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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s660120482
|
p04043
|
u786020649
| 2,000
| 262,144
|
Wrong Answer
| 27
| 9,040
| 173
|
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
import sys
a,b,c=map(int,input().split())
l=[a,b,c,a,b]
for x in range(3):
if l[x]==l[x+1] and l[x]==5 and l[x]==7:
print('YES')
sys.exit()
print('NO')
|
s357153384
|
Accepted
| 27
| 9,136
| 175
|
import sys
a,b,c=map(int,input().split())
l=[a,b,c,a,b]
for x in range(3):
if l[x]==l[x+1] and l[x]==5 and l[x+2]==7:
print('YES')
sys.exit()
print('NO')
|
s513243936
|
p03351
|
u497046426
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,060
| 132
|
Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.
|
a, b, c, d = map(int, input().split())
if (abs(a-c) <= d) or (max(abs(a - b), abs(b-c)) < d):
print('Yes')
else:
print('No')
|
s019673176
|
Accepted
| 18
| 2,940
| 133
|
a, b, c, d = map(int, input().split())
if (abs(a-c) <= d) or (max(abs(a - b), abs(b-c)) <= d):
print('Yes')
else:
print('No')
|
s867448297
|
p03377
|
u931636178
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 86
|
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
A,B,X = map(int,input().split())
if (A+B) >= X:
print("Yes")
else:
print("No")
|
s716233363
|
Accepted
| 17
| 2,940
| 114
|
A,B,X = map(int,input().split())
if A > X:
print("NO")
elif (A+B) >= X:
print("YES")
else:
print("NO")
|
s996419560
|
p03555
|
u813174766
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 60
|
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.
|
a=input()
x='YES' if input()==reversed(a) else 'NO'
print(x)
|
s229987705
|
Accepted
| 18
| 2,940
| 53
|
a=input()[::-1]
print('YES' if input()==a else 'NO')
|
s139992532
|
p00036
|
u868716420
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,344
| 976
|
縦 8、横 8 のマスからなる図 1 のような平面があります。 □| □| □| □| □| □| □| □ ---|---|---|---|---|---|---|--- □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ 図１ --- この平面上に、以下の A から G の図形のどれかが一つだけ置かれています。 | A --- ■| ■| | ---|---|---|--- ■| ■| | | | | | | | | B --- | ■| | ---|---|---|--- | ■| | | ■| | | ■| | | C --- ■| ■| ■| ■ ---|---|---|--- | | | | | | | | | | D --- | ■| | ---|---|---|--- ■| ■| | ■| | | | | | | E --- ■| ■| | ---|---|---|--- | ■| ■| | | | | | | | F --- ■| | | ---|---|---|--- ■| ■| | | ■| | | | | | G --- | ■| ■| ---|---|---|--- ■| ■| | | | | | | | たとえば、次の図 2 の例では E の図形が置かれています。 | □| □| □| □| □| □| □| □ ---|---|---|---|---|---|---|--- □| □| □| □| □| □| □| □ □| ■| ■| □| □| □| □| □ □| □| ■| ■| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ 図２ --- 平面の中で図形が占めているマスを 1、占めていないマスを 0 で表現した数字の列を読み込んで、置かれている図形の種類（A〜G）を出力するプログラムを作成してください。 ただし、ひとつの平面に置かれている図形は必ず１つで、複数の図形が置かれていることはありません。また、A〜G で表される図形以外のものが置かれていることはありません。
|
while True :
try :
switch = 0
for _ in range(9) :
if switch == 0 :
temp = input()
if temp == '00000000' or temp == '' : pass
else :
onetwo, place = 0, 0
if '1111' in temp :
print('C')
switch = 1
else : onetwo, place = temp.count('1'), temp.index('1')
temp = input()
if onetwo == 1 :
if temp.count('1') == 1 : print('B')
elif temp.index('1') == place : print('F')
else : print('D')
elif onetwo == 2 :
if temp.index('1') == place : print('A')
elif temp.index('1') == place + 1 : print('E')
else : print('G')
switch = 1
else : input()
except : break
|
s820400088
|
Accepted
| 30
| 7,336
| 766
|
while True :
try :
temp = input()
if temp == '00000000' or temp == '' : pass
else :
if '1111' in temp : print('C')
else :
onetwo, place = temp.count('1'), temp.index('1')
temp = input()
if onetwo == 1 :
if temp.count('1') == 1 :
print('B')
input()
elif temp.index('1') == place : print('F')
else : print('D')
input()
elif onetwo == 2 :
if temp.index('1') == place : print('A')
elif temp.index('1') == place + 1 : print('E')
else : print('G')
except : break
|
s375250397
|
p03852
|
u003501233
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 124
|
Given a lowercase English letter c, determine whether it is a vowel. Here, there are five vowels in the English alphabet: `a`, `e`, `i`, `o` and `u`.
|
c=str(input())
b=["a","i","u","e","o"]
for i in range(5):
if c == b[i]:
print("vowel")
else:
print("consonant")
|
s192301598
|
Accepted
| 17
| 2,940
| 80
|
c=str(input())
b="aiueo"
if c in b:
print("vowel")
else:
print("consonant")
|
s372076826
|
p03693
|
u713465512
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 85
|
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
r,g,b = map(int, input().split())
if g*b%4==0:
print("YES")
else:
print("NO")
|
s728957404
|
Accepted
| 17
| 2,940
| 90
|
r,g,b = map(int, input().split())
if (g*10+b)%4==0:
print("YES")
else:
print("NO")
|
s288686199
|
p03455
|
u504256702
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 99
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
if any(list(map(lambda x: int(x) % 2 == 0, input().split()))):
print("Odd")
else:
print("Even")
|
s539721854
|
Accepted
| 17
| 2,940
| 99
|
if any(list(map(lambda x: int(x) % 2 == 0, input().split()))):
print("Even")
else:
print("Odd")
|
s207046967
|
p03228
|
u464555934
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 353
|
In the beginning, Takahashi has A cookies, and Aoki has B cookies. They will perform the following operation alternately, starting from Takahashi: * If the number of cookies in his hand is odd, eat one of those cookies; if the number is even, do nothing. Then, give one-half of the cookies in his hand to the other person. Find the numbers of cookies Takahashi and Aoki respectively have after performing K operations in total.
|
a = list(map(int,input().split()))
A = a[0]; B = a[1]; K = a[2]
for i in range(K):
if i % 2 == 0:
if A % 2 == 1:
A -= 1
cokkie = A / 2
A -= cokkie
B += cokkie
else:
if B % 2 == 1:
B -= 1
cokkie = B / 2
B -= cokkie
A += cokkie
print(str(A) + " " + str(B))
|
s890001434
|
Accepted
| 17
| 3,064
| 363
|
a = list(map(int,input().split()))
A = a[0]; B = a[1]; K = a[2]
for i in range(K):
if i % 2 == 0:
if A % 2 == 1:
A -= 1
cokkie = A / 2
A -= cokkie
B += cokkie
else:
if B % 2 == 1:
B -= 1
cokkie = B / 2
B -= cokkie
A += cokkie
print(str(int(A)) + " " + str(int(B)))
|
s649273756
|
p03351
|
u230717961
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 161
|
Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.
|
a,b,c,d = map(int, input().split())
ab = abs(a - b)
bc = abs(b - c)
ca = abs(c - a)
if(ab < d and bc < d and ca < d):
print("Yes")
else:
print("No")
|
s468888382
|
Accepted
| 17
| 2,940
| 155
|
a,b,c,d = map(int, input().split())
ab = abs(a - b)
bc = abs(b - c)
ca = abs(c - a)
if(ab <= d and bc <= d or ca <=d):
print("Yes")
else:
print("No")
|
s965212521
|
p03738
|
u499317876
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 319
|
You are given two positive integers A and B. Compare the magnitudes of these numbers.
|
a=input()
b=input()
A=len(a)
B=len(b)
if A<B:
print("LESS")
elif B<A:
print("GREATER")
else:
for i in range(A):
if int(a[i])<int(b[i]):
print("LESS")
elif int(b[i])<int(a[i]):
print("GREATER")
elif int(a[i])==int(b[i]) and i==A-1:
print("EQUAL")
|
s541137265
|
Accepted
| 18
| 3,064
| 356
|
a=input()
b=input()
A=len(a)
B=len(b)
if A<B:
print("LESS")
elif B<A:
print("GREATER")
else:
for i in range(A):
if int(a[i])<int(b[i]):
print("LESS")
break
elif int(b[i])<int(a[i]):
print("GREATER")
break
elif int(a[i])==int(b[i]) and i==A-1:
print("EQUAL")
|
s275950637
|
p03854
|
u116348130
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,188
| 180
|
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()
sc = ["dreameraser", "dreamerase", "dreamer", "dream", "eracer", "erace"]
for i in sc:
s = s.replace(i, '')
if len(s) == 0:
print("YES")
else:
print("NO")
|
s538450921
|
Accepted
| 18
| 3,188
| 139
|
S = input()
sc = ["eraser", "erase", "dreamer", "dream"]
for i in sc:
S = S.replace(i, "")
if S:
print("NO")
else:
print("YES")
|
s980192057
|
p03658
|
u188244611
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 136
|
Snuke has N sticks. The length of the i-th stick is l_i. Snuke is making a snake toy by joining K of the sticks together. The length of the toy is represented by the sum of the individual sticks that compose it. Find the maximum possible length of the toy.
|
# ABC 67, B - snake toy
n, k = [int(el) for el in input().split(' ')]
L = sorted([int(l) for l in input().split(' ')])
print(L[-k:])
|
s429800485
|
Accepted
| 17
| 2,940
| 141
|
# ABC 67, B - snake toy
n, k = [int(el) for el in input().split(' ')]
L = sorted([int(l) for l in input().split(' ')])
print(sum(L[-k:]))
|
s554371269
|
p03385
|
u627530854
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 50
|
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
print("Yes" if sorted(input()) == "abc" else "No")
|
s326493643
|
Accepted
| 17
| 2,940
| 59
|
print("Yes" if "".join(sorted(input())) == "abc" else "No")
|
s371028885
|
p03469
|
u672316981
| 2,000
| 262,144
|
Wrong Answer
| 26
| 8,892
| 48
|
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
|
s = str(input())
ans = '2018' + s[5:]
print(ans)
|
s477692117
|
Accepted
| 24
| 8,940
| 48
|
s = str(input())
ans = '2018' + s[4:]
print(ans)
|
s394743116
|
p02606
|
u759076129
| 2,000
| 1,048,576
|
Wrong Answer
| 32
| 9,152
| 103
|
How many multiples of d are there among the integers between L and R (inclusive)?
|
L, R, d = map(int, input().split())
c = 0
for i in range(L, R+1):
if i % 3 == 0:
c += 1
print(c)
|
s327331536
|
Accepted
| 34
| 9,148
| 104
|
L, R, d = map(int, input().split())
c = 0
for i in range(L, R+1):
if i % d == 0:
c += 1
print(c)
|
s542637422
|
p03493
|
u642682703
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 218
|
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.
|
evens = [6,4,10,4]
itr = 0
flg = True
while flg:
evens = list(map(lambda x: (x/2), evens))
is_evens = list(map(lambda x: x.is_integer(), evens))
if not all(is_evens):
flg = False
break
itr+=1
print(itr)
|
s816673493
|
Accepted
| 17
| 2,940
| 35
|
num = input()
print(num.count('1'))
|
s748503660
|
p03359
|
u503111914
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 139
|
In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year. For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year. In this country, a date is called _Takahashi_ when the month and the day are equal as numbers. For example, 5-5 is Takahashi. How many days from 2018-1-1 through 2018-a-b are Takahashi?
|
a,b = map(int,input().split())
c = 0
for i in range(1,a):
for j in range(1,b):
if i == j :
c += 1
else:
None
print(c)
|
s602772880
|
Accepted
| 17
| 2,940
| 55
|
a,b = map(int,input().split())
print(a-1 if a>b else a)
|
s012587739
|
p03151
|
u767664985
| 2,000
| 1,048,576
|
Wrong Answer
| 104
| 18,292
| 279
|
A university student, Takahashi, has to take N examinations and pass all of them. Currently, his _readiness_ for the i-th examination is A_{i}, and according to his investigation, it is known that he needs readiness of at least B_{i} in order to pass the i-th examination. Takahashi thinks that he may not be able to pass all the examinations, and he has decided to ask a magician, Aoki, to change the readiness for as few examinations as possible so that he can pass all of them, while not changing the total readiness. For Takahashi, find the minimum possible number of indices i such that A_i and C_i are different, for a sequence C_1, C_2, ..., C_{N} that satisfies the following conditions: * The sum of the sequence A_1, A_2, ..., A_{N} and the sum of the sequence C_1, C_2, ..., C_{N} are equal. * For every i, B_i \leq C_i holds. If such a sequence C_1, C_2, ..., C_{N} cannot be constructed, print -1.
|
N = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
ans, mana = 0, 0
for i in range(N):
if A[i] >= B[i]:
mana += A[i] - B[i]
else:
mana -= B[i] - A[i]
ans += 1
if mana >= 0:
print(ans)
else:
print(-1)
|
s112145629
|
Accepted
| 270
| 42,240
| 483
|
import numpy as np
from bisect import bisect_right
N = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
diff = [0 for i in range(N)]
need = 0
ans = 0
for i in range(N):
diff[i] = A[i] - B[i]
if diff[i] < 0:
need += -diff[i]
ans += 1
diff[i] = 0
diff = sorted(diff, reverse=True)
cumsum = np.cumsum(diff)
if need != 0:
ans += bisect_right(cumsum, need)+1
if ans > N:
ans = -1
print(ans)
|
s898699391
|
p04000
|
u419686324
| 3,000
| 262,144
|
Wrong Answer
| 1,670
| 151,176
| 540
|
We have a grid with H rows and W columns. At first, all cells were painted white. Snuke painted N of these cells. The i-th ( 1 \leq i \leq N ) cell he painted is the cell at the a_i-th row and b_i-th column. Compute the following: * For each integer j ( 0 \leq j \leq 9 ), how many subrectangles of size 3×3 of the grid contains exactly j black cells, after Snuke painted N cells?
|
h, w, n = map(int, input().split())
ab = [list(map(int, input().split())) for _ in range(n)]
sa = {}
for a,b in ab:
for i in range(3):
aa = a + i
if aa < 3 or h < aa:
continue
for j in range(3):
bb = b + j
if bb < 3 or w < bb:
continue
id = "%d,%d" % (aa, bb)
sa[id] = sa.get(id, 0) + 1
s = {}
for x in sa.values():
s[x] = s.setdefault(x, 0) + 1
print((w - 2) * (h - 2) - len(sa.keys()))
for i in range(10):
print(s.get(i, 0))
|
s439030455
|
Accepted
| 1,698
| 151,176
| 543
|
h, w, n = map(int, input().split())
ab = [list(map(int, input().split())) for _ in range(n)]
sa = {}
for a,b in ab:
for i in range(3):
aa = a + i
if aa < 3 or h < aa:
continue
for j in range(3):
bb = b + j
if bb < 3 or w < bb:
continue
id = "%d,%d" % (aa, bb)
sa[id] = sa.get(id, 0) + 1
s = {}
for x in sa.values():
s[x] = s.setdefault(x, 0) + 1
print((w - 2) * (h - 2) - len(sa.keys()))
for i in range(1, 10):
print(s.get(i, 0))
|
s932966662
|
p02972
|
u265506056
| 2,000
| 1,048,576
|
Wrong Answer
| 1,002
| 10,740
| 340
|
There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied: * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2. Does there exist a good set of choices? If the answer is yes, find one good set of choices.
|
N=int(input())
A=[]
B=[]
for i in range(N):
B.append(0)
A=list(map(int,input().split()))
for i in reversed(range(N)):
a=i+1
b=N//a
c=0
for j in range(b):
d=a*(j+1)
c=c+B[d-1]
e=c%2
if e==A[i]:
B[i]=0
else:
B[i]=1
C=sum(B)
if C==0:
print(0)
else:
print(C)
print(*B)
|
s485275607
|
Accepted
| 510
| 14,072
| 348
|
N=int(input())
a=list(map(int,input().split()))
A=[0]*N
B=[]
for i in range(N//2,N):
if a[i]==1:
A[i]=1
B.append(i+1)
for i in range(N//2)[::-1]:
asum=0
x=i+1
for j in range(i,N,x):
asum=asum+A[j]
if asum%2!=a[i]:
A[i]=1
B.append(i+1)
count=sum(A)
print(count)
if count!=0:
print(*B)
|
s721445451
|
p03854
|
u727551259
| 2,000
| 262,144
|
Wrong Answer
| 20
| 4,424
| 273
|
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 re
s = input()
ptn = re.compile(r'maerd|remaerd|sare|resare')
tmp = s[::-1]
while(True):
res = re.match(ptn , tmp)
if res is None:
print(tmp)
break
tmp = tmp.replace(res.group(),'',1)
print(tmp)
if len(tmp) == 0:
print('YES')
else:
print('NO')
|
s258399532
|
Accepted
| 108
| 3,444
| 246
|
import re
s = input()
ptn = re.compile(r'maerd|remaerd|resare|esare')
tmp = s[::-1]
while(True):
res = re.match(ptn , tmp)
if res is None:
break
tmp = tmp.replace(res.group(),'',1)
if len(tmp) == 0:
print('YES')
else:
print('NO')
|
s940968837
|
p03759
|
u368104473
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 88
|
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())
if b-a == b-c:
print('Yes')
else:
print('No')
|
s945269752
|
Accepted
| 17
| 2,940
| 94
|
a, b, c = map(int, input().split())
if b - a == c - b :
print("YES")
else:
print("NO")
|
s683612697
|
p03549
|
u023229441
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,206
| 11,136
| 994
|
Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`. When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases. Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds. Now, he goes through the following process: * Submit the code. * Wait until the code finishes execution on all the cases. * If the code fails to correctly solve some of the M cases, submit it again. * Repeat until the code correctly solve all the cases in one submission. Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).
|
import sys
import math
import heapq
mod=10**9+7
inf=float("inf")
from math import sqrt
from collections import deque
from collections import Counter
from collections import defaultdict
from collections import OrderedDict
from math import ceil
input=lambda: sys.stdin.readline().strip()
sys.setrecursionlimit(11451419)
from decimal import Decimal
from functools import lru_cache
n,m=map(int,input().split())
A=100*(n-m)+1900*m
p=pow(2,m)
@lru_cache(maxsize=10**10)
def per(n):
if n==1:
return 1/p
return (1-sum([per(i) for i in range(1,n)]))*(1/p)
ans=0
for i in range(1,10000):
ans+= i*A*per(i)
print(round(ans))
|
s970463165
|
Accepted
| 212
| 10,272
| 993
|
import sys
import math
import heapq
mod=10**9+7
inf=float("inf")
from math import sqrt
from collections import deque
from collections import Counter
from collections import defaultdict
from collections import OrderedDict
from math import ceil
input=lambda: sys.stdin.readline().strip()
sys.setrecursionlimit(11451419)
from decimal import Decimal
from functools import lru_cache
n,m=map(int,input().split())
A=100*(n-m)+1900*m
p=pow(2,m)
@lru_cache(maxsize=10**10)
def per(n):
if n==1:
return 1/p
return (1-sum([per(i) for i in range(1,n)]))*(1/p)
ans=0
for i in range(1,2000):
ans+= i*A*per(i)
print(round(ans))
|
s170072080
|
p02384
|
u391228754
| 1,000
| 131,072
|
Time Limit Exceeded
| 40,000
| 7,752
| 1,601
|
Construct a dice from a given sequence of integers in the same way as
|
class Dice(object):
def __init__(self, top, south, east, west, north, bottom):
self.top = top
self.south = south
self.east = east
self.west = west
self.north = north
self.bottom = bottom
def get_top(self):
return self.top
def rotate(self, directions):
for direction in directions:
if direction == 'S':
self.prev_top = self.top
self.top = self.north
self.north = self.bottom
self.bottom = self.south
self.south = self.prev_top
elif direction == 'E':
self.prev_top = self.top
self.top = self.west
self.west = self.bottom
self.bottom = self.east
self.east = self.prev_top
elif direction == 'W':
self.prev_top = self.top
self.top = self.east
self.east = self.bottom
self.bottom = self.west
self.west = self.prev_top
elif direction == 'N':
self.prev_top = self.top
self.top = self.south
self.south = self.bottom
self.bottom = self.north
self.north = self.prev_top
dice = Dice(*map(int,input().split()))
n = int(input())
for i in range(n):
a, b = input().split()
if dice.east == b or dice.west == b:
dice.rotate("W")
while a != dice.top:
dice.rotate("N")
while b != dice.south:
dice.rotate("W")
print(dice.east)
|
s740943421
|
Accepted
| 20
| 7,780
| 1,157
|
class Dice:
def __init__(self, faces):
self.faces = tuple(faces)
def roll_north(self):
self.faces = (self.faces[1], self.faces[5], self.faces[2],
self.faces[3], self.faces[0], self.faces[4])
def roll_south(self):
self.faces = (self.faces[4], self.faces[0], self.faces[2],
self.faces[3], self.faces[5], self.faces[1])
def roll_west(self):
self.faces = (self.faces[2], self.faces[1], self.faces[5],
self.faces[0], self.faces[4], self.faces[3])
def roll_east(self):
self.faces = (self.faces[3], self.faces[1], self.faces[0],
self.faces[5], self.faces[4], self.faces[2])
def number(self, face_id):
return self.faces[face_id - 1]
dice = Dice(list(map(int, input().split())))
n = int(input())
for i in range(n):
(face1, face2) = map(int, input().split())
if face2 == dice.number(3) or face2 == dice.number(4):
dice.roll_west()
while dice.number(2) != face2:
dice.roll_north()
while dice.number(1) != face1:
dice.roll_west()
print(dice.number(3))
|
s512560749
|
p03796
|
u223904637
| 2,000
| 262,144
|
Wrong Answer
| 54
| 2,940
| 97
|
Snuke loves working out. He is now exercising N times. Before he starts exercising, his _power_ is 1. After he exercises for the i-th time, his power gets multiplied by i. Find Snuke's power after he exercises N times. Since the answer can be extremely large, print the answer modulo 10^{9}+7.
|
n=int(input())
mod=1e9+7
ans=1
for i in range(1,n+1):
ans*=i%mod
ans=ans%mod
print(ans)
|
s428992045
|
Accepted
| 55
| 2,940
| 99
|
n=int(input())
mod=1e9+7
ans=1
for i in range(1,n+1):
ans*=i
ans=ans%mod
print(round(ans))
|
s418316011
|
p03610
|
u779508975
| 2,000
| 262,144
|
Wrong Answer
| 42
| 9,408
| 138
|
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)
|
s472993330
|
Accepted
| 41
| 9,272
| 140
|
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)
|
s189533629
|
p03997
|
u564060397
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 62
|
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())
c=int(input())
print((a+b)*c/2)
|
s014561064
|
Accepted
| 17
| 2,940
| 63
|
a=int(input())
b=int(input())
c=int(input())
print((a+b)*c//2)
|
s253342494
|
p02646
|
u652104963
| 2,000
| 1,048,576
|
Wrong Answer
| 24
| 9,180
| 173
|
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
a,v = map(int, input().split())
b,w = map(int, input().split())
t = int(input())
x = abs((b-a)/v)
y = abs((w*x)/v)
l = x + y
if l > t:
print('YES')
else:
print('NO')
|
s936701147
|
Accepted
| 23
| 9,168
| 146
|
a,v = map(int, input().split())
b,w = map(int, input().split())
t = int(input())
if v*t >= abs(b-a) + w*t:
print('YES')
else:
print('NO')
|
s345920680
|
p02263
|
u766183517
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,532
| 36
|
An expression is given in a line. Two consequtive symbols (operand or operator) are separated by a space character. You can assume that +, - and * are given as the operator and an operand is a positive integer less than 106
|
line = input().split()
print(line)
|
s780869374
|
Accepted
| 20
| 5,600
| 399
|
line = input().split()
stack = []
ans = 0
for elem in line:
if elem in ['+', '-', '*']:
ele2 = int(stack.pop())
ele1 = int(stack.pop())
if elem == '+':
tmp = ele1 + ele2
if elem == '-':
tmp = ele1 - ele2
if elem == '*':
tmp = ele1 * ele2
stack.append(tmp)
else:
stack.append(elem)
print(stack[0])
|
s328686931
|
p03814
|
u084320347
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,816
| 150
|
Snuke has decided to construct a string that starts with `A` and ends with `Z`, by taking out a substring of a string s (that is, a consecutive part of s). Find the greatest length of the string Snuke can construct. Here, the test set guarantees that there always exists a substring of s that starts with `A` and ends with `Z`.
|
s = input()
print(s[1:-1])
r_s = s[::-1]
if r_s.index("Z")==0:
print(len(s[s.index("A"):]))
else:
print(len(s[s.index("A"):-r_s.index("Z")]))
|
s575362552
|
Accepted
| 18
| 3,560
| 135
|
s = input()
r_s = s[::-1]
if r_s.index("Z")==0:
print(len(s[s.index("A"):]))
else:
print(len(s[s.index("A"):-r_s.index("Z")]))
|
s871207239
|
p03338
|
u027622859
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,064
| 237
|
You are given a string S of length N consisting of lowercase English letters. We will cut this string at one position into two strings X and Y. Here, we would like to maximize the number of different letters contained in both X and Y. Find the largest possible number of different letters contained in both X and Y when we cut the string at the optimal position.
|
def main():
N = int(input())
S = input()
count = 0
max_ = 0
for i in range(N):
x = set(S[:i])
y = set(S[:i])
max_ = max(len(x&y), max_)
print(max_)
if __name__=='__main__':
main()
|
s588905998
|
Accepted
| 17
| 3,064
| 237
|
def main():
N = int(input())
S = input()
count = 0
max_ = 0
for i in range(N):
x = set(S[:i])
y = set(S[i:])
max_ = max(len(x&y), max_)
print(max_)
if __name__=='__main__':
main()
|
s688282147
|
p03160
|
u387774811
| 2,000
| 1,048,576
|
Wrong Answer
| 2,111
| 127,180
| 552
|
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.
|
import sys
sys.setrecursionlimit(10**6)
input = sys.stdin.readline
def cost(l,r):
if (l,r) in memo:
return memo[(l,r)]
else:
ans=float('inf')
for j in range(l,r):
if (l,j) in memo:
a=memo[(l,j)]
else:
a=cost(l,j)
if (j+1) in memo:
b=memo[(j+1,r)]
else:
b=cost(j+1,r)
ans=min(ans,a+b)
memo[(l,r)]=ans+size[r+1]-size[l]
return(ans+size[r+1]-size[l])
memo={}
N=int(input())
size=[0]*(N+1)
a = list(map(int,input().split()))
for i in range(N):
memo[(i,i)] = 0
size[i+1]=size[i]+a[i]
print(cost(0,N-1))
|
s822812075
|
Accepted
| 138
| 13,928
| 204
|
N=int(input())
h=list(map(int,input().split()))
ans=[0]*(N)
ans[0]=0
ans[1]=abs(h[1]-h[0])
if N>2 :
for i in range(2,N):
ans[i]=min(ans[i-2]+abs(h[i]-h[i-2]),ans[i-1]+abs(h[i]-h[i-1]))
print(ans[N-1])
|
s418018684
|
p02612
|
u920977317
| 2,000
| 1,048,576
|
Wrong Answer
| 33
| 9,152
| 150
|
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.
|
def main():
N=int(input())
mod=N%1000
if N==0:
print("0")
else:
print(1000-N)
if __name__=="__main__":
main()
|
s017771919
|
Accepted
| 27
| 9,156
| 152
|
def main():
N=int(input())
mod=N%1000
if mod==0:
print(0)
else:
print(1000-mod)
if __name__=="__main__":
main()
|
s065233473
|
p03377
|
u007550226
| 2,000
| 262,144
|
Wrong Answer
| 21
| 3,060
| 78
|
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
A,B,X = map(int,input().split())
if A<= X <= A+B:print('Yes')
else:print('No')
|
s020135695
|
Accepted
| 17
| 2,940
| 78
|
A,B,X = map(int,input().split())
if A<= X <= A+B:print('YES')
else:print('NO')
|
s834793417
|
p03998
|
u363407238
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 196
|
Alice, Bob and Charlie are playing _Card Game for Three_ , as below: * At first, each of the three players has a deck consisting of some number of cards. Each card has a letter `a`, `b` or `c` written on it. The orders of the cards in the decks cannot be rearranged. * The players take turns. Alice goes first. * If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says `a`, Alice takes the next turn.) * If the current player's deck is empty, the game ends and the current player wins the game. You are given the initial decks of the players. More specifically, you are given three strings S_A, S_B and S_C. The i-th (1≦i≦|S_A|) letter in S_A is the letter on the i-th card in Alice's initial deck. S_B and S_C describes Bob's and Charlie's initial decks in the same way. Determine the winner of the game.
|
a = 'abcb'
b = 'aacb'
c = 'bccc'
d = {
'a': a,
'b': b,
'c': c
}
n = 'a'
while (True):
if d[n] == '':
break
nx = d[n][0]
d[n] = d[n][1:]
n = nx
print(n.upper())
|
s700814027
|
Accepted
| 18
| 2,940
| 95
|
d = {x : list(input()) for x in "abc"}
x = "a"
while d[x]:
x = d[x].pop(0)
print(x.upper())
|
s495959372
|
p03567
|
u314902127
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 65
|
Snuke built an online judge to hold a programming contest. When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.) Determine whether the judge can return the string `AC` as the verdict to a program.
|
s = input()
if s in "AC":
print("Yes")
else:
print("No")
|
s466040393
|
Accepted
| 17
| 2,940
| 66
|
s = input()
if "AC" in s :
print("Yes")
else:
print("No")
|
s676728760
|
p03711
|
u811000506
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 216
|
Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group.
|
x,y = map(int,input().split())
li1 = [1,3,5,7,8,10,12]
li2 = [4,6,9,11]
li3 = [2]
if li1.count(x)==1 and li1.count(y)==1 or li2.count(x)==1 and li2.count(y)==1 or x==y and y==2:
print("YES")
else:
print("NO")
|
s809772988
|
Accepted
| 18
| 3,064
| 217
|
x,y = map(int,input().split())
li1 = [1,3,5,7,8,10,12]
li2 = [4,6,9,11]
li3 = [2]
if li1.count(x)==1 and li1.count(y)==1 or li2.count(x)==1 and li2.count(y)==1 or x==y and y==2:
print("Yes")
else:
print("No")
|
s695118513
|
p02420
|
u957840591
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,520
| 377
|
Your task is to shuffle a deck of n cards, each of which is marked by a alphabetical letter. A single shuffle action takes out h cards from the bottom of the deck and moves them to the top of the deck. The deck of cards is represented by a string as follows. abcdeefab The first character and the last character correspond to the card located at the bottom of the deck and the card on the top of the deck respectively. For example, a shuffle with h = 4 to the above deck, moves the first 4 characters "abcd" to the end of the remaining characters "eefab", and generates the following deck: eefababcd You can repeat such shuffle operations. Write a program which reads a deck (a string) and a sequence of h, and prints the final state (a string).
|
Str=[]
H=[]
h=[]
while True:
x=input()
if x.isalpha():
if h!=[]:
H.append(h)
h=[]
Str.append(x)
elif x=='-':
H.append(h)
break
else:
h.append(int(x))
for i in range(len(Str)):
a=Str[i]
for j in range(len(H[i])):
a=a[H[i][j]:]+a[:H[i][j]]
Str[i]=a
for str in Str:
print(str)
|
s446196401
|
Accepted
| 20
| 7,624
| 358
|
Str=[]
H=[]
h=[]
while True:
x=input()
if x=='-':
break
Str.append(x)
m=int(input())
for i in range(m):
x=int(input())
h.append(x)
H.append(h)
h=[]
for i in range(len(Str)):
a=Str[i]
for j in range(len(H[i])):
a=''.join([a[H[i][j]:],a[:H[i][j]]])
Str[i]=a
for str in Str:
print(str)
|
s200108620
|
p03379
|
u545368057
| 2,000
| 262,144
|
Wrong Answer
| 291
| 25,556
| 332
|
When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l. You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i. Find B_i for each i = 1, 2, ..., N.
|
N = int(input())
Xs = list(map(int, input().split()))
sXs = sorted(Xs)
m1 = sXs[N//2]
m2 = sXs[N//2 - 1]
print(m1,m2)
if m1 == m2:
for i in range(N):
print(m1)
else:
for x in Xs:
if x <= m2:
print(m1)
else:
print(m2)
|
s045833076
|
Accepted
| 311
| 25,228
| 334
|
N = int(input())
Xs = list(map(int, input().split()))
sXs = sorted(Xs)
m1 = sXs[N//2]
m2 = sXs[N//2 - 1]
# print(m1,m2)
if m1 == m2:
for i in range(N):
print(m1)
else:
for x in Xs:
if x <= m2:
print(m1)
else:
print(m2)
|
s173747761
|
p02749
|
u291278680
| 2,000
| 1,048,576
|
Wrong Answer
| 1,472
| 198,844
| 1,635
|
We have a tree with N vertices. The vertices are numbered 1 to N, and the i-th edge connects Vertex a_i and Vertex b_i. Takahashi loves the number 3. He is seeking a permutation p_1, p_2, \ldots , p_N of integers from 1 to N satisfying the following condition: * For every pair of vertices (i, j), if the distance between Vertex i and Vertex j is 3, the sum or product of p_i and p_j is a multiple of 3. Here the distance between Vertex i and Vertex j is the number of edges contained in the shortest path from Vertex i to Vertex j. Help Takahashi by finding a permutation that satisfies the condition.
|
import sys
sys.setrecursionlimit(700000)
def s_in():
return input()
def n_in():
return int(input())
def l_in():
return list(map(int, input().split()))
def print_l(l):
print(' '.join(map(str, l)))
class Interval():
def __init__(self, li):
self.li = li
self.n = len(li)
self.sum_li = [li[0]]
for i in range(1, self.n):
self.sum_li.append(self.sum_li[i-1] + li[i])
def sum(self, a, b=None):
if b is None:
return self.sum(0, a)
res = self.sum_li[min(self.n-1, b-1)]
if a > 0:
res -= self.sum_li[a-1]
return res
N = n_in()
edges = [[] for _ in range(N)]
for _ in range(N-1):
a,b = l_in()
edges[a-1].append(b-1)
edges[b-1].append(a-1)
S = []
F = [0]*N
depth = [0]*N
visited = set()
def dfs(v, d):
visited.add(v)
F[v] = len(S)
depth[v] = d
S.append(v)
for w in edges[v]:
if w in visited:
continue
dfs(w, d+1)
S.append(v)
dfs(0, 0)
res = [0 for _ in range(N)]
from collections import deque
q = deque([i for i in range(1,N+1) if i%3 == 0 ] + [i for i in range(1,N+1) if i%3 != 0 ])
if len(list(filter(lambda d: d%2 == 0, depth))) < N//2:
for i, d in enumerate(depth):
if d%2 == 0:
res[i] = q.popleft()
for i, d in enumerate(depth):
if d%2 != 0:
res[i] = q.popleft()
else:
for i, d in enumerate(depth):
if d%2 != 0:
res[i] = q.popleft()
for i, d in enumerate(depth):
if d%2 == 0:
res[i] = q.popleft()
print_l(res)
|
s866092871
|
Accepted
| 1,457
| 193,596
| 2,346
|
import sys
sys.setrecursionlimit(700000)
def s_in():
return input()
def n_in():
return int(input())
def l_in():
return list(map(int, input().split()))
def print_l(l):
print(' '.join(map(str, l)))
class Interval():
def __init__(self, li):
self.li = li
self.n = len(li)
self.sum_li = [li[0]]
for i in range(1, self.n):
self.sum_li.append(self.sum_li[i-1] + li[i])
def sum(self, a, b=None):
if b is None:
return self.sum(0, a)
res = self.sum_li[min(self.n-1, b-1)]
if a > 0:
res -= self.sum_li[a-1]
return res
N = n_in()
edges = [[] for _ in range(N)]
for _ in range(N-1):
a,b = l_in()
edges[a-1].append(b-1)
edges[b-1].append(a-1)
S = []
depth = [0]*N
visited = set()
def dfs(v, d):
visited.add(v)
depth[v] = d
S.append(v)
for w in edges[v]:
if w in visited:
continue
dfs(w, d+1)
S.append(v)
root = 0
dfs(root, 0)
res = [0 for _ in range(N)]
from collections import deque
q0 = deque(i for i in range(1,N+1) if i%3 == 0)
q1 = deque(i for i in range(1,N+1) if i%3 == 1)
q2 = deque(i for i in range(1,N+1) if i%3 == 2)
even_depth = len(list(filter(lambda d: d%2 == 0, depth)))
odd_depth = len(list(filter(lambda d: d%2 != 0, depth)))
if even_depth > N//3 and odd_depth > N//3:
for i, d in enumerate(depth):
if d%2 == 0:
if len(q1) > 0:
res[i] = q1.popleft()
else:
res[i] = q0.popleft()
else:
if len(q2) > 0:
res[i] = q2.popleft()
else:
res[i] = q0.popleft()
elif even_depth <= N//3:
for i, d in enumerate(depth):
if d%2 == 0:
res[i] = q0.popleft()
else:
if len(q1) > 0:
res[i] = q1.popleft()
elif len(q2) > 0:
res[i] = q2.popleft()
else:
res[i] = q0.popleft()
else:
for i, d in enumerate(depth):
if d%2 != 0:
res[i] = q0.popleft()
else:
if len(q1) > 0:
res[i] = q1.popleft()
elif len(q2) > 0:
res[i] = q2.popleft()
else:
res[i] = q0.popleft()
print_l(res)
|
s811346928
|
p02601
|
u034194513
| 2,000
| 1,048,576
|
Wrong Answer
| 32
| 9,464
| 384
|
M-kun has the following three cards: * A red card with the integer A. * A green card with the integer B. * A blue card with the integer C. He is a genius magician who can do the following operation at most K times: * Choose one of the three cards and multiply the written integer by 2. His magic is successful if both of the following conditions are satisfied after the operations: * The integer on the green card is **strictly** greater than the integer on the red card. * The integer on the blue card is **strictly** greater than the integer on the green card. Determine whether the magic can be successful.
|
import random
nums = input().split()
print(nums)
#A = nums[0]
#B = nums[1]
#C = nums[2]
a = int(nums[0])
b = int(nums[1])
c = int(nums[2])
print(a)
K = input()
k = int(K)
#print(type(a))
print(k)
cnt = 0
#print(type(cnt))
while a >= b:
cnt +=1
b *= 2
print(b)
while b >= c:
cnt += 1
c *= 2
print(c)
#if cnt
if cnt <= k:
print("Yes")
else:
print("No")
|
s501388727
|
Accepted
| 33
| 9,484
| 471
|
import random
nums = input().split()
#print(nums)
#A = nums[0]
#B = nums[1]
#C = nums[2]
a = int(nums[0])
b = int(nums[1])
c = int(nums[2])
#print(a)
K = input()
k = int(K)
#print(type(a))
#print(k)
cnt = 0
#print(type(cnt))
while a >= b:
cnt +=1
b *= 2
#print(b)
while b >= c:
cnt += 1
c *= 2
#print(c)
#if cnt
if cnt <= k:
print("Yes")
else:
print("No")
#for hoge in range(1,k +1):
# print(hoge)
#print(random.choice(nums))
#print(A)
|
s157287517
|
p03623
|
u094565093
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 98
|
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
x,a,b=map(int,input().split())
if abs(x-a)>abs(x-b):
print(abs(x-a))
else:
print(abs(x-b))
|
s416906512
|
Accepted
| 17
| 2,940
| 88
|
x,a,b=map(int,input().split())
if abs(x-a)>abs(x-b):
print('B')
else:
print('A')
|
s209920980
|
p03399
|
u777846995
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 140
|
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.
|
number = 4
fare = [int(input()) for a in range(number)]
fare_train = min(fare[0:1])
fare_bus = min(fare[2:3])
print(fare_train + fare_bus)
|
s924362239
|
Accepted
| 17
| 3,064
| 394
|
number = 4
fare = [int(input()) for a in range(number)]
fare_added = []
fare_added.append(fare[0] + fare[2]) #A + C
fare_added.append(fare[0] + fare[3]) #A + D
fare_added.append(fare[1] + fare[2]) #B + C
fare_added.append(fare[1] + fare[3]) #B + D
for i, elem in enumerate(fare_added):
if i == 0:
fare_min = elem
if fare_min > elem:
fare_min = elem
print(fare_min)
|
s230860439
|
p02865
|
u735912743
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 15
|
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
int(input())//2
|
s208960436
|
Accepted
| 17
| 2,940
| 26
|
print((int(input())-1)//2)
|
s941786315
|
p03067
|
u071211072
| 2,000
| 1,048,576
|
Wrong Answer
| 28
| 9,096
| 122
|
There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise.
|
A,B,C = list(map(int,input().split(' ')))
[A,B] = [max(A,B),min(A,B)]
if C>=A and C<=B:
print('Yes')
else:
print('No')
|
s165060162
|
Accepted
| 25
| 9,080
| 110
|
A,B,C = list(map(int,input().split(' ')))
if C>min(A,B) and C<max(A,B):
print('Yes')
else:
print('No')
|
s500470554
|
p03386
|
u619458041
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 296
|
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
import sys
def main():
input = sys.stdin.readline
A, B, K = map(int, input().split())
i = 0
while i < K:
a, b = A + i, B - i
i += 1
if a == b:
print(a)
break
elif a > b:
break
else:
print(a)
print(b)
if __name__ == '__main__':
main()
|
s351025132
|
Accepted
| 18
| 3,060
| 260
|
import sys
def main():
input = sys.stdin.readline
A, B, K = map(int, input().split())
ans = list(range(A, min(A+K, B))) + list(range(max(B-(K-1), A), B+1))
ans = sorted(list(set(ans)))
for i in ans:
print(i)
if __name__ == '__main__':
main()
|
s207780767
|
p03433
|
u864711178
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 75
|
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
#coding: utf-8
n = int(input())
a = int(input())
rest = n % 500
print(rest)
|
s558287263
|
Accepted
| 17
| 2,940
| 96
|
n = int(input())
a = int(input())
if a - (n % 500) >= 0:
print("Yes")
else:
print("No")
|
s204094099
|
p03377
|
u941022948
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 109
|
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, c= map(int, input().split())
if a+b>c:
print('No')
elif a>c:
print('No')
else:
print('Yes')
|
s275851867
|
Accepted
| 17
| 2,940
| 109
|
a, b, c= map(int, input().split())
if a+b<c:
print('NO')
elif a>c:
print('NO')
else:
print('YES')
|
s962411834
|
p03469
|
u727787724
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 54
|
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
|
s=input().split('/')
s[0]='2018'
print(s[0]+s[1]+s[2])
|
s239758509
|
Accepted
| 18
| 2,940
| 63
|
s=input().split('/')
s[0]='2018'
print(s[0]+'/'+s[1]+'/'+s[2])
|
s383212962
|
p03998
|
u778720350
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 234
|
Alice, Bob and Charlie are playing _Card Game for Three_ , as below: * At first, each of the three players has a deck consisting of some number of cards. Each card has a letter `a`, `b` or `c` written on it. The orders of the cards in the decks cannot be rearranged. * The players take turns. Alice goes first. * If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says `a`, Alice takes the next turn.) * If the current player's deck is empty, the game ends and the current player wins the game. You are given the initial decks of the players. More specifically, you are given three strings S_A, S_B and S_C. The i-th (1≦i≦|S_A|) letter in S_A is the letter on the i-th card in Alice's initial deck. S_B and S_C describes Bob's and Charlie's initial decks in the same way. Determine the winner of the game.
|
a = input()
b = input()
c = input()
cards = {'a': a, 'b': b, 'c': c}
turn = 'a'
while True:
if len(cards[turn]) == 0:
print(turn)
break
draw = cards[turn][0]
cards[turn] = cards[turn][1:]
turn = draw
|
s711157376
|
Accepted
| 17
| 3,060
| 242
|
a = input()
b = input()
c = input()
cards = {'a': a, 'b': b, 'c': c}
turn = 'a'
while True:
if len(cards[turn]) == 0:
print(turn.upper())
break
draw = cards[turn][0]
cards[turn] = cards[turn][1:]
turn = draw
|
s835943992
|
p03377
|
u601082779
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 59
|
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,c=map(int,input().split());print("YNEOS"[a<=c<=a+b::2])
|
s408669111
|
Accepted
| 17
| 2,940
| 62
|
a,b,c=map(int,input().split());print("YNEOS"[a>c or c>a+b::2])
|
s389766777
|
p02646
|
u266113953
| 2,000
| 1,048,576
|
Wrong Answer
| 19
| 9,204
| 166
|
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
A,V = map(int, input().split())
B,W = map(int, input().split())
T = int(input())
if (V-W) != 0 and T >= (abs(A-B)/(V-W)) >= 0:
print("Yes")
else:
print("No")
|
s070532315
|
Accepted
| 21
| 9,180
| 166
|
A,V = map(int, input().split())
B,W = map(int, input().split())
T = int(input())
if (V-W) != 0 and T >= (abs(A-B)/(V-W)) >= 0:
print("YES")
else:
print("NO")
|
s786281263
|
p02389
|
u556014061
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,488
| 49
|
Write a program which calculates the area and perimeter of a given rectangle.
|
a, b = map(int, input().split())
print(a*2 + b*2)
|
s525742136
|
Accepted
| 20
| 7,528
| 54
|
a, b = map(int, input().split())
print(a*b, a*2 + b*2)
|
s113763537
|
p04030
|
u853900545
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 102
|
Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key. To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string: * The `0` key: a letter `0` will be inserted to the right of the string. * The `1` key: a letter `1` will be inserted to the right of the string. * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted. Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?
|
s = input()
a = []
for _ in s:
if _ == 'B':
a = a[:-1]
else:
a += _
print(a)
|
s025737609
|
Accepted
| 17
| 2,940
| 102
|
s = input()
a = ''
for _ in s:
if _ == 'B':
a = a[:-1]
else:
a += _
print(a)
|
s978936135
|
p03680
|
u433380437
| 2,000
| 262,144
|
Wrong Answer
| 2,265
| 49,644
| 142
|
Takahashi wants to gain muscle, and decides to work out at AtCoder Gym. The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up. These buttons are numbered 1 through N. When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It is possible that i=a_i. When Button i is not lighten up, nothing will happen by pressing it. Initially, Button 1 is lighten up. Takahashi wants to quit pressing buttons when Button 2 is lighten up. Determine whether this is possible. If the answer is positive, find the minimum number of times he needs to press buttons.
|
n = int(input())
a=[0]
for i in range(n):
a.append(int(input()))
s=1
cnt=0
while s !=2:
s=a[s]
cnt +=1
print(s)
print(cnt)
|
s970199219
|
Accepted
| 171
| 12,760
| 174
|
n = int(input())
a=[0]
for i in range(n):
a.append(int(input()))
s=1
cnt=0
while s !=2:
s=a[s]
cnt +=1
if cnt ==n:
print(-1)
exit()
print(cnt)
|
s716572617
|
p03162
|
u494058663
| 2,000
| 1,048,576
|
Wrong Answer
| 570
| 36,132
| 425
|
Taro's summer vacation starts tomorrow, and he has decided to make plans for it now. The vacation consists of N days. For each i (1 \leq i \leq N), Taro will choose one of the following activities and do it on the i-th day: * A: Swim in the sea. Gain a_i points of happiness. * B: Catch bugs in the mountains. Gain b_i points of happiness. * C: Do homework at home. Gain c_i points of happiness. As Taro gets bored easily, he cannot do the same activities for two or more consecutive days. Find the maximum possible total points of happiness that Taro gains.
|
N = int(input())
A = []
B = []
C = []
for i in range(N):
tmp_a,tmp_b,tmp_c = map(int,input().split())
A.append(tmp_a)
B.append(tmp_b)
C.append(tmp_c)
dp = [[0 for j in range(3)] for i in range(N)]
dp[0][0] = A[0]
dp[0][1] = B[0]
dp[0][2] = C[0]
for i in range(1,N):
dp[i][0] = max(dp[i-1][1],dp[i-1][2])+A[i]
dp[i][1] = max(dp[i-1][0],dp[i-1][2])+B[i]
dp[i][2] = max(dp[i-1][0],dp[i-1][1])+C[i]
|
s995269982
|
Accepted
| 569
| 36,208
| 473
|
N = int(input())
A = []
B = []
C = []
for i in range(N):
tmp_a,tmp_b,tmp_c = map(int,input().split())
A.append(tmp_a)
B.append(tmp_b)
C.append(tmp_c)
dp = [[0 for j in range(3)] for i in range(N)]
dp[0][0] = A[0]
dp[0][1] = B[0]
dp[0][2] = C[0]
for i in range(1,N):
dp[i][0] = max(dp[i-1][1],dp[i-1][2])+A[i]
dp[i][1] = max(dp[i-1][0],dp[i-1][2])+B[i]
dp[i][2] = max(dp[i-1][0],dp[i-1][1])+C[i]
print(max(dp[-1][0],dp[-1][1],dp[-1][2]))
|
s617682818
|
p02613
|
u127126063
| 2,000
| 1,048,576
|
Wrong Answer
| 151
| 9,204
| 277
|
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
n=int(input())
ac=0
wa=0
tle=0
re=0
for i in range(n):
c=input()
if c=='AC':
ac+=1
elif c=='WA':
wa+=1
elif c=='TLE':
tle+=1
else:
re+=1
print("AC X {}".format(ac))
print("WA X {}".format(wa))
print("TLE X {}".format(tle))
print("RE X {}".format(re))
|
s714326638
|
Accepted
| 151
| 9,232
| 277
|
n=int(input())
ac=0
wa=0
tle=0
re=0
for i in range(n):
c=input()
if c=='AC':
ac+=1
elif c=='WA':
wa+=1
elif c=='TLE':
tle+=1
else:
re+=1
print("AC x {}".format(ac))
print("WA x {}".format(wa))
print("TLE x {}".format(tle))
print("RE x {}".format(re))
|
s342631378
|
p03778
|
u117193815
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 83
|
AtCoDeer the deer found two rectangles lying on the table, each with height 1 and width W. If we consider the surface of the desk as a two-dimensional plane, the first rectangle covers the vertical range of AtCoDeer will move the second rectangle horizontally so that it connects with the first rectangle. Find the minimum distance it needs to be moved.
|
w,a,b= map(int ,input().split())
if b>=w+a:
print((w+a)-b)
else:
print(w-b)
|
s355518291
|
Accepted
| 17
| 3,060
| 133
|
w,a,b= map(int ,input().split())
if w>=a and b<=w+a:
print(0)
elif a+w<=b:
print(b-(a+w))
elif b+w<=a:
print(a-(b+w))
|
s918909630
|
p03564
|
u403331159
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 101
|
Square1001 has seen an electric bulletin board displaying the integer 1. He can perform the following operations A and B to change this value: * Operation A: The displayed value is doubled. * Operation B: The displayed value increases by K. Square1001 needs to perform these operations N times in total. Find the minimum possible value displayed in the board after N operations.
|
n=int(input())
k=int(input())
d=1
for i in range(n):
if k>2*d:
d=d*2
else:
d=d+k
print(d)
|
s495343376
|
Accepted
| 17
| 2,940
| 101
|
n=int(input())
k=int(input())
d=1
for i in range(n):
if k>=d:
d=d*2
else:
d=d+k
print(d)
|
s365418262
|
p00004
|
u811841526
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,628
| 206
|
Write a program which solve a simultaneous equation: ax + by = c dx + ey = f The program should print x and y for given a, b, c, d, e and f (-1,000 ≤ a, b, c, d, e, f ≤ 1,000). You can suppose that given equation has a unique solution.
|
while True:
try:
a,b,c,d,e,f = map(int, input().split())
except EOFError:
break
x = c/a + (f*a-c*d)*b / (e*a-d*b) / a
y = (f*a - c*d) / (e*a - b*d)
print('{:.3f} {:.3f}')
|
s651615219
|
Accepted
| 20
| 7,568
| 221
|
while True:
try:
a,b,c,d,e,f = map(int, input().split())
except EOFError:
break
x = c/a - (b/a) * (a*f-c*d) / (a*e-b*d)
y = (f*a - c*d) / (e*a - b*d)
print('{:.3f} {:.3f}'.format(x, y))
|
s278723034
|
p03338
|
u953379577
| 2,000
| 1,048,576
|
Wrong Answer
| 26
| 9,044
| 132
|
You are given a string S of length N consisting of lowercase English letters. We will cut this string at one position into two strings X and Y. Here, we would like to maximize the number of different letters contained in both X and Y. Find the largest possible number of different letters contained in both X and Y when we cut the string at the optimal position.
|
n = int(input())
s = list(input())
ans = 0
for i in range(n):
x = s[:i+1]
y = s[i:]
ans = max(ans,len(set(x)&set(y)))
|
s048108019
|
Accepted
| 26
| 9,020
| 146
|
n = int(input())
s = list(input())
ans = 0
for i in range(n):
x = s[:i+1]
y = s[i+1:]
ans = max(ans,len(set(x)&set(y)))
print(ans)
|
s487996193
|
p04029
|
u202877219
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 169
|
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?
|
s = input()
z = ""
for i in range(len(s)):
if s[i] == "0":
z += "0"
elif s[i] == "1":
z+= "1"
elif s[i] == "B":
z = z[:-1]
print (z)
|
s081663292
|
Accepted
| 17
| 2,940
| 66
|
n = int(input())
x = 0
for i in range(n+1):
x = x+i
print (x)
|
s851556987
|
p02388
|
u464802001
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,476
| 32
|
Write a program which calculates the cube of a given integer x.
|
a=int(input())
s=a**3
print("s")
|
s549653880
|
Accepted
| 30
| 7,608
| 30
|
a=int(input())
s=a**3
print(s)
|
s008358329
|
p03502
|
u698919163
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 132
|
An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10. Given an integer N, determine whether it is a Harshad number.
|
N = input()
tmp = 0
for i in range(len(N)):
tmp += int(N[i])
if tmp % int(N) == 0:
print('Yes')
else:
print('No')
|
s590607008
|
Accepted
| 18
| 2,940
| 132
|
N = input()
tmp = 0
for i in range(len(N)):
tmp += int(N[i])
if int(N) % tmp == 0:
print('Yes')
else:
print('No')
|
s845132368
|
p03673
|
u858136677
| 2,000
| 262,144
|
Wrong Answer
| 95
| 26,180
| 72
|
You are given an integer sequence of length n, a_1, ..., a_n. Let us consider performing the following n operations on an empty sequence b. The i-th operation is as follows: 1. Append a_i to the end of b. 2. Reverse the order of the elements in b. Find the sequence b obtained after these n operations.
|
n = int(input())
a = list(map(int,input().split()))
a.reverse()
print(a)
|
s799895809
|
Accepted
| 360
| 26,180
| 325
|
n = int(input())
a = list(map(int,input().split()))
a.insert(0,0)
b = []
if n%2==0:
i=n
while i>=2:
b.append(a[i])
i += -2
k=1
while k<=n-1:
b.append(a[k])
k += 2
else:
i=n
while i>=1:
b.append(a[i])
i += -2
k=2
while k<=n-1:
b.append(a[k])
k+=2
for i in range(n-1):
print(b[i],end=' ')
print(b[n-1])
|
s105665734
|
p03836
|
u088552457
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,064
| 384
|
Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him.
|
sx, sy, tx, ty = map(int, input().split())
dx = tx - sx
dy = ty - sy
r = ''
for i in range(dy):
r += 'U'
for i in range(dx):
r += 'R'
for i in range(dy):
r += 'D'
for i in range(dx):
r += 'L'
r += 'L'
for i in range(dy + 1):
r += 'U'
for i in range(dx + 1):
r += 'R'
r += 'R'
for i in range(dy + 1):
r += 'D'
for i in range(dx + 1):
r += 'L'
r += 'U'
print(r)
|
s863609377
|
Accepted
| 19
| 3,064
| 393
|
sx, sy, tx, ty = map(int, input().split())
dx = tx - sx
dy = ty - sy
r = ''
for i in range(dy):
r += 'U'
for i in range(dx):
r += 'R'
for i in range(dy):
r += 'D'
for i in range(dx):
r += 'L'
r += 'L'
for i in range(dy + 1):
r += 'U'
for i in range(dx + 1):
r += 'R'
r += 'D'
r += 'R'
for i in range(dy + 1):
r += 'D'
for i in range(dx + 1):
r += 'L'
r += 'U'
print(r)
|
s588028637
|
p03485
|
u598229387
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 55
|
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(int((a+b)//2 +1))
|
s041199729
|
Accepted
| 17
| 2,940
| 51
|
a,b = map(int, input().split())
print(-(-(a+b)//2))
|
s235130312
|
p02578
|
u007637377
| 2,000
| 1,048,576
|
Wrong Answer
| 102
| 32,108
| 129
|
N persons are standing in a row. The height of the i-th person from the front is A_i. We want to have each person stand on a stool of some heights - at least zero - so that the following condition is satisfied for every person: Condition: Nobody in front of the person is taller than the person. Here, the height of a person includes the stool. Find the minimum total height of the stools needed to meet this goal.
|
N = int(input())
A = list(map(int, input().split()))
ans = 0
h = A[0]
for a in A:
if h > a:
ans += h - a
h = a
print(ans)
|
s440665121
|
Accepted
| 100
| 32,164
| 139
|
N = int(input())
A = list(map(int, input().split()))
ans = 0
h = A[0]
for a in A:
if h > a:
ans += h - a
else:
h = a
print(ans)
|
s975799717
|
p02255
|
u917432951
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,548
| 379
|
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.
|
if __name__ == '__main__':
N = (int)(input())
A = list(map(int,input().split()))
for i in range(1,N):
v = A[i]
j = i - 1
while j >= 0 and A[j] > v:
A[j+1] = A[j]
j = j -1
A[j+1] = v
if i == N - 1:
print(' '.join(map(str,A)),end = '')
else:
print(' '.join(map(str,A)))
|
s591519542
|
Accepted
| 20
| 7,724
| 318
|
if __name__ == '__main__':
N = (int)(input())
A = list(map(int,input().split()))
print(' '.join(map(str,A)))
for i in range(1,N):
v = A[i]
j = i - 1
while j >= 0 and A[j] > v:
A[j+1] = A[j]
j = j -1
A[j+1] = v
print(' '.join(map(str,A)))
|
s206137100
|
p03607
|
u826263061
| 2,000
| 262,144
|
Wrong Answer
| 216
| 14,312
| 139
|
You are playing the following game with Joisino. * Initially, you have a blank sheet of paper. * Joisino announces a number. If that number is written on the sheet, erase the number from the sheet; if not, write the number on the sheet. This process is repeated N times. * Then, you are asked a question: How many numbers are written on the sheet now? The numbers announced by Joisino are given as A_1, ... ,A_N in the order she announces them. How many numbers will be written on the sheet at the end of the game?
|
n = int(input())
a = set([])
for i in range(n):
k = int(input())
if k in a:
a.remove(k)
else:
a.add(k)
ans = len(a)
print(a)
|
s000931349
|
Accepted
| 201
| 11,884
| 141
|
n = int(input())
a = set([])
for i in range(n):
k = int(input())
if k in a:
a.remove(k)
else:
a.add(k)
ans = len(a)
print(ans)
|
s072576884
|
p03229
|
u405256066
| 2,000
| 1,048,576
|
Wrong Answer
| 251
| 8,912
| 395
|
You are given N integers; the i-th of them is A_i. Find the maximum possible sum of the absolute differences between the adjacent elements after arranging these integers in a row in any order you like.
|
from sys import stdin
N = int(stdin.readline().rstrip())
A = []
for i in range(N):
A.append(int(input()))
A.sort()
ans = 0
if N % 2 == 0:
tmp = [-1,1] + [2]*((N-2)//2) + [-2]*((N-2)//2)
for i,j in zip(sorted(tmp),A):
ans += (i*j)
print(ans)
else:
tmp = [-1,1] + [2]*((N-1)//2) + [-2]*((N-2)//2)
for i,j in zip(sorted(tmp),A):
ans += (i*j)
print(ans)
|
s359791204
|
Accepted
| 279
| 9,656
| 552
|
from sys import stdin
N = int(stdin.readline().rstrip())
A = []
for i in range(N):
A.append(int(input()))
A.sort()
if N % 2 != 0:
ans1 = 0
tmp = [-1,-1] + [2]*((N-1)//2) + [-2]*((N-2)//2)
for i,j in zip(sorted(tmp),A):
ans1 += (i*j)
ans2 = 0
tmp = [1,1] + [2]*((N-2)//2) + [-2]*((N-1)//2)
for i,j in zip(sorted(tmp),A):
ans2 += (i*j)
print(max(ans1,ans2))
else:
ans = 0
tmp = [-1,1] + [2]*((N-2)//2) + [-2]*((N-2)//2)
for i,j in zip(sorted(tmp),A):
ans += (i*j)
print(ans)
|
s019173241
|
p03473
|
u629540524
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 28
|
How many hours do we have until New Year at M o'clock (24-hour notation) on 30th, December?
|
m = int(input())
print(m+24)
|
s771943609
|
Accepted
| 17
| 2,940
| 28
|
m = int(input())
print(48-m)
|
s011285646
|
p03473
|
u759412327
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 22
|
How many hours do we have until New Year at M o'clock (24-hour notation) on 30th, December?
|
print(int(input())+24)
|
s033459545
|
Accepted
| 27
| 9,144
| 22
|
print(48-int(input()))
|
s942982292
|
p03024
|
u249447366
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 108
|
Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament. The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day. Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility.
|
s = list(input())
cnt = s.count('o')
rest = 15 - len(s)
ans = 'Yes' if cnt + rest >= 8 else 'No'
print(ans)
|
s882463146
|
Accepted
| 17
| 2,940
| 108
|
s = list(input())
cnt = s.count('o')
rest = 15 - len(s)
ans = 'YES' if cnt + rest >= 8 else 'NO'
print(ans)
|
s408282187
|
p03079
|
u174603263
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 103
|
You are given three integers A, B and C. Determine if there exists an equilateral triangle whose sides have lengths A, B and C.
|
a, b, c = input().split()
if(a+b > c and a+c > b and b+c > a):
print('yes')
else:
print('no')
|
s658290406
|
Accepted
| 17
| 2,940
| 89
|
a, b, c = input().split()
if(a == b and b == c):
print('Yes')
else:
print('No')
|
s237572740
|
p03503
|
u609738635
| 2,000
| 262,144
|
Wrong Answer
| 80
| 3,064
| 718
|
Joisino is planning to open a shop in a shopping street. Each of the five weekdays is divided into two periods, the morning and the evening. For each of those ten periods, a shop must be either open during the whole period, or closed during the whole period. Naturally, a shop must be open during at least one of those periods. There are already N stores in the street, numbered 1 through N. You are given information of the business hours of those shops, F_{i,j,k}. If F_{i,j,k}=1, Shop i is open during Period k on Day j (this notation is explained below); if F_{i,j,k}=0, Shop i is closed during that period. Here, the days of the week are denoted as follows. Monday: Day 1, Tuesday: Day 2, Wednesday: Day 3, Thursday: Day 4, Friday: Day 5. Also, the morning is denoted as Period 1, and the afternoon is denoted as Period 2. Let c_i be the number of periods during which both Shop i and Joisino's shop are open. Then, the profit of Joisino's shop will be P_{1,c_1}+P_{2,c_2}+...+P_{N,c_N}. Find the maximum possible profit of Joisino's shop when she decides whether her shop is open during each period, making sure that it is open during at least one period.
|
# -*- coding: utf-8 -*-
def main(N,F,P):
inf = float("INF")
ans = -inf
print(F,P)
for i in range(1,1<<10):
count = [0]*N
tmp = 0
for j in range(10):
if((i>>j)&1):
for k in range(N):
if F[k][j]==1:
count[k] += 1
else:
continue
for l in range(N):
tmp += P[l][count[l]]
ans = max(ans,tmp)
print(ans)
if __name__ == '__main__':
N = int(input())
F,P = [],[]
for i in range(N):
F.append([int(j) for j in input().split()])
for i in range(N):
P.append([int(j) for j in input().split()])
main(N,F,P)
|
s058637217
|
Accepted
| 80
| 3,064
| 703
|
# -*- coding: utf-8 -*-
def main(N,F,P):
inf = float("INF")
ans = -inf
for i in range(1,1<<10):
count = [0]*N
tmp = 0
for j in range(10):
if((i>>j)&1):
for k in range(N):
if F[k][j]==1:
count[k] += 1
else:
continue
for l in range(N):
tmp += P[l][count[l]]
ans = max(ans,tmp)
print(ans)
if __name__ == '__main__':
N = int(input())
F,P = [],[]
for i in range(N):
F.append([int(j) for j in input().split()])
for i in range(N):
P.append([int(j) for j in input().split()])
main(N,F,P)
|
s823087745
|
p03229
|
u518042385
| 2,000
| 1,048,576
|
Wrong Answer
| 242
| 8,284
| 350
|
You are given N integers; the i-th of them is A_i. Find the maximum possible sum of the absolute differences between the adjacent elements after arranging these integers in a row in any order you like.
|
n=int(input())
l=[]
for i in range(n):
l.append(int(input()))
l1=sorted(l)
if n%2==0:
k=n//2
sum=0
for i in range(0,k):
sum-=2*l1[i]
for i in range(k,n):
sum+=2*l1[i]
sum-=l1[k]
sum+=l1[k-1]
else:
k=n//2
sum=0
for i in range(k,n):
sum+=2*l1[i]
for i in range(k):
sum-=2*l1[i]
sum-=l1[0]
sum-=l1[1]
print(sum)
|
s036303971
|
Accepted
| 263
| 8,296
| 500
|
n=int(input())
l=[]
for i in range(n):
l.append(int(input()))
l1=sorted(l)
if n%2==0:
k=n//2
sum=0
for i in range(0,k):
sum-=2*l1[i]
for i in range(k,n):
sum+=2*l1[i]
sum-=l1[k]
sum+=l1[k-1]
print(sum)
else:
k=n//2
sum=0
for i in range(k,n):
sum+=2*l1[i]
for i in range(k):
sum-=2*l1[i]
sum-=l1[k]
sum-=l1[k+1]
sum1=0
for i in range(k+1,n):
sum1+=2*l1[i]
for i in range(k+1):
sum1-=2*l1[i]
sum1+=l1[k]
sum1+=l1[k-1]
print(max(sum,sum1))
|
s653846116
|
p02260
|
u995963067
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,592
| 471
|
Write a program of the Selection Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: SelectionSort(A) 1 for i = 0 to A.length-1 2 mini = i 3 for j = i to A.length-1 4 if A[j] < A[mini] 5 mini = j 6 swap A[i] and A[mini] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 6 of the pseudocode in the case where i ≠ mini.
|
def sel_sort(A, N):
count = 0
for n in range(N):
minm = n
for m in range(n, N):
if A[m] < A[minm]:
minm = m
if minm != n:
A[n], A[minm] = A[minm], A[n]
count += 1
return (A, count)
if __name__ == '__main__':
N = int(input())
A = list(map(int, input().split(' ')))
ans = sel_sort(A, N)
print(ans[0])
print(ans[1])
|
s137056551
|
Accepted
| 20
| 5,608
| 499
|
def sel_sort(A, N):
count = 0
for n in range(N):
minm = n
for m in range(n, N):
if A[m] < A[minm]:
minm = m
if minm != n:
A[n], A[minm] = A[minm], A[n]
count += 1
return (A, count)
if __name__ == '__main__':
N = int(input())
A = list(map(int, input().split(' ')))
ans = sel_sort(A, N)
print(' '.join([str(x) for x in ans[0]]))
print(ans[1])
|
s560786132
|
p03719
|
u341087021
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 124
|
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
import sys
a, b, c = [int(x) for x in sys.stdin.readline().split()]
if a <= c and b >= c:
print('YES')
else:
print('NO')
|
s136754106
|
Accepted
| 17
| 2,940
| 124
|
import sys
a, b, c = [int(x) for x in sys.stdin.readline().split()]
if a <= c and b >= c:
print('Yes')
else:
print('No')
|
s710619111
|
p02831
|
u557750361
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 126
|
Takahashi is organizing a party. At the party, each guest will receive one or more snack pieces. Takahashi predicts that the number of guests at this party will be A or B. Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted. We assume that a piece cannot be divided and distributed to multiple guests.
|
A, B = map(int, input().split())
if A > B and 0 == A % B:
print(A)
elif B > A and 0 == B % B:
print(B)
else:
print(A*B)
|
s025341537
|
Accepted
| 45
| 2,940
| 186
|
num = [int(i) for i in input().split()]
A = max(num)
B = min(num)
X = A*B
num = 0
while X >= num:
num += B
if 0 == num % A:
print(num)
break
elif X == num:
print(num)
|
s478773564
|
p03494
|
u854093727
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 226
|
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
N = input()
A = list(map(int,input().split()))
print(A)
res = 0
while True:
result = 0
for i in A:
if i%2:
result = 1
if result:
break
A = [i/2 for i in A]
res += 1
print(res)
|
s108354961
|
Accepted
| 22
| 2,940
| 217
|
N = input()
A = list(map(int,input().split()))
res = 0
while True:
result = 0
for i in A:
if i%2:
result = 1
if result:
break
A = [i/2 for i in A]
res += 1
print(res)
|
s002015691
|
p04029
|
u138486156
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 34
|
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)
|
s743978685
|
Accepted
| 17
| 2,940
| 35
|
n = int(input())
print(n*(n+1)//2)
|
s586914147
|
p03854
|
u359358631
| 2,000
| 262,144
|
Wrong Answer
| 37
| 9,148
| 393
|
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`.
|
def main():
S = input()
border = len(S)
while True:
if S[border - 5: border] == "dream":
border -= 5
elif S[border - 7: border] == "dreamer":
border -= 7
elif S[border - 5: border] == "erase":
border -= 5
elif S[border - 6: border] == "eraser":
border -= 6
else:
break
if border == 0:
print("Yes")
else:
print("No")
if __name__ == "__main__":
main()
|
s351832832
|
Accepted
| 37
| 9,244
| 393
|
def main():
S = input()
border = len(S)
while True:
if S[border - 5: border] == "dream":
border -= 5
elif S[border - 7: border] == "dreamer":
border -= 7
elif S[border - 5: border] == "erase":
border -= 5
elif S[border - 6: border] == "eraser":
border -= 6
else:
break
if border == 0:
print("YES")
else:
print("NO")
if __name__ == "__main__":
main()
|
s134823749
|
p03160
|
u670133596
| 2,000
| 1,048,576
|
Wrong Answer
| 145
| 14,724
| 357
|
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.
|
N = int(input())
H = list(map(int, input().split()))
dp = []
for i, h in enumerate(H) :
if (i==0) :
dp.append(0)
continue
if (i==1) :
dp.append(abs(H[0]-H[1]))
continue
one = abs(dp[i-2] + (H[i]-H[i-1]))
two = abs(H[i]-H[i-2])
if (one < two):
dp.append(dp[i-1] + one)
else :
dp.append(dp[i-2] + two)
print(dp)
|
s396896998
|
Accepted
| 186
| 13,928
| 403
|
N = int(input())
H = list(map(int, input().split()))
f_inf = float('inf')
dp = [f_inf] * (10**5+10)
dp[0] = 0
for i in range(N-1):
dp[i+1] = min(dp[i+1], dp[i] + abs(H[i+1] - H[i]))
if (i != N-2):
dp[i+2] = min(dp[i+2], dp[i] + abs(H[i+2] - H[i]))
print(dp[N-1])
|
s179877093
|
p03555
|
u844196583
| 2,000
| 262,144
|
Wrong Answer
| 23
| 9,096
| 312
|
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.
|
data1 = list(map(str, input().split()))
data2 = list(map(str, input().split()))
count = len(data1) - 1
sw = 0
i = 0
j = len(data2)-1
while i <= count:
if data1[i] == data2[j]:
sw += 1
print(data1[i],data2[j])
i += 1
j -= 1
if sw == len(data1):
print("Yes")
else:
print("No")
|
s839584674
|
Accepted
| 23
| 8,824
| 211
|
data1 = input()
data2 = input()
count = 3 -1
sw = 0
i = 0
j = len(data2)-1
while i <= count:
if data1[i] == data2[j]:
sw += 1
i += 1
j -= 1
if sw == 3:
print("YES")
else:
print("NO")
|
s049066853
|
p03828
|
u599547273
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 248
|
You are given an integer N. Find the number of the positive divisors of N!, modulo 10^9+7.
|
def is_prime(q):
q = abs(q)
if q == 2:
return True
if q < 2 or q&1 == 0:
return False
return pow(2, q-1, q) == 1
n = int(input())
count = 1
for i in range(1, n+1):
if is_prime(i):
count += 1
print(count)
|
s120507967
|
Accepted
| 125
| 3,188
| 447
|
def is_prime(q):
q = abs(q)
if q == 2: return True
if q < 2 or q&1 == 0: return False
return pow(2, q-1, q) == 1
N = int(input())
p = [0]*(N+1)
for i in range(2, N+1):
for j in range(2, i+1):
if is_prime(j) == False:
continue
while i%j==0:
i //= j
p[j] += 1
if i == 1:
break
r = 1
for q in p:
r = r * (q+1)
print(r%(10**9+7))
|
s041926637
|
p03635
|
u339922532
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 58
|
In _K-city_ , there are n streets running east-west, and m streets running north-south. Each street running east-west and each street running north-south cross each other. We will call the smallest area that is surrounded by four streets a block. How many blocks there are in K-city?
|
n, m = map(int, input().split())
print((n - 2) * (m - 2))
|
s898949509
|
Accepted
| 17
| 2,940
| 58
|
n, m = map(int, input().split())
print((n - 1) * (m - 1))
|
s272519188
|
p03814
|
u584529823
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,812
| 87
|
Snuke has decided to construct a string that starts with `A` and ends with `Z`, by taking out a substring of a string s (that is, a consecutive part of s). Find the greatest length of the string Snuke can construct. Here, the test set guarantees that there always exists a substring of s that starts with `A` and ends with `Z`.
|
s = input()
A_index = s.find("A")
Z_index = s.rfind("Z")
print(s[A_index:Z_index + 1])
|
s737621314
|
Accepted
| 17
| 3,500
| 86
|
s = input()
A_index = s.find("A")
Z_index = s.rfind("Z")
print(Z_index - A_index + 1)
|
s378320128
|
p02612
|
u725359833
| 2,000
| 1,048,576
|
Wrong Answer
| 29
| 9,032
| 34
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n = int(input())
print(n % 1000)
|
s256914048
|
Accepted
| 29
| 9,036
| 84
|
n = int(input())
if n % 1000 == 0:
print(0)
else:
print(1000 - (n % 1000))
|
s050646773
|
p02401
|
u508732591
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,484
| 124
|
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.
|
import sys
for i in sys.stdin:
a,op,b = i.split()
if op == "?":
break
else:
print(eval(a+op+b))
|
s551991364
|
Accepted
| 40
| 7,408
| 113
|
import sys
for i in sys.stdin:
if "?" in i:
break
else:
print(eval(i.replace("/","//")))
|
s879887077
|
p03470
|
u084949493
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 80
|
An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?
|
N = int(input())
d = list(map(int, input().split()))
ds = set(d)
print(len(ds))
|
s274850333
|
Accepted
| 18
| 2,940
| 98
|
N = int(input())
d = []
for ni in range(N):
d.append(int(input()))
ds = set(d)
print(len(ds))
|
s284367705
|
p03478
|
u808537778
| 2,000
| 262,144
|
Wrong Answer
| 30
| 3,060
| 278
|
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
n,a,b = map(int, input().split())
c = 0
for i in range(1,n+1):
n4 = n//10**4
n3 = n//10**3 - n4*10
n2 = n//10**2 - n4*100 - n3*10
n1 = n//10**1 - n4*1000 - n3*100 -n2*10
n0 = n - n4*10000 -n3*1000-n2*100-n1*10
sum = n4+n3+n2+n1+n0
if sum>=a and sum<=b:
c+=sum
|
s369072047
|
Accepted
| 30
| 3,060
| 291
|
n,a,b = map(int, input().split())
c = 0
for i in range(1,n+1):
n4 = i//10**4
n3 = i//10**3 - n4*10
n2 = i//10**2 - n4*100 - n3*10
n1 = i//10**1 - n4*1000 - n3*100 -n2*10
n0 = i - n4*10000 -n3*1000-n2*100-n1*10
total = n4+n3+n2+n1+n0
if total>=a and total<=b:
c+= i
print(c)
|
s263119754
|
p03730
|
u018591138
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 203
|
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 = map(int, input().split())
flag = False
print(C)
for i in range(B):
t = i*A%B
if t == C:
flag = True
if flag == True:
print("YES")
else:
print("NO")
|
s969736391
|
Accepted
| 17
| 2,940
| 193
|
A,B,C = map(int, input().split())
flag = False
for i in range(B):
t = i*A%B
if t == C:
flag = True
if flag == True:
print("YES")
else:
print("NO")
|
s407376829
|
p04029
|
u019584841
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 31
|
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+1)*n/2)
|
s826460435
|
Accepted
| 17
| 2,940
| 37
|
n=int(input())
print(int((n+1)*n/2))
|
s200918096
|
p03380
|
u625963200
| 2,000
| 262,144
|
Wrong Answer
| 276
| 25,324
| 203
|
Let {\rm comb}(n,r) be the number of ways to choose r objects from among n objects, disregarding order. From n non-negative integers a_1, a_2, ..., a_n, select two numbers a_i > a_j so that {\rm comb}(a_i,a_j) is maximized. If there are multiple pairs that maximize the value, any of them is accepted.
|
from scipy.misc import comb
n=int(input())
A=list(map(int,input().split()))
A.sort()
max_,num=0,0
for a in A:
if max_<comb(A[-1],a):
max_=comb(A[-1],a)
num=a
else:
break
print(A[-1],num)
|
s771292168
|
Accepted
| 129
| 14,060
| 166
|
n=int(input())
A=list(map(int,input().split()))
A.sort()
num=float('inf')
for i in range(n-1):
if abs(num-A[-1]/2)>abs(A[i]-A[-1]/2):
num=A[i]
print(A[-1],num)
|
s017846001
|
p00002
|
u995990363
| 1,000
| 131,072
|
Wrong Answer
| 30
| 5,688
| 223
|
Write a program which computes the digit number of sum of two integers a and b.
|
import sys, math
def run():
N = []
for n in sys.stdin:
a,b = map(int, n.split())
N.append((a,b))
for a,b in N:
print(math.ceil(math.log10(a+b)))
if __name__ == '__main__':
run()
|
s225111746
|
Accepted
| 30
| 5,688
| 228
|
import sys, math
def run():
N = []
for n in sys.stdin:
a,b = map(int, n.split())
N.append((a,b))
for a,b in N:
print(math.floor(math.log10(a+b)) + 1)
if __name__ == '__main__':
run()
|
s995056235
|
p03997
|
u045953894
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 66
|
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())
S=h*(a+b)/2
print(S)
|
s275624702
|
Accepted
| 17
| 2,940
| 74
|
a=int(input());b=int(input());h=int(input())
S=h*(a+b)/2
s=int(S)
print(s)
|
s198351207
|
p03681
|
u027002244
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 3,480
| 346
|
Snuke has N dogs and M monkeys. He wants them to line up in a row. As a Japanese saying goes, these dogs and monkeys are on bad terms. _("ken'en no naka", literally "the relationship of dogs and monkeys", means a relationship of mutual hatred.)_ Snuke is trying to reconsile them, by arranging the animals so that there are neither two adjacent dogs nor two adjacent monkeys. How many such arrangements there are? Find the count modulo 10^9+7 (since animals cannot understand numbers larger than that). Here, dogs and monkeys are both distinguishable. Also, two arrangements that result from reversing each other are distinguished.
|
a=list(map(int,input().split()))
ans=0
i=0
if max(a)-min(a)>=2:
pass
else:
ans=1
i=max(a)
if a[0]==a[1]:
while i!=0:
ans=ans*i
i=i-1
ans=ans*ans
else:
while i!=0:
ans=ans*i
i=i-1
ans=ans*ans/max(a)
#print(ans)
print(int(ans%1000000007 ))
|
s457481200
|
Accepted
| 444
| 4,376
| 356
|
import math
a=list(map(int,input().split()))
ans=0
if max(a)-min(a)>=2:
print(0)
else:
if a[0]==a[1]:
ans=math.factorial(max(a))
ans=(ans%1000000007)*(ans%1000000007)*2
else:
ans=math.factorial(max(a))
ans1 = math.factorial(min(a))
ans=(ans%1000000007)*((ans1)%1000000007)
print(int(ans%1000000007 ))
|
s677016338
|
p03796
|
u257974487
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,188
| 33
|
Snuke loves working out. He is now exercising N times. Before he starts exercising, his _power_ is 1. After he exercises for the i-th time, his power gets multiplied by i. Find Snuke's power after he exercises N times. Since the answer can be extremely large, print the answer modulo 10^{9}+7.
|
print(int(input()) % 1000000007)
|
s032021754
|
Accepted
| 40
| 2,940
| 86
|
N = int(input())
p = 1
for i in range(1,N+1):
p *= i
p %= 1000000007
print(p)
|
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