elems that are less than k
2-->if some x is necessary all elems grtr than x are also necessary
Base for some x to be necessary::
atleast one of the sums in range [sm-x,sm) must be possible
excluding x
now at any instance after traversing x elements
we make sure that we have max posible sm <k
now
if x+sm <k
we cannot include this anyway
otherwise
we can make a set with {sm,x} now x is important
BOOYAH (:
'''
|
s133794208
|
p03854
|
u920543723
| 2,000
| 262,144
|
Wrong Answer
| 234
| 13,704
| 880
|
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`.
|
import numpy as np
import math
import sys
def sinput(): return sys.stdin.readline()
def iinput(): return int(sinput())
def imap(): return map(int, sinput().split())
def fmap(): return map(float, sinput().split())
def iarr(): return list(imap())
def farr(): return list(fmap())
def sarr(): return sinput().split()
s = list(input())
s.reverse()
v = []
v.append(list("dream"))
v.append(list("erase"))
v.append(list("eraser"))
v.append(list("dreamer"))
for i in range(4):
v[i].reverse()
now = 0
while(1):
check = False
for i in range(4):
if now+len(v[i]) > len(s): continue
tmp = ''.join(s[now:now+len(v[i])])
x = ''.join(v[i])
if tmp == x:
print(tmp, x)
check = True
now += len(v[i])
if check: break
if not check: break
if now == len(s): break
if check: print("YES")
else: print("NO")
|
s598134561
|
Accepted
| 210
| 13,404
| 854
|
import numpy as np
import math
import sys
def sinput(): return sys.stdin.readline()
def iinput(): return int(sinput())
def imap(): return map(int, sinput().split())
def fmap(): return map(float, sinput().split())
def iarr(): return list(imap())
def farr(): return list(fmap())
def sarr(): return sinput().split()
s = list(input())
s.reverse()
v = []
v.append(list("dream"))
v.append(list("erase"))
v.append(list("eraser"))
v.append(list("dreamer"))
for i in range(4):
v[i].reverse()
now = 0
while(1):
check = False
for i in range(4):
if now+len(v[i]) > len(s): continue
tmp = ''.join(s[now:now+len(v[i])])
x = ''.join(v[i])
if tmp == x:
check = True
now += len(v[i])
if check: break
if not check: break
if now == len(s): break
if check: print("YES")
else: print("NO")
|
s075888958
|
p03434
|
u512099209
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 124
|
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())
a = list(map(int, input().split()))
a.sort()
alice = sum(a[0::2])
bob = sum(a[1::2])
print(alice - bob)
|
s049587980
|
Accepted
| 19
| 2,940
| 136
|
N = int(input())
a = list(map(int, input().split()))
a.sort(reverse=True)
alice = sum(a[0::2])
bob = sum(a[1::2])
print(alice - bob)
|
s766478742
|
p02613
|
u529706830
| 2,000
| 1,048,576
|
Wrong Answer
| 143
| 9,072
| 325
|
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.
|
s = int(input())
ac = 0
wa = 0
tle = 0
re = 0
for i in range(1,s):
ch = input()
if ch == "AC":
ac += 1
elif ch == "WA":
wa += 1
elif ch == "TLE":
tle +=1
elif ch == "RE":
re += 1
print("AC x "+str(ac))
print("WA x "+str(wa))
print("TLE x "+str(tle))
print("RE x "+str(re))
|
s392366567
|
Accepted
| 147
| 9,196
| 327
|
s = int(input())
ac = 0
wa = 0
tle = 0
re = 0
for i in range(1,s+1):
ch = input()
if ch == "AC":
ac += 1
elif ch == "WA":
wa += 1
elif ch == "TLE":
tle +=1
elif ch == "RE":
re += 1
print("AC x "+str(ac))
print("WA x "+str(wa))
print("TLE x "+str(tle))
print("RE x "+str(re))
|
s173546883
|
p03501
|
u600261652
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 54
|
You are parking at a parking lot. You can choose from the following two fee plans: * Plan 1: The fee will be A×T yen (the currency of Japan) when you park for T hours. * Plan 2: The fee will be B yen, regardless of the duration. Find the minimum fee when you park for N hours.
|
N, A, B = map(int, input().split())
print(min(N-A, B))
|
s219421596
|
Accepted
| 17
| 2,940
| 54
|
N, A, B = map(int, input().split())
print(min(N*A, B))
|
s078605516
|
p03399
|
u019578976
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 48
|
You planned a trip using trains and buses. The train fare will be A yen (the currency of Japan) if you buy ordinary tickets along the way, and B yen if you buy an unlimited ticket. Similarly, the bus fare will be C yen if you buy ordinary tickets along the way, and D yen if you buy an unlimited ticket. Find the minimum total fare when the optimal choices are made for trains and buses.
|
print(min(input(),input())+min(input(),input()))
|
s160868467
|
Accepted
| 17
| 2,940
| 68
|
print(min(int(input()),int(input()))+min(int(input()),int(input())))
|
s624148056
|
p03457
|
u065099501
| 2,000
| 262,144
|
Wrong Answer
| 216
| 17,684
| 512
|
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
N = int(input())
T = [];X = [];Y = []
for _ in range(N):
t,x,y = map(int, input().split())
T.append(t)
X.append(x)
Y.append(y)
if N == 1:
print('Yes' if T[0] == sum(X+Y) else 'No')
exit()
for i in range(N):
d = abs(X[i] + Y[i] - X[i-1] - Y[i-1])
t = T[i] - T[i-1]
if t >= d:
t = t % 2
d = d % 2
if (t != 0 and d == 0) or (t == 0 and d != 0):
print('No')
exit()
else:
print('No')
exit()
else:
print('Yes')
|
s609952862
|
Accepted
| 269
| 17,596
| 406
|
N = int(input())
T = [];X = [];Y = []
for _ in range(N):
t,x,y = map(int, input().split())
T.append(t)
X.append(x)
Y.append(y)
if N == 1:
print('Yes' if T[0] == sum(X+Y) else 'No')
exit()
for i in range(N):
if i == 0: continue
d = abs(X[i] + Y[i] - X[i-1] - Y[i-1])
t = T[i] - T[i-1]
if (t + d) % 2 or t < d:
print('No')
exit()
else:
print('Yes')
|
s637240849
|
p02748
|
u623687794
| 2,000
| 1,048,576
|
Wrong Answer
| 589
| 38,952
| 319
|
You are visiting a large electronics store to buy a refrigerator and a microwave. The store sells A kinds of refrigerators and B kinds of microwaves. The i-th refrigerator ( 1 \le i \le A ) is sold at a_i yen (the currency of Japan), and the j-th microwave ( 1 \le j \le B ) is sold at b_j yen. You have M discount tickets. With the i-th ticket ( 1 \le i \le M ), you can get a discount of c_i yen from the total price when buying the x_i-th refrigerator and the y_i-th microwave together. Only one ticket can be used at a time. You are planning to buy one refrigerator and one microwave. Find the minimum amount of money required.
|
a,b,m=map(int,input().split())
fri=list(map(int,input().split()))
mic=list(map(int,input().split()))
wari=[list(map(int,input().split())) for i in range(m)]
fri.sort()
mic.sort()
mi=fri[0]+mic[0]
for i in range(m):
if fri[wari[i][0]-1]+mic[wari[i][1]-1]<mi:
mi=fri[wari[i][0]-1]+mic[wari[i][1]-1]
print(mi)
|
s037104689
|
Accepted
| 466
| 38,568
| 323
|
a,b,m=map(int,input().split())
fri=list(map(int,input().split()))
mic=list(map(int,input().split()))
wari=[list(map(int,input().split())) for i in range(m)]
mi=min(fri)+min(mic)
for i in range(m):
if fri[wari[i][0]-1]+mic[wari[i][1]-1]-wari[i][2]<mi:
mi=fri[wari[i][0]-1]+mic[wari[i][1]-1]-wari[i][2]
print(mi)
|
s499415267
|
p03415
|
u422272120
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 57
|
We have a 3×3 square grid, where each square contains a lowercase English letters. The letter in the square at the i-th row from the top and j-th column from the left is c_{ij}. Print the string of length 3 that can be obtained by concatenating the letters in the squares on the diagonal connecting the top-left and bottom-right corner of the grid, from the top-left to bottom-right.
|
print (input()[0])
print (input()[1])
print (input()[2])
|
s646844144
|
Accepted
| 18
| 2,940
| 75
|
ans = ""
ans += input()[0]
ans += input()[1]
ans += input()[2]
print (ans)
|
s945613306
|
p03387
|
u354638986
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,104
| 4,592
| 530
|
You are given three integers A, B and C. Find the minimum number of operations required to make A, B and C all equal by repeatedly performing the following two kinds of operations in any order: * Choose two among A, B and C, then increase both by 1. * Choose one among A, B and C, then increase it by 2. It can be proved that we can always make A, B and C all equal by repeatedly performing these operations.
|
import random
data = list(map(int, input().split()))
lst = data[:]
n = 0
minimum = 1000000
for i in range(1000):
while True:
if data[0] == data[1] and data[1] == data[2]:
data = lst[:]
if n < minimum:
minimum = n
n = 0
break
f = random.randint(0, 1)
if f == 0:
lst[random.randint(0, 2)] += 1
lst[random.randint(0, 2)] += 1
else:
lst[random.randint(0, 2)] += 2
n += 1
print(minimum)
|
s074097948
|
Accepted
| 17
| 2,940
| 143
|
a, b, c = map(int, input().split())
m = max(a, b, c)
if 3*m%2 == (a+b+c)%2:
print((3*m-(a+b+c))//2)
else:
print((3*(m+1)-(a+b+c))//2)
|
s938687985
|
p03455
|
u345136423
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 90
|
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%2 !=0 or b%2!=0:
print('Odd')
else:
print('Even')
|
s141416918
|
Accepted
| 17
| 2,940
| 95
|
a,b = map(int,input().split())
if a%2 !=0 and b%2!=0 :
print('Odd')
else:
print('Even')
|
s593163799
|
p03044
|
u642012866
| 2,000
| 1,048,576
|
Wrong Answer
| 706
| 47,768
| 512
|
We have a tree with N vertices numbered 1 to N. The i-th edge in the tree connects Vertex u_i and Vertex v_i, and its length is w_i. Your objective is to paint each vertex in the tree white or black (it is fine to paint all vertices the same color) so that the following condition is satisfied: * For any two vertices painted in the same color, the distance between them is an even number. Find a coloring of the vertices that satisfies the condition and print it. It can be proved that at least one such coloring exists under the constraints of this problem.
|
from collections import deque
N = int(input())
ki = [[] for _ in range(N)]
for _ in range(N-1):
u, v, w = map(int, input().split())
ki[u-1].append([v-1, w])
ki[v-1].append([u-1, w])
f = [False]*N
f[0] = True
c = [0]*N
q = deque([0])
while q:
now = q.popleft()
for nxt, w in ki[now]:
if f[nxt]:
continue
if w&1:
c[nxt] = c[now]^1
else:
c[nxt] = c[now]
q.append(nxt)
f[nxt] = True
print(f)
for a in c:
print(a)
|
s712583503
|
Accepted
| 741
| 46,820
| 503
|
from collections import deque
N = int(input())
ki = [[] for _ in range(N)]
for _ in range(N-1):
u, v, w = map(int, input().split())
ki[u-1].append([v-1, w])
ki[v-1].append([u-1, w])
f = [False]*N
f[0] = True
c = [0]*N
q = deque([0])
while q:
now = q.popleft()
for nxt, w in ki[now]:
if f[nxt]:
continue
if w&1:
c[nxt] = c[now]^1
else:
c[nxt] = c[now]
q.append(nxt)
f[nxt] = True
for a in c:
print(a)
|
s429949703
|
p00069
|
u024715419
| 1,000
| 131,072
|
Wrong Answer
| 40
| 5,616
| 1,073
|
縦線が n 本のあみだくじがあります。このあみだくじは以下の条件を満たしています。 * 横線は真横に引きます。斜めに引くことはありません。 * 横線は必ず隣り合った縦線同士をつなぎます。つまり、横線が縦線をまたぐことはありません。 * どの縦線についても同じ点から左右同時に横線が出ることはありません。つまり、横線が縦線を横切ることはありません。 * 当りはひとつだけです。 下図 に n = 5 のときの、あみだくじの例を示します。上側の数字は縦線の番号(左から1, 2, 3, 4, 5 ) を表します。☆が当たりです。 縦線の本数 n、選んだ縦線の番号 m、あみだくじの当りの場所、各段における横線の有無を読み込んで、当りにたどり着けるかどうかの判定を出力するプログラムを作成してください。ただし、与えられたあみだくじの任意の位置に1 本だけ横線を付け加えることができるものとします（付け加えなくてもかまいません）。横線を1 本付け加えた後のあみだくじも前述の条件を満たしていなければなりません。
|
def check_lots(start, goal, line):
for i in range(len(line)):
for j in range(len(line[i])):
if line[i][j] == 1:
if j == start - 1:
start -= 1
elif j == start:
start += 1
return start == goal
while True:
n = int(input())
if n == 0:
break
m = int(input())
s = int(input())
d = int(input())
l = []
for i in range(d):
l.append([0] + list(map(int, list(input()))) + [0])
d_ans = 0
n_ans = 0
if check_lots(m, s, l):
print(0)
else:
for i in range(d):
for j in range(1, n):
l_tmp = l[:]
if sum(l_tmp[i][j-1:j+2]) > 0:
continue
l_tmp[i][j] = 1
if check_lots(m, s, l_tmp):
d_ans = i + 1
n_ans = j
break
else:
continue
break
if d_ans:
print(d_ans, n_ans)
else:
print(1)
|
s202637414
|
Accepted
| 190
| 6,340
| 1,097
|
import copy
def check_lots(start, goal, line):
for i in range(len(line)):
for j in range(len(line[i])):
if line[i][j] == 1:
if j == start - 1:
start -= 1
elif j == start:
start += 1
return start == goal
while True:
n = int(input())
if n == 0:
break
m = int(input())
s = int(input())
d = int(input())
l = []
for i in range(d):
l.append([0] + list(map(int, list(input()))) + [0])
d_ans = 0
n_ans = 0
if check_lots(m, s, l):
print(0)
else:
for i in range(d):
for j in range(1, n):
l_tmp = copy.deepcopy(l[:])
if sum(l[i][j-1:j+2]) > 0:
continue
l_tmp[i][j] = 1
if check_lots(m, s, l_tmp):
d_ans = i + 1
n_ans = j
break
else:
continue
break
if d_ans:
print(d_ans, n_ans)
else:
print(1)
|
s583988606
|
p03543
|
u121732701
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 201
|
We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**?
|
N = input()
j = -1
count=0
count1=0
for i in N:
if i == j:
count+=1
elif count==3:
print("Yes")
break
j = i
count1 += 1
if count1 == 4:
print("No")
|
s489181128
|
Accepted
| 18
| 2,940
| 93
|
N = input()
if N[0]==N[1]==N[2] or N[1]==N[2]==N[3]:
print("Yes")
else:
print("No")
|
s312450699
|
p03963
|
u912862653
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 53
|
There are N balls placed in a row. AtCoDeer the deer is painting each of these in one of the K colors of his paint cans. For aesthetic reasons, any two adjacent balls must be painted in different colors. Find the number of the possible ways to paint the balls.
|
N, K = map(int, input().split())
print(K*(K-1)^(N-1))
|
s377667665
|
Accepted
| 18
| 2,940
| 54
|
N, K = map(int, input().split())
print(K*(K-1)**(N-1))
|
s700055692
|
p03407
|
u327248573
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,316
| 114
|
An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.
|
# -- coding: utf-8 --
A, B, C = map(int, input().split(' '))
if C <= A + B:
print('yes')
else:
print('no')
|
s958457647
|
Accepted
| 17
| 2,940
| 114
|
# -- coding: utf-8 --
A, B, C = map(int, input().split(' '))
if C <= A + B:
print('Yes')
else:
print('No')
|
s317612177
|
p02747
|
u442877951
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 122
|
A Hitachi string is a concatenation of one or more copies of the string `hi`. For example, `hi` and `hihi` are Hitachi strings, while `ha` and `hii` are not. Given a string S, determine whether S is a Hitachi string.
|
S = str(input())
if "hi" in S and S.replace('hi', '') == "":
print("Yes")
else:
print("No")
print(S.replace('hi', ''))
|
s253528820
|
Accepted
| 21
| 3,316
| 335
|
S = str(input())
ans = 'Yes'
from collections import Counter
SC = Counter(list(S))
h,i = 0,0
for a,b in SC.items():
if a == 'h':
h = b
elif a == 'i':
i = b
else:
ans = 'No'
break
if h == i:
for i in range(0,len(S)-1,2):
if S[i]+S[i+1] != 'hi':
ans = 'No'
break
else:
ans = 'No'
print(ans)
|
s213348980
|
p03608
|
u297109012
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 6,692
| 918
|
There are N towns in the State of Atcoder, connected by M bidirectional roads. The i-th road connects Town A_i and B_i and has a length of C_i. Joisino is visiting R towns in the state, r_1,r_2,..,r_R (not necessarily in this order). She will fly to the first town she visits, and fly back from the last town she visits, but for the rest of the trip she will have to travel by road. If she visits the towns in the order that minimizes the distance traveled by road, what will that distance be?
|
from itertools import permutations
def warshal(N, ABCs):
dists = [[10 ** 10 for _ in range(N)] for __ in range(N)]
for i in range(N):
dists[i][i] = 0
for a, b, c in ABCs:
dists[a - 1][b - 1] = c
dists[b - 1][a - 1] = c
for s in range(N):
for e in range(N):
for i in range(N):
dists[s][e] = min(dists[s][e], dists[s][i] + dists[i][e])
return dists
def solve(N, M, R, Rs, ABCs):
print(N, ",", M, ",", R, ",", Rs, ", ", ABCs)
dists = warshal(N, ABCs)
ans = 10 ** 10
for way in permutations(Rs):
ans = min(ans, sum([dists[i - 1][j - 1] for i, j in zip(way, way[1:])]))
return ans
if __name__ == "__main__":
N, M, R = tuple(map(int, input().split(" ")))
Rs = list(map(int, input().split(" ")))
ABCs = [tuple(map(int, input().split(" "))) for _ in range(M)]
print(solve(N, M, R, Rs, ABCs))
|
s750604869
|
Accepted
| 578
| 6,052
| 2,528
|
from itertools import permutations
import heapq
import sys
INF = 2 ** sys.int_info.bits_per_digit
def warshall(N, ABCs):
dists = [[INF for _ in range(N)] for __ in range(N)]
for i in range(N):
dists[i][i] = 0
for a, b, c in ABCs:
dists[a - 1][b - 1] = c
dists[b - 1][a - 1] = c
for s in range(N):
for e in range(N):
for i in range(N):
dists[s][e] = min(dists[s][e], dists[s][i] + dists[i][e])
return dists
def dijkstra(s, e, N, ABCs):
adjlist = [[] for _ in range(N)]
for a, b, c in ABCs:
adjlist[a - 1].append((b, c))
adjlist[b - 1].append((a, c))
q = [(0, s)]
while len(q):
d, i = heapq.heappop(q)
if i == e:
return d
for nexti, nextd in adjlist[i - 1]:
heapq.heappush(q, (d + nextd, nexti))
return None
def dense_dijkstra(s, e, N, ABCs):
distsmap = [[-1 for _ in range(N)] for __ in range(N)]
for a, b, c in ABCs:
distsmap[a - 1][b - 1] = c
distsmap[b - 1][a - 1] = c
dists = [INF for _ in range(N)]
visited = [False for _ in range(N)]
dists[s - 1] = 0
while True:
minW, index = min([(dists[i], i) for i in range(N) if not visited[i]])
visited[index] = True
if minW == INF:
return -1
if index + 1 == e:
return minW
for rindex, w in enumerate(distsmap[index]):
if w == -1:
continue
dists[rindex] = min(dists[rindex], minW + w)
# dists = warshall(N, ABCs)
# ans = INF
# for way in permutations(Rs):
# return ans
def solve(N, M, R, Rs, ABCs):
dists = [[-1 for _ in range(N)] for __ in range(N)]
ans = INF
for way in permutations(Rs):
tans = 0
for s, e in zip(way, way[1:]):
if dists[s - 1][e - 1] == -1:
d = dense_dijkstra(s, e, N, ABCs)
dists[s - 1][e - 1] = d
dists[e - 1][s - 1] = d
else:
d = dists[s - 1][e - 1]
tans += d
ans = min(ans, tans)
return ans
if __name__ == "__main__":
N, M, R = tuple(map(int, input().split(" ")))
Rs = list(map(int, input().split(" ")))
ABCs = [tuple(map(int, input().split(" "))) for _ in range(M)]
print(solve(N, M, R, Rs, ABCs))
|
s556761272
|
p03573
|
u556477263
| 2,000
| 262,144
|
Wrong Answer
| 29
| 9,028
| 86
|
You are given three integers, A, B and C. Among them, two are the same, but the remaining one is different from the rest. For example, when A=5,B=7,C=5, A and C are the same, but B is different. Find the one that is different from the rest among the given three integers.
|
a,b,c = map(int,input().split())
if a != b and a == c:
print(a)
else:
print(b)
|
s044916852
|
Accepted
| 27
| 9,000
| 123
|
a,b,c = map(int,input().split())
if a != b and a == c:
print(b)
elif a != b and a != c:
print(a)
else:
print(c)
|
s327926076
|
p03671
|
u486773779
| 2,000
| 262,144
|
Wrong Answer
| 24
| 9,008
| 67
|
Snuke is buying a bicycle. The bicycle of his choice does not come with a bell, so he has to buy one separately. He has very high awareness of safety, and decides to buy two bells, one for each hand. The store sells three kinds of bells for the price of a, b and c yen (the currency of Japan), respectively. Find the minimum total price of two different bells.
|
price=list(input().split())
price.sort()
print(price[0]+price[1])
|
s940180744
|
Accepted
| 28
| 9,036
| 55
|
a,b,c=map(int,input().split())
print(a+b+c-max(a,b,c))
|
s021119765
|
p02406
|
u313089641
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,592
| 215
|
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())
def three(n):
result = [" ", 3]
while result[-1] < n:
result.append(result[-1] + 3)
result = map(str, result)
result = " ".join(result)
return(result)
print(three(n))
|
s558744215
|
Accepted
| 20
| 5,632
| 243
|
n = int(input())
def three(n):
result = ""
counter = 3
while counter < n+1:
if "3" in str(counter) or counter % 3 == 0:
result += " {}".format(counter)
counter += 1
return(result)
print(three(n))
|
s184593284
|
p02262
|
u588555117
| 6,000
| 131,072
|
Wrong Answer
| 20
| 7,664
| 510
|
Shell Sort is a generalization of [Insertion Sort](http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ALDS1_1_A) to arrange a list of $n$ elements $A$. 1 insertionSort(A, n, g) 2 for i = g to n-1 3 v = A[i] 4 j = i - g 5 while j >= 0 && A[j] > v 6 A[j+g] = A[j] 7 j = j - g 8 cnt++ 9 A[j+g] = v 10 11 shellSort(A, n) 12 cnt = 0 13 m = ? 14 G[] = {?, ?,..., ?} 15 for i = 0 to m-1 16 insertionSort(A, n, G[i]) A function shellSort(A, n) performs a function insertionSort(A, n, g), which considers every $g$-th elements. Beginning with large values of $g$, it repeats the insertion sort with smaller $g$. Your task is to complete the above program by filling ?. Write a program which reads an integer $n$ and a sequence $A$, and prints $m$, $G_i (i = 0, 1, ..., m − 1)$ in the pseudo code and the sequence $A$ in ascending order. The output of your program must meet the following requirements: * $1 \leq m \leq 100$ * $0 \leq G_i \leq n$ * cnt does not exceed $\lceil n^{1.5}\rceil$
|
m = 3
G = [7,4,1]
def insertion_sort(array,g):
for i in range(g,len(array)-1):
v = array[i]
j = i-g
while j>=0 and array[j] > v:
array[j+g] = array[j]
j = j-g
array[-1] += 1
array[j+g] = v
def shell_sort(array,G):
for i in G:
insertion_sort(array,i)
n = int(input())
arr = [int(input()) for i in range(n)] + [0]
shell_sort(arr,G)
print(m)
print(' '.join([str(i) for i in G]))
print(arr[-1])
[print(i) for i in arr[:-1]]
|
s431815899
|
Accepted
| 17,190
| 62,964
| 594
|
G = [120001,60001,30001,15001,3001,901,601,301,91,58,31,7,4,1]
def insertion_sort(array,g):
for i in range(g,len(array)-1):
v = array[i]
j = i-g
while j>=0 and array[j] > v:
array[j+g] = array[j]
j = j-g
array[-1] += 1
array[j+g] = v
def shell_sort(array,G):
for i in G:
insertion_sort(array,i)
n = int(input())
arr = [int(input()) for i in range(n)] + [0]
Gn = [i for i in G if i <= n]
shell_sort(arr,Gn)
print(len(Gn))
print(' '.join([str(i) for i in Gn]))
print(arr[-1])
[print(i) for i in arr[:-1]]
|
s269292870
|
p03920
|
u619819312
| 2,000
| 262,144
|
Wrong Answer
| 22
| 3,316
| 178
|
The problem set at _CODE FESTIVAL 20XX Finals_ consists of N problems. The score allocated to the i-th (1≦i≦N) problem is i points. Takahashi, a contestant, is trying to score exactly N points. For that, he is deciding which problems to solve. As problems with higher scores are harder, he wants to minimize the highest score of a problem among the ones solved by him. Determine the set of problems that should be solved.
|
n=int(input())
p=int(n**0.5)
t=0
for i in range(p,10**5):
if i*(i+1)//2>n:
t=i
l=t*(t+1)//2-n
break
for i in range(1,t):
if i!=l:
print(i)
|
s332438146
|
Accepted
| 22
| 3,316
| 178
|
n=int(input())
p=int(n**0.5)
t=0
for i in range(10**5):
if i*(i+1)//2>n:
t=i
l=t*(t+1)//2-n
break
for i in range(1,t+1):
if i!=l:
print(i)
|
s253164287
|
p02402
|
u239524997
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,532
| 157
|
Write a program which reads a sequence of $n$ integers $a_i (i = 1, 2, ... n)$, and prints the minimum value, maximum value and sum of the sequence.
|
a = input()
num = input()
num_1 = num.split()
s = 0
cnt = 0
for i in range(len(num_1)):
s += int(num_1[cnt])
cnt += 1
print(min(num_1),max(num_1),s)
|
s442910420
|
Accepted
| 60
| 8,716
| 236
|
a = input()
num = input()
num_1 = num.split()
s = 0
cnt = 0
for i in range(len(num_1)):
s += int(num_1[cnt])
cnt += 1
b = []
cnt = 0
for i in range(len(num_1)):
b.append(int(num_1[cnt]))
cnt += 1
print(min(b),max(b),s)
|
s248809457
|
p03608
|
u670180528
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 4,648
| 626
|
There are N towns in the State of Atcoder, connected by M bidirectional roads. The i-th road connects Town A_i and B_i and has a length of C_i. Joisino is visiting R towns in the state, r_1,r_2,..,r_R (not necessarily in this order). She will fly to the first town she visits, and fly back from the last town she visits, but for the rest of the trip she will have to travel by road. If she visits the towns in the order that minimizes the distance traveled by road, what will that distance be?
|
from itertools import permutations
n,m,r=map(int,input().split())
*R,=map(int,input().split())
INF=float("inf")
dist=[[INF]*n for _ in range(n)]
for i in range(n):
dist[i][i]=0
for _ in range(m):
a,b,c=map(int,input().split())
dist[a-1][b-1]=c
dist[b-1][a-1]=c
for k in range(n):
for i in range(n):
for j in range(n):
dist[i][j]=min(dist[i][j],dist[i][k]+dist[k][j])
v=[]
for i in R:
t=[]
for j in R:
t.append(dist[i-1][j-1])
v.append(t)
mn=INF
print(v)
for t in permutations(range(r)):
a=0;f=1
for i,j in zip(t,t[1:]):
if v[i][j]==INF:
f=0
break
else:
a+=v[i][j]
if f:
mn=min(mn,a)
print(mn)
|
s159432829
|
Accepted
| 504
| 22,224
| 313
|
from scipy.sparse import csr_matrix
from scipy.sparse.csgraph import floyd_warshall
from itertools import permutations
n,m,r,*L=map(int,open(0).read().split())
f=floyd_warshall(csr_matrix((L[r+2::3],(L[r::3],L[r+1::3])),(n+1,n+1)),0)
print(int(min(sum(f[s,t]for s,t in zip(o,o[1:]))for o in permutations(L[:r]))))
|
s996878164
|
p03545
|
u546198000
| 2,000
| 262,144
|
Wrong Answer
| 26
| 9,020
| 289
|
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.
|
A = list(input())
sign = ['+','-']
for i in range(2):
for j in range(2):
for k in range(2):
if eval(A[0]+sign[i]+A[1]+sign[j]+A[2]+sign[k]+A[3]) == 7:
print('{}{}{}{}{}{}{}'.format(A[0],sign[i],A[1],sign[j],A[2],sign[k],A[3]))
exit()
|
s091358068
|
Accepted
| 28
| 9,068
| 291
|
A = list(input())
sign = ['+','-']
for i in range(2):
for j in range(2):
for k in range(2):
if eval(A[0]+sign[i]+A[1]+sign[j]+A[2]+sign[k]+A[3]) == 7:
print('{}{}{}{}{}{}{}=7'.format(A[0],sign[i],A[1],sign[j],A[2],sign[k],A[3]))
exit()
|
s144358796
|
p03997
|
u045408189
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 64
|
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*1/2)
|
s538550772
|
Accepted
| 17
| 2,940
| 65
|
a=int(input())
b=int(input())
h=int(input())
print((a+b)*h*1//2)
|
s942968784
|
p03795
|
u120026978
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 47
|
Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.
|
n = int(input())
print(n * 800 - 200 * n // 15)
|
s763074134
|
Accepted
| 17
| 2,940
| 49
|
n = int(input())
print(n * 800 - 200 * (n // 15))
|
s239392533
|
p03598
|
u019584841
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 165
|
There are N balls in the xy-plane. The coordinates of the i-th of them is (x_i, i). Thus, we have one ball on each of the N lines y = 1, y = 2, ..., y = N. In order to collect these balls, Snuke prepared 2N robots, N of type A and N of type B. Then, he placed the i-th type-A robot at coordinates (0, i), and the i-th type-B robot at coordinates (K, i). Thus, now we have one type-A robot and one type-B robot on each of the N lines y = 1, y = 2, ..., y = N. When activated, each type of robot will operate as follows. * When a type-A robot is activated at coordinates (0, a), it will move to the position of the ball on the line y = a, collect the ball, move back to its original position (0, a) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. * When a type-B robot is activated at coordinates (K, b), it will move to the position of the ball on the line y = b, collect the ball, move back to its original position (K, b) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. Snuke will activate some of the 2N robots to collect all of the balls. Find the minimum possible total distance covered by robots.
|
n=int(input())
k=int(input())
l=[int(x) for x in range(n)]
s=0
for i in range(n):
if abs(l[i]-k)>abs(l[i]):
s+=abs(l[i])
else:
s+=abs(l[i]-k)
print(s)
|
s495708831
|
Accepted
| 17
| 2,940
| 132
|
n = int(input())
k = int(input())
x = [int(x) for x in input().split()]
ans = 0
for i in x:
ans += min(i*2, (k-i)*2)
print(ans)
|
s484372357
|
p02612
|
u566371932
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 9,180
| 30
|
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.
|
s = int(input())
print(s%1000)
|
s661394576
|
Accepted
| 34
| 9,048
| 92
|
s = int(input())
change = s % 1000
if change == 0:
print(0)
else:
print(1000-change)
|
s509308205
|
p02418
|
u506705885
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,452
| 142
|
Write a program which finds a pattern $p$ in a ring shaped text $s$.
|
letters=input()
letters=letters+letters
want_find=input()
print(letters)
if letters.find(want_find)>=0:
print('Yes')
else:
print('No')
|
s804659478
|
Accepted
| 50
| 7,460
| 121
|
letters=input()
letters=letters+letters
want_find=input()
if want_find in letters:
print('Yes')
else:
print('No')
|
s785754309
|
p03433
|
u748488158
| 2,000
| 262,144
|
Wrong Answer
| 27
| 9,164
| 87
|
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
N = int(input())
A = int(input())
if(N%5 <= A):
print("YES")
else:
print("NO")
|
s727383581
|
Accepted
| 26
| 8,996
| 89
|
N = int(input())
A = int(input())
if(N%500 <= A):
print("Yes")
else:
print("No")
|
s551111984
|
p03377
|
u598009172
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 90
|
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=(int(i) for i in input().split())
if a+b >= x:
print("Yes")
else :
print("No")
|
s411368674
|
Accepted
| 19
| 2,940
| 100
|
a,b,x=(int(i) for i in input().split())
if a+b > x and a <= x:
print("YES")
else :
print("NO")
|
s548416366
|
p03610
|
u801247169
| 2,000
| 262,144
|
Wrong Answer
| 34
| 9,356
| 145
|
You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.
|
s = input ()
odd_num = list()
for i in range(len(s)):
if i % 2 == 0:
odd_num.append(s[i])
answer = " ".join(odd_num)
print(answer)
|
s640698906
|
Accepted
| 27
| 9,048
| 25
|
s = input()
print(s[::2])
|
s093235346
|
p03761
|
u759412327
| 2,000
| 262,144
|
Wrong Answer
| 22
| 3,316
| 145
|
Snuke loves "paper cutting": he cuts out characters from a newspaper headline and rearranges them to form another string. He will receive a headline which contains one of the strings S_1,...,S_n tomorrow. He is excited and already thinking of what string he will create. Since he does not know the string on the headline yet, he is interested in strings that can be created regardless of which string the headline contains. Find the longest string that can be created regardless of which string among S_1,...,S_n the headline contains. If there are multiple such strings, find the lexicographically smallest one among them.
|
from collections import Counter
N = int(input())
C = Counter(input())
for n in range(N-1):
C&=Counter(input())
print(*C.elements(),sep="")
|
s621120230
|
Accepted
| 28
| 9,404
| 146
|
from collections import *
N = int(input())
C = Counter(input())
for n in range(N-1):
C&=Counter(input())
print(*sorted(C.elements()),sep="")
|
s468190352
|
p03760
|
u275685840
| 2,000
| 262,144
|
Wrong Answer
| 27
| 8,812
| 166
|
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions. You are given two strings O and E. O contains the characters at the odd- numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
|
O = input()
E = input()
num = 0
num_2 = 0
pw = ""
for i in list(O):
if num % 2 == 0:
pw+=i
num+=1
else:
pw+=E[num_2]
num+=1
num_2+=1
print(pw)
|
s277931080
|
Accepted
| 27
| 8,984
| 206
|
O = input()
E = input()
Ol = len(O)
El = len(E)
alllen = Ol + El
num = 0
num_2 = 0
pw = ""
for i in range(alllen):
if i % 2 == 0:
pw+=O[num]
num+=1
else:
pw+=E[num_2]
num_2+=1
print(pw)
|
s628175621
|
p03672
|
u842747358
| 2,000
| 262,144
|
Wrong Answer
| 22
| 3,064
| 504
|
We will call a string that can be obtained by concatenating two equal strings an _even_ string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not. You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
|
def func(x):
for i_ in range(len(x)):
y = x[:i_+1]
if len(x) % len(y) != 0 or len(x) % 2 == 1:
continue
count = 0
for j_ in range(len(x)//len(y)):
if x[len(y)*j_:len(y)*(j_+1)] != y or x == y:
break
count += 1
if count == len(x)//len(y):
return x
return ''
S = input()
ans = ' '
for i in range(len(S)):
ans = func(S[:-i-1])
if ans != '':
break
print(ans, len(ans))
|
s335763503
|
Accepted
| 18
| 2,940
| 234
|
S = input()
S = S[:-2]
flg = True
while flg:
flg = False
x = 0
for i in range(len(S)//2):
if S[x] != S[len(S)//2 + x]:
flg = True
S = S[:-2]
break
x += 1
print(len(S))
|
s356991525
|
p03777
|
u394482932
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 51
|
Two deer, AtCoDeer and TopCoDeer, are playing a game called _Honest or Dishonest_. In this game, an honest player always tells the truth, and an dishonest player always tell lies. You are given two characters a and b as the input. Each of them is either `H` or `D`, and carries the following information: If a=`H`, AtCoDeer is honest; if a=`D`, AtCoDeer is dishonest. If b=`H`, AtCoDeer is saying that TopCoDeer is honest; if b=`D`, AtCoDeer is saying that TopCoDeer is dishonest. Given this information, determine whether TopCoDeer is honest.
|
for s in input().upper().split():print(s[0],end='')
|
s532470202
|
Accepted
| 17
| 2,940
| 33
|
s=input();print('DH'[s[0]==s[2]])
|
s671271278
|
p03854
|
u661977789
| 2,000
| 262,144
|
Wrong Answer
| 77
| 3,188
| 210
|
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()[::-1]
res = "Yes"
while s:
start = len(s)
for t in ['maerd', 'remaerd', 'esare', 'resare']:
if t == s[:len(t)]:
s = s[len(t):]
if start == len(s):
res = "No"
break
print(res)
|
s706564869
|
Accepted
| 82
| 3,188
| 200
|
s = input()[::-1]
while s:
start = len(s)
for t in ['maerd', 'remaerd', 'esare', 'resare']:
if t == s[:len(t)]:
s = s[len(t):]
if start == len(s):
break
print("YNEOS"[len(s)>0::2])
|
s368155497
|
p02748
|
u891945807
| 2,000
| 1,048,576
|
Wrong Answer
| 2,114
| 124,780
| 558
|
You are visiting a large electronics store to buy a refrigerator and a microwave. The store sells A kinds of refrigerators and B kinds of microwaves. The i-th refrigerator ( 1 \le i \le A ) is sold at a_i yen (the currency of Japan), and the j-th microwave ( 1 \le j \le B ) is sold at b_j yen. You have M discount tickets. With the i-th ticket ( 1 \le i \le M ), you can get a discount of c_i yen from the total price when buying the x_i-th refrigerator and the y_i-th microwave together. Only one ticket can be used at a time. You are planning to buy one refrigerator and one microwave. Find the minimum amount of money required.
|
import numpy as np
A,B,M = map(int,input().rstrip().split())
a = list(map(int,input().rstrip().split()))
b = list(map(int,input().rstrip().split()))
suml = [[0] * len(b)] * len(a)
print(suml)
for i in range(len(a)):
for j in range(len(b)):
suml[i][j] = a[i]+b[j]
print(suml)
sumk = [[0] * len(b)] * len(a)
for k in range(M):
i,j,c = list(map(int,input().rstrip().split()))
if sumk[i-1][j-1] != 0:
sumk[i-1][j-1] = max(sumk[i-1][j-1],c)
else:
sumk[i-1][j-1] = c
x = np.array(suml)
y = np.array(sumk)
z = x - y
print(np.amin(z))
|
s319448189
|
Accepted
| 602
| 27,492
| 311
|
import numpy as np
A,B,M = map(int,input().rstrip().split())
a = list(map(int,input().rstrip().split()))
b = list(map(int,input().rstrip().split()))
minp = min(a) + min(b)
minpp = 10**6
for k in range(M):
i,j,c = map(int,input().rstrip().split())
minpp = min(a[i-1]+b[j-1]-c,minpp)
print(min(minp,minpp))
|
s815154340
|
p03730
|
u478417863
| 2,000
| 262,144
|
Wrong Answer
| 26
| 9,112
| 137
|
We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`.
|
A,B,C= list(map(int, input().strip().split()))
a=A%B
f="No"
for i in range(B):
if (a*i)%B==C:
f="Yes"
break
print(f)
|
s525729008
|
Accepted
| 30
| 9,096
| 137
|
A,B,C= list(map(int, input().strip().split()))
a=A%B
f="NO"
for i in range(B):
if (a*i)%B==C:
f="YES"
break
print(f)
|
s408875937
|
p03495
|
u845620905
| 2,000
| 262,144
|
Wrong Answer
| 246
| 24,748
| 254
|
Takahashi has N balls. Initially, an integer A_i is written on the i-th ball. He would like to rewrite the integer on some balls so that there are at most K different integers written on the N balls. Find the minimum number of balls that Takahashi needs to rewrite the integers on them.
|
k,n=map(int,input().split())
a = list(map(int,input().split()))
a=sorted(a)
d=[]
cnt=1
for i in range(k-1):
if a[i]==a[i+1]:
cnt+=1
else:
d.append(cnt)
cnt=0
d.append(cnt)
d=sorted(d)
ans=0
for i in range(len(d)-n):
ans+=d[i]
print(ans)
|
s056623205
|
Accepted
| 249
| 24,748
| 255
|
k,n=map(int,input().split())
a = list(map(int,input().split()))
a=sorted(a)
d=[]
cnt=1
for i in range(k-1):
if a[i]==a[i+1]:
cnt+=1
else:
d.append(cnt)
cnt=1
d.append(cnt)
d=sorted(d)
ans=0
for i in range(len(d)-n):
ans+=d[i]
print(ans)
|
s304108262
|
p03485
|
u226779434
| 2,000
| 262,144
|
Wrong Answer
| 26
| 9,076
| 50
|
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
a,b = map(int,input().split())
print(((a+b)//2)+1)
|
s554006511
|
Accepted
| 27
| 8,884
| 91
|
a,b =map(int,input().split())
if (a+b)%2 ==0:
print((a+b)//2)
else:
print(((a+b)//2)+1)
|
s102748810
|
p03672
|
u277802731
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 216
|
We will call a string that can be obtained by concatenating two equal strings an _even_ string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not. You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
|
#66b
s = input()
for _ in range(len(s)):
if len(s)%2==0:
if s[len(s)//2:]==s[:len(s)//2]:
print(len(s)//2)
exit()
else:
s=s[:-2]
else:
s=s[:-1]
|
s494989334
|
Accepted
| 18
| 2,940
| 118
|
#66b
s = input()
while True:
s = s[:-2]
i = len(s)//2
if s[:i] == s[i:]:
break
print(len(s))
|
s151027528
|
p02418
|
u400765446
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,556
| 286
|
Write a program which finds a pattern $p$ in a ring shaped text $s$.
|
def main():
sentence = input()
phrase = input()
for i in range(len(sentence)):
if phrase in sentence:
print("Yes")
break
else:
sentence = sentence[1:] + sentence[0]
print("No")
if __name__ == '__main__':
main()
|
s845062903
|
Accepted
| 20
| 5,560
| 188
|
s = input()
p = input()
cnt = 0
for i in range(len(p)):
if p in s:
cnt = 1
break
else:
s = s[1:]+s[0]
if cnt >= 1:
print('Yes')
else:
print('No')
|
s055188373
|
p03494
|
u396890425
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,104
| 2,940
| 164
|
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()
Numbers = list(map(int,input().split()))
ans=0
while True:
if all([x%2==0 for x in Numbers]):
ans+=1
[t/2 for t in Numbers]
print(ans)
|
s647154497
|
Accepted
| 18
| 3,060
| 150
|
n=input()
Numbers = list(map(int,input().split()))
ans=0
while all(x%2==0 for x in Numbers):
ans+=1
Numbers=[t/2 for t in Numbers]
print(ans)
|
s279682271
|
p02401
|
u992449685
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,564
| 80
|
Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part.
|
while(True):
a = input()
if '?' in a:
break
print(eval(a))
|
s424748577
|
Accepted
| 20
| 5,552
| 85
|
while(True):
a = input()
if '?' in a:
break
print(int(eval(a)))
|
s877345397
|
p02255
|
u755362870
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,600
| 321
|
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.
|
input_len = int(input())
l = input().split()
sort_line = [0 for i in range(input_len)]
for i in range(input_len):
sort_line[i] = int(l[i])
for i in range(1, input_len):
v = sort_line[i]
j = i - 1
while (j >= 0 and v < sort_line[j]):
sort_line[j+1] = sort_line[j]
j -= 1
sort_line[j + 1] = v
print(sort_line)
|
s204901168
|
Accepted
| 30
| 5,984
| 540
|
import sys
input_len = int(input())
l = input().split()
sort_line = [0 for i in range(input_len)]
for i in range(input_len):
sort_line[i] = int(l[i])
for k in range(input_len):
if (k > 0):
sys.stdout.write(" ")
sys.stdout.write(str(sort_line[k]))
print()
for i in range(1, input_len):
v = sort_line[i]
j = i - 1
while (j >= 0 and v < sort_line[j]):
sort_line[j+1] = sort_line[j]
j -= 1
sort_line[j + 1] = v
for k in range(input_len):
if (k > 0):
sys.stdout.write(" ")
sys.stdout.write(str(sort_line[k]))
print()
|
s636410678
|
p04029
|
u763963344
| 2,000
| 262,144
|
Wrong Answer
| 27
| 9,140
| 121
|
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
candy=int(input('子供の人数にたいして必要なキャンディーの数は？>'))
print(int(candy*(candy+1)/2))
|
s661459698
|
Accepted
| 30
| 9,104
| 52
|
N=int(input())
answer=int(N*(N+1)/2)
print(answer)
|
s463944620
|
p00711
|
u661290476
| 1,000
| 131,072
|
Wrong Answer
| 220
| 8,268
| 806
|
There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles. Write a program to count the number of black tiles which he can reach by repeating the moves described above.
|
mx = [0, 1, 0, -1]
my = [-1, 0, 1, 0]
while True:
w, h = map(int, input().split())
if w == h == 0:
break
tiles = [[i for i in input()] for j in range(h)]
ax = ay = 0
for i in range(h):
for j in range(w):
if tiles[i][j] == '@':
ax, ay = j, i
que = [(ax, ay)]
tiles[ay][ax] = '#'
cnt = 1
while 0 < len(que):
new_que = []
for x, y in que:
for i in range(4):
nx, ny = x + mx[i], y + my[i]
print(nx, ny)
if 0 <= nx <= w - 1 and 0 <= ny <= h - 1:
if tiles[ny][nx] == '.':
cnt += 1
new_que.append((nx, ny))
tiles[ny][nx] = '#'
que = new_que
print(cnt)
|
s924515397
|
Accepted
| 160
| 7,700
| 776
|
mx = [0, 1, 0, -1]
my = [-1, 0, 1, 0]
while True:
w, h = map(int, input().split())
if w == h == 0:
break
tiles = [[i for i in input()] for j in range(h)]
ax = ay = 0
for i in range(h):
for j in range(w):
if tiles[i][j] == '@':
ax, ay = j, i
que = [(ax, ay)]
tiles[ay][ax] = '#'
cnt = 1
while 0 < len(que):
new_que = []
for x, y in que:
for i in range(4):
nx, ny = x + mx[i], y + my[i]
if 0 <= nx <= w - 1 and 0 <= ny <= h - 1:
if tiles[ny][nx] == '.':
cnt += 1
new_que.append((nx, ny))
tiles[ny][nx] = '#'
que = new_que
print(cnt)
|
s272015057
|
p03644
|
u733738237
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 119
|
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 = 1
while True:
n =2**count
if n <= N:
count += 1
continue
else:
break
print(count - 1)
|
s803991069
|
Accepted
| 17
| 2,940
| 122
|
N = int(input())
count = 1
while True:
n =2**count
if n <= N:
count += 1
continue
else:
break
print(2**(count-1))
|
s015963591
|
p02610
|
u978313283
| 2,000
| 1,048,576
|
Wrong Answer
| 1,074
| 39,476
| 936
|
We have N camels numbered 1,2,\ldots,N. Snuke has decided to make them line up in a row. The happiness of Camel i will be L_i if it is among the K_i frontmost camels, and R_i otherwise. Snuke wants to maximize the total happiness of the camels. Find the maximum possible total happiness of the camel. Solve this problem for each of the T test cases given.
|
import heapq
T=int(input())
for t in range(T):
N=int(input())
ListL=[]
ListR=[]
ans=0
for n in range(N):
k,l,r=map(int,input().split())
ans+=min(l,r)
if min(l,r)==r:
ListL.append([k-1,abs(l-r)])
else:
ListR.append([k-1,abs(l-r)])
ListL=sorted(ListL,key=lambda x: x[0])
ListR=sorted(ListR,key=lambda x: x[0])
S=[]
for j in range(N):
for s in range(len(S),len(ListL)):
if ListL[s][0]==j:
heapq.heappush(S,ListL[s][1])
else:
break
while len(S)>j+1:
heapq.heappop(S)
ans+=sum(S)
S=[]
for j in range(N):
for s in range(len(S),len(ListR)):
if ListR[s][0]==j:
heapq.heappush(S,ListR[s][1])
else:
break
while len(S)>j+1:
heapq.heappop(S)
ans+=sum(S)
print(ans)
|
s057231599
|
Accepted
| 813
| 49,648
| 756
|
import heapq
for _ in range(int(input())):
N=int(input())
ListL=[[] for _ in range(N)]
ListR=[[] for _ in range(N)]
ans=0
for _ in range(N):
k,l,r=map(int,input().split())
ans+=min(l,r)
if l>=r:
ListL[k-1].append(abs(l-r))
else:
ListR[k-1].append(abs(l-r))
S=[]
for n,L in enumerate(ListL,1):
for l in L:
if len(S)<n:
heapq.heappush(S,l)
else:
heapq.heappushpop(S,l)
ans+=sum(S)
S=[]
for n,L in enumerate(ListR[::-1][1:],1):
for l in L:
if len(S)<n:
heapq.heappush(S,l)
else:
heapq.heappushpop(S,l)
ans+=sum(S)
print(ans)
|
s724908376
|
p04045
|
u540290227
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 342
|
Iroha is very particular about numbers. There are K digits that she dislikes: D_1, D_2, ..., D_K. She is shopping, and now paying at the cashier. Her total is N yen (the currency of Japan), thus she has to hand at least N yen to the cashier (and possibly receive the change). However, as mentioned before, she is very particular about numbers. When she hands money to the cashier, the decimal notation of the amount must not contain any digits that she dislikes. Under this condition, she will hand the minimum amount of money. Find the amount of money that she will hand to the cashier.
|
n, k = map(int, input().split())
d = list(map(int, input().split()))
like = [1, 2, 3, 4, 5, 6, 7, 8, 9, 0]
ans = ''
result_list = list(set(like) - set(d))
for i in str(n):
for result in result_list:
if int(i) == result == 0:
ans = ans + '0'
elif int(i) <= result:
ans = ans + str(result)
print(ans)
|
s561273728
|
Accepted
| 46
| 2,940
| 181
|
n, k = map(int,input().split())
d = set(input().split())
while True:
for c in str(n):
if c in d:
break
else:
print(n)
exit()
n += 1
|
s227197565
|
p03601
|
u001024152
| 3,000
| 262,144
|
Wrong Answer
| 36
| 5,048
| 618
|
Snuke is making sugar water in a beaker. Initially, the beaker is empty. Snuke can perform the following four types of operations any number of times. He may choose not to perform some types of operations. * Operation 1: Pour 100A grams of water into the beaker. * Operation 2: Pour 100B grams of water into the beaker. * Operation 3: Put C grams of sugar into the beaker. * Operation 4: Put D grams of sugar into the beaker. In our experimental environment, E grams of sugar can dissolve into 100 grams of water. Snuke will make sugar water with the highest possible density. The beaker can contain at most F grams of substances (water and sugar combined), and there must not be any undissolved sugar in the beaker. Find the mass of the sugar water Snuke will make, and the mass of sugar dissolved in it. If there is more than one candidate, any of them will be accepted. We remind you that the sugar water that contains a grams of water and b grams of sugar is \frac{100b}{a + b} percent. Also, in this problem, pure water that does not contain any sugar is regarded as 0 percent density sugar water.
|
from fractions import gcd
A,B,C,D,E,F = map(int, input().split())
ans = {"w":114514, "s":0}
def density(water, sugar):
if water+sugar==0: return -1
return 100*sugar/(water+sugar)
G = gcd(C,D)
for a in range(F//(100*A)+1):
for b in range(F//(100*B)+1):
max_sugar = (a*A+b*B)*E
sugar = G*(max_sugar//G)
water = (a*A+b*B)*100
if density(sugar,water)<0:
continue
if water+sugar>F:
continue
if density(sugar,water)>density(ans["w"], ans["s"]):
ans["w"] = water
ans["s"] = sugar
print(ans["w"]+ans["s"], ans["s"])
|
s446662736
|
Accepted
| 25
| 3,064
| 622
|
A,B,C,D,E,F = map(int, input().split())
ans = {"w":A*100, "s":0}
def density(water, sugar):
return 100*sugar/(water+sugar)
for a in range(31):
for b in range(31):
water = (a*A+b*B)*100
if water>F:
continue
max_sugar = min((a*A+b*B)*E, F-water)
for c in range(max_sugar//C+1):
sugar = c*C
sugar += ((max_sugar-sugar)//D)*D
if water+sugar==0:
continue
if density(water, sugar)>density(ans["w"], ans["s"]):
ans["w"] = water
ans["s"] = sugar
print(ans["w"]+ans["s"], ans["s"])
|
s246187621
|
p02936
|
u704284486
| 2,000
| 1,048,576
|
Wrong Answer
| 2,107
| 57,708
| 400
|
Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i. Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0. Now, the following Q operations will be performed: * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j. Find the value of the counter on each vertex after all operations.
|
n,q = map(int,input().split(" "))
g = [[i] for i in range(n)]
counter = [0]*n
for i in range(n-1):
s, t = map(int, input().split(" "))
g[s-1].append(t-1)
g[t-1].append(s-1)
for j in range(q):
root, plus = map(int, input().split(" "))
if root == 1:
for k in range(n):
counter[k]+=plus
else:
for k in range(n):
if root-1 in g[k]:
counter[k]+=plus
print(*counter)
|
s822625565
|
Accepted
| 1,878
| 81,536
| 552
|
from collections import deque
N,Q = map(int,input().split(" "))
cnt = [0 for _ in range(N)]
edge = [set() for i in range(N)]
for i in range(N-1):
a,b = map(lambda x: int(x)-1, input().split(" "))
edge[a].add(b)
edge[b].add(a)
for i in range(Q):
p,x = map(int,input().split(" "))
p -= 1
cnt[p] += x
plug = [0 for _ in range(N)]
stack = deque([0])
while stack:
v = stack.pop()
plug[v] = 1
for e in edge[v]:
if plug[e] != 1 and e != v:
cnt[e] += cnt[v]
stack.append(e)
print(*cnt)
|
s724208871
|
p02314
|
u463990569
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,576
| 253
|
Find the minimum number of coins to make change for n cents using coins of denominations d1, d2,.., dm. The coins can be used any number of times.
|
n, m = [int(el) for el in input().split(' ')]
c = [int(el) for el in input().split(' ')]
t = [0] + [float('inf') for _ in range(n)]
for i in range(m):
for j in range(c[i], n+1):
t[j] = min(t[j], t[j-c[i]] + 1)
print(t)
print(t[n])
|
s103796803
|
Accepted
| 630
| 9,660
| 235
|
n, m = [int(el) for el in input().split(' ')]
c = [int(el) for el in input().split(' ')]
t = [0] + [float('inf') for _ in range(n)]
for i in range(m):
for j in range(c[i], n+1):
t[j] = min(t[j], t[j-c[i]] + 1)
print(t[n])
|
s333791830
|
p04029
|
u442030035
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 33
|
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
n = int(input())
print(n*(n+1)/2)
|
s331332401
|
Accepted
| 17
| 2,940
| 35
|
n = int(input())
print(n*(n+1)//2)
|
s659795303
|
p00001
|
u215939156
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,560
| 133
|
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.
|
data = []
for i in range(0,10):
data.append(input())
print()
data.sort()
data.reverse()
for i in range(3):
print(data[i])
|
s627573591
|
Accepted
| 20
| 5,600
| 133
|
data = []
for i in range(0,10):
data.append(int(input()))
data.sort()
data.reverse()
for i in range(0,3):
print(data[i])
|
s902441475
|
p03713
|
u665415433
| 2,000
| 262,144
|
Wrong Answer
| 127
| 3,064
| 522
|
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 = [int(i) for i in input().split()]
s = [0] * 3
ans = H * W
for h in range(H // 2 + 1, 0, -1):
s[0] = W * h
s[1] = (H - h) * (W//2)
s[2] = H * W - s[0] - s[1]
if max(s) - min(s) < ans:
ans = max(s) - min(s)
if ans == 0:
print(ans)
exit()
for w in range(W // 2 + 1, 0, -1):
s[0] = H * w
s[1] = (W - w) * (h//2)
s[2] = H * W - s[0] - s[1]
if max(s) - min(s) < ans:
ans = max(s) - min(s)
if ans == 0:
print(ans)
exit()
print(ans)
|
s410977617
|
Accepted
| 406
| 3,064
| 645
|
def solve(H, W):
s = [0]*3
ans = H*W
for h in range(H):
s[0] = W * h
s[1] = (H - h) * (W // 2)
s[2] = H * W - s[0] - s[1]
if max(s) - min(s) < ans:
ans = max(s) - min(s)
if ans == 0:
print(ans)
exit()
for h in range(H):
s[0] = h * W
s[1] = (H - h) // 2 * W
s[2] = H * W - s[0] - s[1]
if max(s) - min(s) < ans:
ans = max(s) - min(s)
if ans == 0:
print(ans)
exit()
return ans
H, W = [int(i) for i in input().split()]
ans = solve(H,W)
ans2 = solve(W,H)
print(min(ans,ans2))
|
s900549085
|
p04029
|
u336979293
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 50
|
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
import math
N = int(input())
print((1 + N)/2*N)
|
s208873691
|
Accepted
| 17
| 2,940
| 43
|
N = int(input())
print(int((1 + N)/2*N))
|
s250286233
|
p02392
|
u073709667
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,520
| 79
|
Write a program which reads three integers a, b and c, and prints "Yes" if a < b < c, otherwise "No".
|
a,b,c=map(int,input().split())
if a<b<c:
print("Yse")
else:
print("No")
|
s411116280
|
Accepted
| 20
| 7,632
| 79
|
a,b,c=map(int,input().split())
if a<b<c:
print("Yes")
else:
print("No")
|
s164691992
|
p00613
|
u471400255
| 1,000
| 131,072
|
Wrong Answer
| 60
| 5,772
| 147
|
In the city, there are two pastry shops. One shop was very popular because its cakes are pretty tasty. However, there was a man who is displeased at the shop. He was an owner of another shop. Although cause of his shop's unpopularity is incredibly awful taste of its cakes, he never improved it. He was just growing hate, ill, envy, and jealousy. Finally, he decided to vandalize the rival. His vandalize is to mess up sales record of cakes. The rival shop sells _K_ kinds of cakes and sales quantity is recorded for each kind. He calculates sum of sales quantities for all pairs of cakes. Getting _K_ ( _K_ -1)/2 numbers, then he rearranges them randomly, and replace an original sales record with them. An owner of the rival shop is bothered. Could you write, at least, a program that finds total sales quantity of all cakes for the pitiful owner?
|
while True:
K = int(input())
if K == 0:
break
else:
po = [int(i) for i in input().split()]
print(sum(po)/(K-1))
|
s955218501
|
Accepted
| 50
| 5,772
| 197
|
while True:
K = int(input())
if K == 0:
break
elif K == 1:
print(int(input()))
else:
po = [int(i) for i in input().split()]
print(int(sum(po)/(K-1)))
|
s373611291
|
p02841
|
u436982376
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 9,056
| 118
|
In this problem, a date is written as Y-M-D. For example, 2019-11-30 means November 30, 2019. Integers M_1, D_1, M_2, and D_2 will be given as input. It is known that the date 2019-M_2-D_2 follows 2019-M_1-D_1. Determine whether the date 2019-M_1-D_1 is the last day of a month.
|
m1,d1 = map(int,input().split())
m2,d2 = map(int,input().split())
if d2 == 1:
print('Yes')
else:
print('No')
|
s268306588
|
Accepted
| 31
| 9,160
| 111
|
m1,d1 = map(int,input().split())
m2,d2 = map(int,input().split())
if d2 == 1:
print(1)
else:
print(0)
|
s505021889
|
p03166
|
u080475902
| 2,000
| 1,048,576
|
Wrong Answer
| 998
| 138,044
| 575
|
There is a directed graph G with N vertices and M edges. The vertices are numbered 1, 2, \ldots, N, and for each i (1 \leq i \leq M), the i-th directed edge goes from Vertex x_i to y_i. G **does not contain directed cycles**. Find the length of the longest directed path in G. Here, the length of a directed path is the number of edges in it.
|
import sys
from functools import lru_cache
sys.setrecursionlimit(10**6)
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
def read():
return int(readline())
def reads():
return map(int, readline().split())
def mp(arg):
return map(int,arg.split())
n,m=reads()
link=[[]for i in range(n)]
for i in range(m):
x,y=reads()
link[x-1].append(y-1)
print(link)
@lru_cache(None)
def dp(k):
res=0
for ele in link[k]:
res=max(res,dp(ele)+1)
return res
res=1
for i in range(n) :
res=max(res,dp(i))
print(res)
|
s997292478
|
Accepted
| 877
| 137,100
| 576
|
import sys
from functools import lru_cache
sys.setrecursionlimit(10**6)
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
def read():
return int(readline())
def reads():
return map(int, readline().split())
def mp(arg):
return map(int,arg.split())
n,m=reads()
link=[[]for i in range(n)]
for i in range(m):
x,y=reads()
link[x-1].append(y-1)
#print(link)
@lru_cache(None)
def dp(k):
res=0
for ele in link[k]:
res=max(res,dp(ele)+1)
return res
res=1
for i in range(n) :
res=max(res,dp(i))
print(res)
|
s672748265
|
p03565
|
u680851063
| 2,000
| 262,144
|
Wrong Answer
| 28
| 3,956
| 254
|
E869120 found a chest which is likely to contain treasure. However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters. He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`. One more thing he found is a sheet of paper with the following facts written on it: * Condition 1: The string S contains a string T as a contiguous substring. * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1. Print the string S. If such a string does not exist, print `UNRESTORABLE`.
|
s = input()
t = input()
import fnmatch
for i in range(len(s) - len(t)+1):
if fnmatch.fnmatch(t, s[i:i+len(t)]):
x = s[0:i] + t + s[i+len(t):]
x = x.replace('?', 'a')
print(x.replace('?', 'a'))
else:
print('UNRESTORABLE')
|
s718120582
|
Accepted
| 27
| 3,828
| 398
|
s = input()
t = input()
import fnmatch
if len(s) < len(t):
print('UNRESTORABLE')
else:
l = []
for i in range(len(s) - len(t)+1):
if fnmatch.fnmatch(t, s[i:i+len(t)]):
x = s[0:i] + t + s[i+len(t):]
l += [x.replace('?', 'a')]
else:
if len(l) == 0:
print('UNRESTORABLE')
else:
l.sort()
print(l[0])
|
s076929789
|
p03555
|
u699296734
| 2,000
| 262,144
|
Wrong Answer
| 24
| 8,988
| 126
|
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.
|
c1 = input()
c2 = input()
if c1[0] == c2[-1] and c1[1] == c2[-2] and c1[2] == c2[-3]:
print("Yes")
else:
print("No")
|
s549873554
|
Accepted
| 26
| 8,928
| 126
|
c1 = input()
c2 = input()
if c1[0] == c2[-1] and c1[1] == c2[-2] and c1[2] == c2[-3]:
print("YES")
else:
print("NO")
|
s865623372
|
p03455
|
u485349322
| 2,000
| 262,144
|
Wrong Answer
| 30
| 9,056
| 93
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a,b=input().split()
a=int(a)
b=int(b)
if (a*b)%2==0:
print('Odd')
else:
print('Even')
|
s670178548
|
Accepted
| 27
| 9,040
| 82
|
a,b=map(int,input().split())
if a*b%2==0:
print("Even")
else:
print("Odd")
|
s954140976
|
p03471
|
u086612293
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 739
|
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.
|
#!/usr/bin/python3
# -*- coding: utf-8 -*-
import sys
def out(i, j, k):
print(i, j, k)
sys.exit()
def main():
n, y = map(int, sys.stdin.readline().split())
for i in range(y // 10000, -1, -1):
z = y - 10000 * i
if i > n:
out(-1, -1, -1)
elif z == 0:
out(i, 0, 0)
for j in range(z // 5000, -1, -1):
w = z - 5000 * j
if i + j > n:
out(-1, -1, -1)
elif w == 0:
out(i, j, 0)
k, r = divmod(w, 1000)
if i + j + k > n:
out(-1, -1, -1)
elif r == 0:
out(i, j, k)
else:
out(-1, -1, -1)
if __name__ == '__main__':
main()
|
s722664052
|
Accepted
| 717
| 3,060
| 428
|
#!/usr/bin/python3
# -*- coding: utf-8 -*-
import sys
def main():
n, y = map(int, sys.stdin.readline().split())
for i in range(y // 10000, -1, -1):
z = y - 10000 * i
for j in range(z // 5000, -1, -1):
k = n - i - j
if z == 5000 * j + 1000 * k:
print(i, j, k)
sys.exit()
else:
print('-1 -1 -1')
if __name__ == '__main__':
main()
|
s918018009
|
p02394
|
u316246166
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,592
| 292
|
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
|
W, H, x, y, r = map(int, input().split())
if x - r < 0:
if x + r > W:
if y - r < 0:
if y + r > H:
print('Yes')
else:
print('No')
else:
print('No')
else:
print('No')
else:
print('No')
|
s068336534
|
Accepted
| 20
| 5,596
| 146
|
W, H, x, y, r = map(int, input().split())
if x - r >= 0 and x + r <= W and y - r >= 0 and y + r <= H:
print('Yes')
else:
print('No')
|
s744703677
|
p02402
|
u839008951
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,588
| 230
|
Write a program which reads a sequence of $n$ integers $a_i (i = 1, 2, ... n)$, and prints the minimum value, maximum value and sum of the sequence.
|
length = int(input())
arr = input().split()
iniV = int(arr[0])
mi = iniV
mx = iniV
su = iniV
for s in arr:
val = int(s)
su += val
if(val < mi):
mi = val
if(val > mx):
mx = val
print(mi, mx, su)
|
s117619767
|
Accepted
| 20
| 6,336
| 227
|
length = int(input())
arr = input().split()
iniV = int(arr[0])
mi = iniV
mx = iniV
su = 0
for s in arr:
val = int(s)
su += val
if(val < mi):
mi = val
if(val > mx):
mx = val
print(mi, mx, su)
|
s240480617
|
p03197
|
u257974487
| 2,000
| 1,048,576
|
Wrong Answer
| 648
| 3,572
| 142
|
There is an apple tree that bears apples of N colors. The N colors of these apples are numbered 1 to N, and there are a_i apples of Color i. You and Lunlun the dachshund alternately perform the following operation (starting from you): * Choose one or more apples from the tree and eat them. Here, the apples chosen at the same time must all have different colors. The one who eats the last apple from the tree will be declared winner. If both you and Lunlun play optimally, which will win?
|
n = int(input())
for i in range(n):
p = int(input())
if p % 2 == 0:
continue
else:
print("first")
print("second")
|
s815918600
|
Accepted
| 198
| 3,060
| 157
|
n = int(input())
for i in range(n):
p = int(input())
if p % 2 == 0:
continue
else:
print("first")
exit()
print("second")
|
s477038122
|
p03044
|
u001024152
| 2,000
| 1,048,576
|
Wrong Answer
| 593
| 39,144
| 502
|
We have a tree with N vertices numbered 1 to N. The i-th edge in the tree connects Vertex u_i and Vertex v_i, and its length is w_i. Your objective is to paint each vertex in the tree white or black (it is fine to paint all vertices the same color) so that the following condition is satisfied: * For any two vertices painted in the same color, the distance between them is an even number. Find a coloring of the vertices that satisfies the condition and print it. It can be proved that at least one such coloring exists under the constraints of this problem.
|
from collections import deque
N = int(input())
adj = [[] for _ in range(N)]
for _ in range(N-1):
u,v,w = map(lambda x:int(x)-1, input().split())
w += 1
adj[u].append((v,w))
adj[v].append((u,w))
visited = [False]*N
q = deque([0])
while q:
cur = q.popleft()
if visited[cur]:
continue
visited[cur] = True
for nex, w in adj[cur]:
if (not visited[nex]) and (w%2==0):
q.append(nex)
for v in visited:
print(1 if v else 0)
|
s823235842
|
Accepted
| 747
| 43,872
| 520
|
from collections import deque
N = int(input())
adj = [[] for _ in range(N)]
for _ in range(N-1):
u,v,w = map(lambda x:int(x)-1, input().split())
w += 1
adj[u].append((v,w))
adj[v].append((u,w))
visited = [False]*N
q = deque([0])
dist = [0]*N
while q:
cur = q.popleft()
if visited[cur]:
continue
visited[cur] = True
for nex, w in adj[cur]:
if (not visited[nex]):
q.append(nex)
dist[nex] = dist[cur] + w
# print(dist)
for d in dist:
print(d%2)
|
s656128094
|
p02402
|
u297342993
| 1,000
| 131,072
|
Wrong Answer
| 30
| 6,724
| 170
|
Write a program which reads a sequence of $n$ integers $a_i (i = 1, 2, ... n)$, and prints the minimum value, maximum value and sum of the sequence.
|
input()
sum = 0
min = 9999
max = 0
for x in [int(y) for y in input().split()]:
sum += x
if min > x:
min = x
if max < x:
max = x
print(sum, min, max)
|
s267500743
|
Accepted
| 40
| 8,028
| 88
|
input()
data = [int(x) for x in input().split()]
print(min(data), max(data), sum(data))
|
s724532888
|
p03339
|
u316386814
| 2,000
| 1,048,576
|
Wrong Answer
| 328
| 22,164
| 214
|
There are N people standing in a row from west to east. Each person is facing east or west. The directions of the people is given as a string S of length N. The i-th person from the west is facing east if S_i = `E`, and west if S_i = `W`. You will appoint one of the N people as the leader, then command the rest of them to face in the direction of the leader. Here, we do not care which direction the leader is facing. The people in the row hate to change their directions, so you would like to select the leader so that the number of people who have to change their directions is minimized. Find the minimum number of people who have to change their directions.
|
import numpy
n = int(input())
s = input()
li = numpy.asarray([1 if i == 'W' else 0 for i in s])
w = numpy.cumsum(li)
e = numpy.cumsum(1 - li[::-1])[::-1]
su = w + e
ans = min(su) - 1
print(li, w, e)
print(ans)
|
s870342976
|
Accepted
| 308
| 22,144
| 198
|
import numpy
n = int(input())
s = input()
li = numpy.asarray([1 if i == 'W' else 0 for i in s])
w = numpy.cumsum(li)
e = numpy.cumsum(1 - li[::-1])[::-1]
su = w + e
ans = min(su) - 1
print(ans)
|
s990577340
|
p03160
|
u580697892
| 2,000
| 1,048,576
|
Wrong Answer
| 134
| 15,032
| 207
|
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
#coding: utf-8
N = int(input())
dp = [0 for _ in range(N)]
h = list(map(int, input().split()))
for i in range(2, N):
dp[i] = min(dp[i-2] + abs(h[i-2] - h[i]), dp[i-1] + abs(h[i-1] - h[i]))
print(dp[N-1])
|
s978363702
|
Accepted
| 179
| 13,980
| 258
|
#coding: utf-8
N = int(input())
h = list(map(int, input().split()))
dp = [float("inf") for _ in range(N)]
dp[0] = 0
for i in range(1, N):
dp[i] = min(dp[i], dp[i-1] + abs(h[i] - h[i-1]))
dp[i] = min(dp[i], dp[i-2] + abs(h[i] - h[i-2]))
print(dp[N-1])
|
s101222454
|
p03149
|
u330152685
| 2,000
| 1,048,576
|
Wrong Answer
| 19
| 3,060
| 322
|
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".
|
def checkInput(inputList):
numberList = [1,9,7,4]
result = all(num in numberList for num in inputList)
if(result):
return True
else:
return False
if __name__ == '__main__':
inp = list(map(int,input().split()))
if(checkInput(inp)):
print("Yes")
else:
print("No")
|
s963981789
|
Accepted
| 18
| 3,064
| 504
|
def checkInput(inputList):
a = 0
b = 0
c = 0
d = 0
for num in inputList:
if(num == 1):
a = 1
elif(num == 9):
b = 1
elif(num == 7):
c = 1
elif(num == 4):
d = 1
if(a == 1 and b == 1 and c == 1 and d == 1):
return True
else:
return False
if __name__ == '__main__':
inp = list(map(int,input().split()))
if(checkInput(inp)):
print("YES")
else:
print("NO")
|
s624731439
|
p03780
|
u785578220
| 2,000
| 262,144
|
Wrong Answer
| 26
| 4,340
| 196
|
AtCoDeer the deer has N cards with positive integers written on them. The number on the i-th card (1≤i≤N) is a_i. Because he loves big numbers, he calls a subset of the cards _good_ when the sum of the numbers written on the cards in the subset, is K or greater. Then, for each card i, he judges whether it is _unnecessary_ or not, as follows: * If, for any good subset of the cards containing card i, the set that can be obtained by eliminating card i from the subset is also good, card i is unnecessary. * Otherwise, card i is NOT unnecessary. Find the number of the unnecessary cards. Here, he judges each card independently, and he does not throw away cards that turn out to be unnecessary.
|
N,K=map(int,input().split())
x=[int(p) for p in input().split()]
x.sort()
x = x[::-1]
pp=0
ans=0
print(x)
for a in x:
print(ans)
if pp+a < K:
pp+=a
ans+=1
else:
ans=0
print(ans)
|
s093817847
|
Accepted
| 20
| 3,572
| 174
|
N,K=map(int,input().split())
x=[int(p) for p in input().split()]
x.sort()
x = x[::-1]
pp=0
ans=0
for a in x:
if pp+a < K:
pp+=a
ans+=1
else:
ans=0
print(ans)
|
s438585507
|
p03251
|
u644778646
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 191
|
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,X,Y = map(int,input().split())
x = list(map(int,input().split()))
y = list(map(int,input().split()))
mx = max(x)
my = min(y)
if my - mx > 1:
print("No War")
else:
print("War")
|
s484709414
|
Accepted
| 17
| 3,064
| 314
|
N,M,X,Y = map(int,input().split())
x = list(map(int,input().split()))
y = list(map(int,input().split()))
mx = max(x)
my = min(y)
if my - mx > 0:
Z = mx
for i in range(my - mx):
Z += 1
if X < Z and Z <= Y and mx < Z and Z <= my:
print("No War")
exit()
print("War")
|
s771635560
|
p02694
|
u258325541
| 2,000
| 1,048,576
|
Wrong Answer
| 25
| 9,128
| 141
|
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())
money = 100
year = 0
while True:
if money > X:
print(year)
break
else:
money = int(money*1.01)
year += 1
|
s013510115
|
Accepted
| 23
| 9,172
| 150
|
X = int(input())
money = 100
year = 0
while True:
if money >= X:
print(year)
break
else:
money = money + int(money*0.01)
year += 1
|
s965290820
|
p03493
|
u511421299
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 3
|
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.
|
101
|
s219093657
|
Accepted
| 17
| 2,940
| 78
|
a = input()
total = 0
for b in a:
if b == '1':
total += 1
print(total)
|
s120623834
|
p03139
|
u128927017
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 108
|
We conducted a survey on newspaper subscriptions. More specifically, we asked each of the N respondents the following two questions: * Question 1: Are you subscribing to Newspaper X? * Question 2: Are you subscribing to Newspaper Y? As the result, A respondents answered "yes" to Question 1, and B respondents answered "yes" to Question 2. What are the maximum possible number and the minimum possible number of respondents subscribing to both newspapers X and Y? Write a program to answer this question.
|
num = list(map(int, input().split()))
a = min(num[1],num[2])
b = max(0, num[0] - (num[1]+num[2]))
print(a,b)
|
s814257011
|
Accepted
| 18
| 2,940
| 107
|
num = list(map(int, input().split()))
a = min(num[1],num[2])
b = max(0,(num[1]+num[2]) - num[0])
print(a,b)
|
s122323060
|
p03161
|
u195054737
| 2,000
| 1,048,576
|
Wrong Answer
| 2,109
| 24,056
| 753
|
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to one of the following: Stone i + 1, i + 2, \ldots, i + K. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
import sys
import itertools
from operator import itemgetter
from fractions import gcd
from math import ceil, floor, sqrt, isinf
from copy import deepcopy
from collections import Counter, deque
import heapq
import numpy as np
from functools import reduce
sys.setrecursionlimit(200000)
input = sys.stdin.readline
# template
n, k = map(int, input().split())
h = [int(i) for i in input().split()]
dp = [0] * n
for i in range(1, n):
j = max(0, i-k)
for l in range(j, i):
dp[i] = min(dp[i], dp[l] + abs(h[j] - h[i]))
print(dp[-1])
|
s271165732
|
Accepted
| 1,792
| 23,092
| 276
|
import numpy as np
n, k = map(int, input().split())
h = np.array([int(i) for i in input().split()])
dp = np.array([0] * n)
for i in range(1, n):
j = max(0, i-k)
dp[i] = np.min(dp[j:i] + np.abs(h[i] - h[j:i]))
print(dp[-1])
|
s172515744
|
p02278
|
u918276501
| 1,000
| 131,072
|
Time Limit Exceeded
| 40,000
| 7,700
| 778
|
You are given $n$ integers $w_i (i = 0, 1, ..., n-1)$ to be sorted in ascending order. You can swap two integers $w_i$ and $w_j$. Each swap operation has a cost, which is the sum of the two integers $w_i + w_j$. You can perform the operations any number of times. Write a program which reports the minimal total cost to sort the given integers.
|
def swap(A,i,j):
A[i],A[j] = A[j],A[i]
return A
def mincost(A):
B = sorted(A)
cost = 0
for j, bj in enumerate(B):
t = 0
i = A.index(bj)
tmp_cost = 0
# swap in a cyclic group
while j != i:
t += 1
bi = B[i]
k = A.index(bi)
tmp_cost += bi + bj #= ak + ai
swap(A,j,k)
j = k
# diff. to swap with min(B)
dec = t * (bj - B[0])
inc = 2 * (bj + B[0])
if dec < inc:
cost += tmp_cost
else:
cost += tmp_cost - dec + inc
return cost
if __name__ == "__main__":
input()
A = list(map(int, input().strip().split()))
cost = mincost(A)
print(cost)
|
s287242087
|
Accepted
| 40
| 7,828
| 777
|
def swap(A,i,j):
A[i],A[j] = A[j],A[i]
return A
def mincost(A):
B = sorted(A)
cost = 0
b0 = B[0]
for i, bi in enumerate(B):
t = 0
j = A.index(bi)
tmp_cost = 0
# swap in a cyclic group
while j != i:
t += 1
bj = B[j]
k = A.index(bj)
tmp_cost += bi + bj
swap(A,j,k)
j = k
# diff. to swap with min(B)
dec = t * (bi - b0)
inc = 2 * (bi + b0)
if dec < inc:
cost += tmp_cost
else:
cost += tmp_cost - dec + inc
return cost
if __name__ == "__main__":
input()
A = list(map(int, input().strip().split()))
cost = mincost(A)
print(cost)
|
s412595445
|
p03679
|
u975676823
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 135
|
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though. Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious. Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it. Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
|
x, a, b = map(int, input().split())
if(a - b < 0):
print("delicious")
elif(abs(a - b) <= x):
print("safe")
else:
print("dangerous")
|
s934006785
|
Accepted
| 17
| 2,940
| 157
|
x, a, b = map(int, input().split())
if(a - b >= 0):
print("delicious")
elif(abs(a - b) <= x):
print("safe")
else:
print("dangerous")
|
s187563706
|
p03139
|
u760961723
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 93
|
We conducted a survey on newspaper subscriptions. More specifically, we asked each of the N respondents the following two questions: * Question 1: Are you subscribing to Newspaper X? * Question 2: Are you subscribing to Newspaper Y? As the result, A respondents answered "yes" to Question 1, and B respondents answered "yes" to Question 2. What are the maximum possible number and the minimum possible number of respondents subscribing to both newspapers X and Y? Write a program to answer this question.
|
N, A, B = map(int,input().split())
print(str(min(A, B)) + " " + str(max(N, A+B)-min(N,A+B)))
|
s824344492
|
Accepted
| 17
| 2,940
| 120
|
N, A, B = map(int,input().split())
if N > A+B:
mini = 0
else:
mini = A+B-N
print(str(min(A, B)) + " " + str(mini))
|
s430244896
|
p03493
|
u703059820
| 2,000
| 262,144
|
Wrong Answer
| 20
| 3,316
| 73
|
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.
|
from collections import Counter
s = input()
count = Counter(s)
count["1"]
|
s943182259
|
Accepted
| 20
| 3,316
| 80
|
from collections import Counter
s = input()
count = Counter(s)
print(count["1"])
|
s245042732
|
p03694
|
u687574784
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 81
|
It is only six months until Christmas, and AtCoDeer the reindeer is now planning his travel to deliver gifts. There are N houses along _TopCoDeer street_. The i-th house is located at coordinate a_i. He has decided to deliver gifts to all these houses. Find the minimum distance to be traveled when AtCoDeer can start and end his travel at any positions.
|
# -*- coding: utf-8 -*-
a = list(map(int, input().split()))
print(max(a)-min(a))
|
s945536801
|
Accepted
| 17
| 2,940
| 92
|
# -*- coding: utf-8 -*-
_ = input()
a = list(map(int, input().split()))
print(max(a)-min(a))
|
s531908258
|
p03943
|
u950376354
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 153
|
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
a,b,c = map(int, input().split())
l = [a,b,c]
ld = sorted(l, reverse=True)
max = ld[0]
mix = ld[1]+ld[2]
if max==mix :
print("YES")
else:
print("NO")
|
s358011952
|
Accepted
| 18
| 3,060
| 153
|
a,b,c = map(int, input().split())
l = [a,b,c]
ld = sorted(l, reverse=True)
max = ld[0]
mix = ld[1]+ld[2]
if max==mix :
print("Yes")
else:
print("No")
|
s052379070
|
p03759
|
u373047809
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 60
|
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])
|
s928037196
|
Accepted
| 18
| 2,940
| 63
|
a, b, c = map(int, input().split())
print("YNEOS"[b-a!=c-b::2])
|
s808492697
|
p02936
|
u753803401
| 2,000
| 1,048,576
|
Wrong Answer
| 2,112
| 130,548
| 562
|
Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i. Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0. Now, the following Q operations will be performed: * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j. Find the value of the counter on each vertex after all operations.
|
n, q = map(int, input().split())
ab = [list(map(int, input().split())) for _ in range(n-1)]
px = [list(map(int, input().split())) for _ in range(q)]
dp = [0] * n
ls = [[] for i in range(n)]
for i in ab:
ls[i[0]-1].append(i[1]-1)
for i in px:
dp[i[0]-1] += i[1]
qq = []
for z in ls[i[0]-1]:
qq.append(z)
while True:
if len(qq) == 0:
break
else:
qn = qq.pop()
dp[qn] += i[1]
if len(ls[qn]) != 0:
for k in ls[qn]:
qq.append(k)
print(dp)
|
s156672881
|
Accepted
| 1,677
| 128,488
| 894
|
def slove():
import sys
import collections
input = sys.stdin.readline
n, q = list(map(int, input().rstrip('\n').split()))
d = collections.defaultdict(list)
for i in range(n-1):
a, b = list(map(int, input().rstrip('\n').split()))
d[a] += [b]
d[b] += [a]
score = collections.defaultdict(int)
for i in range(q):
p, x = list(map(int, input().rstrip('\n').split()))
score[p] += x
ql = collections.deque()
ql.append(1)
fq = collections.defaultdict(list)
fq[1]
while True:
if len(ql) != 0:
p = ql.popleft()
for v in d[p]:
if v not in fq:
score[v] += score[p]
fq[v]
ql.append(v)
else:
break
print(*[score[i] for i in range(1, n + 1)])
if __name__ == '__main__':
slove()
|
s234359737
|
p03433
|
u492447501
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 66
|
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
N = int(input())
A = int(input())
print(((N//500)*500) + (N%500))
|
s877496155
|
Accepted
| 18
| 2,940
| 106
|
N = int(input())
A = int(input())
print("Yes" if N==(((N//500)*500) + (N%500)) and N%500 <= A else "No")
|
s048152714
|
p02743
|
u660018248
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 189
|
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
|
def c():
a, b,c = map(int, input().split())
ab = a*b
a_b_c = (a+b-c)**2
if ab < a_b_c:
print("Yes")
else:
print("No")
if __name__ == '__main__':
c()
|
s548578346
|
Accepted
| 18
| 2,940
| 273
|
if __name__ == '__main__':
a, b, c =map(int, input().split())
ab = 4 * a * b
a_b_c = (c - (a + b))
a_b_c_2 = a_b_c ** 2
if a_b_c <= 0:
print("No")
else:
if ab < a_b_c_2:
print("Yes")
else:
print("No")
|
s432458308
|
p02936
|
u603958124
| 2,000
| 1,048,576
|
Wrong Answer
| 572
| 28,724
| 961
|
Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i. Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0. Now, the following Q operations will be performed: * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j. Find the value of the counter on each vertex after all operations.
|
from math import ceil,floor,factorial,gcd,sqrt,log2,cos,sin,tan,acos,asin,atan,degrees,radians,pi,inf,comb
from itertools import accumulate,groupby,permutations,combinations,product,combinations_with_replacement
from collections import deque,defaultdict,Counter
from bisect import bisect_left,bisect_right
from operator import itemgetter
from heapq import heapify,heappop,heappush
from queue import Queue,LifoQueue,PriorityQueue
from copy import deepcopy
from time import time
from functools import reduce
import string
import sys
sys.setrecursionlimit(10 ** 7)
def input() : return sys.stdin.readline().strip()
def INT() : return int(input())
def MAP() : return map(int,input().split())
def LIST() : return list(MAP())
n, q = MAP()
g = [0]*(n+1)
ans = [0]*(n+1)
for i in range(n-1):
a, b = MAP()
g[b] = a
for i in range(q):
p, x = MAP()
ans[p] += x
for i in range(1,n+1):
ans[i] += ans[g[i]]
for i in range(1, n+1):
print(ans[i])
|
s068167969
|
Accepted
| 1,123
| 283,540
| 1,462
|
from math import ceil,floor,factorial,gcd,sqrt,log2,cos,sin,tan,acos,asin,atan,degrees,radians,pi,inf,comb
from itertools import accumulate,groupby,permutations,combinations,product,combinations_with_replacement
from collections import deque,defaultdict,Counter
from bisect import bisect_left,bisect_right
from operator import itemgetter
from heapq import heapify,heappop,heappush
from queue import Queue,LifoQueue,PriorityQueue
from copy import deepcopy
from time import time
from functools import reduce
import string
import sys
sys.setrecursionlimit(10 ** 7)
def input() : return sys.stdin.readline().strip()
def INT() : return int(input())
def MAP() : return map(int,input().split())
def LIST() : return list(MAP())
n, q = MAP()
graph = [[] for i in range(n)]
point = [0]*n
for i in range(n-1):
a, b = MAP()
graph[a-1].append(b-1)
graph[b-1].append(a-1)
for i in range(q):
p, x = MAP()
point[p-1] += x
def dfs(now, prev = -1):
for next in graph[now]:
if next == prev:
continue
point[next] += point[now]
dfs(next, now)
dfs(0)
print(*point)
|
s705442548
|
p02922
|
u094932051
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 316
|
Takahashi's house has only one socket. Takahashi wants to extend it with some number of power strips, each with A sockets, into B or more empty sockets. One power strip with A sockets can extend one empty socket into A empty sockets. Find the minimum number of power strips required.
|
while True:
try:
A, B = map(int, input().split())
strip = 1
socket = A
for i in range(A+1):
if (socket < B):
strip += 1
socket += (A-1)
else:
break
print(strip)
except:
break
|
s914162816
|
Accepted
| 17
| 2,940
| 144
|
import math
while True:
try:
A, B = map(int, input().split())
print(math.ceil((B-A) / (A-1)) + 1)
except:
break
|
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