output_description stringlengths 15 956 | submission_id stringlengths 10 10 | status stringclasses 3 values | problem_id stringlengths 6 6 | input_description stringlengths 9 2.55k | attempt stringlengths 1 13.7k | problem_description stringlengths 7 5.24k | samples stringlengths 2 2.72k |
|---|---|---|---|---|---|---|---|
Print an integer representing the number of choices of rows and columns
satisfying the condition.
* * * | s235100387 | Wrong Answer | p02614 | Input is given from Standard Input in the following format:
H W K
c_{1,1}c_{1,2}...c_{1,W}
c_{2,1}c_{2,2}...c_{2,W}
:
c_{H,1}c_{H,2}...c_{H,W} | import numpy as np
def HandV():
num = list(map(int, input().split()))
list_result = []
nun_black = 0
nun_white = 0
for n1 in range(num[0]):
gyou = str(input())
list_gyou = []
for n2 in range(num[1]):
list_gyou.append(gyou[n2])
list_result.append(list_gyou)
nun_black = nun_black + list_gyou.count("#")
nun_white = nun_white + list_gyou.count(".")
array = np.array(list_result)
count_num = 0
for i in range(num[0] + 1):
array_test = array
if i != 0:
if nun_black - np.count_nonzero(array_test[i - 1, :] == "#") == num[2]:
count_num = count_num + 1
for i2 in range(i + 1, num[0] + 1):
if (
nun_black
- np.count_nonzero(array_test[i - 1, :] == "#")
- np.count_nonzero(array_test[i2 - 1, :] == "#")
== num[2]
):
count_num = count_num + 1
for j in range(num[1] + 1):
if i == 0 and j == 0:
if np.count_nonzero(array == "#") == num[2]:
count_num = count_num + 1
elif i == 0 and j != 0:
if nun_black - np.count_nonzero(array_test[:, j - 1] == "#") == num[2]:
count_num = count_num + 1
for j2 in range(j + 1, num[1] + 1):
if (
nun_black
- np.count_nonzero(array_test[:, j - 1] == "#")
- np.count_nonzero(array_test[:, j2 - 1] == "#")
== num[2]
):
count_num = count_num + 1
elif i != 0 and j != 0:
if (
nun_black
- np.count_nonzero(array_test[i - 1, :] == "#")
- np.count_nonzero(array_test[:, j - 1] == "#")
+ np.count_nonzero(array_test[i - 1, j - 1] == "#")
== num[2]
):
count_num = count_num + 1
print(count_num)
if __name__ == "__main__":
HandV()
| Statement
We have a grid of H rows and W columns of squares. The color of the square at
the i-th row from the top and the j-th column from the left (1 \leq i \leq H,
1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white
if c_{i,j} is `.`, and black if c_{i,j} is `#`.
Consider doing the following operation:
* Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns.
You are given a positive integer K. How many choices of rows and columns
result in exactly K black squares remaining after the operation? Here, we
consider two choices different when there is a row or column chosen in only
one of those choices. | [{"input": "2 3 2\n ..#\n ###", "output": "5\n \n\nFive choices below satisfy the condition.\n\n * The 1-st row and 1-st column\n * The 1-st row and 2-nd column\n * The 1-st row and 3-rd column\n * The 1-st and 2-nd column\n * The 3-rd column\n\n* * *"}, {"input": "2 3 4\n ..#\n ###", "output": "1\n \n\nOne choice, which is choosing nothing, satisfies the condition.\n\n* * *"}, {"input": "2 2 3\n ##\n ##", "output": "0\n \n\n* * *"}, {"input": "6 6 8\n ..##..\n .#..#.\n #....#\n ######\n #....#\n #....#", "output": "208"}] |
Print an integer representing the number of choices of rows and columns
satisfying the condition.
* * * | s278056108 | Accepted | p02614 | Input is given from Standard Input in the following format:
H W K
c_{1,1}c_{1,2}...c_{1,W}
c_{2,1}c_{2,2}...c_{2,W}
:
c_{H,1}c_{H,2}...c_{H,W} | from copy import deepcopy
def check(S):
n = 0
for ch in S:
if ch == "#":
n += 1
return n
h, w, k = map(int, input().split())
tmp = [list(input()) for _ in range(h)]
c = deepcopy(tmp)
y = 0
for l in c:
x = check("".join(l))
y += x
n = 0
for i in range(h + w):
n += 2**i
n += 1
m = 0
for i in range(n):
i = format(i, str(h + w).zfill(2) + "b")
H = i[:h]
W = i[h:]
ans = 0
x = 0
for j in H:
if j == "1":
ans += check("".join(c[x]))
# print(''.join(c[x]))
c[x] = ["."] * w
x += 1
x = 0
for j in W:
if j == "1":
s = str()
for z in range(h):
s += c[z][x]
ans += check(s)
for z in range(h):
c[z][x] = "."
x += 1
# print(H,W,y-ans,k)
if k == y - ans:
m += 1
c = deepcopy(tmp)
print(m)
| Statement
We have a grid of H rows and W columns of squares. The color of the square at
the i-th row from the top and the j-th column from the left (1 \leq i \leq H,
1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white
if c_{i,j} is `.`, and black if c_{i,j} is `#`.
Consider doing the following operation:
* Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns.
You are given a positive integer K. How many choices of rows and columns
result in exactly K black squares remaining after the operation? Here, we
consider two choices different when there is a row or column chosen in only
one of those choices. | [{"input": "2 3 2\n ..#\n ###", "output": "5\n \n\nFive choices below satisfy the condition.\n\n * The 1-st row and 1-st column\n * The 1-st row and 2-nd column\n * The 1-st row and 3-rd column\n * The 1-st and 2-nd column\n * The 3-rd column\n\n* * *"}, {"input": "2 3 4\n ..#\n ###", "output": "1\n \n\nOne choice, which is choosing nothing, satisfies the condition.\n\n* * *"}, {"input": "2 2 3\n ##\n ##", "output": "0\n \n\n* * *"}, {"input": "6 6 8\n ..##..\n .#..#.\n #....#\n ######\n #....#\n #....#", "output": "208"}] |
Print an integer representing the number of choices of rows and columns
satisfying the condition.
* * * | s485617693 | Wrong Answer | p02614 | Input is given from Standard Input in the following format:
H W K
c_{1,1}c_{1,2}...c_{1,W}
c_{2,1}c_{2,2}...c_{2,W}
:
c_{H,1}c_{H,2}...c_{H,W} | from sys import stdin
def readline_from_stdin() -> str:
return stdin.readline().rstrip
def cT(c) -> int:
ccc = [cc.count(True) for cc in c]
return sum(ccc)
def sortFn(c) -> list:
nl = sorted(c, key=lambda x: x.count(True), reverse=True)
return nl
def compFn(c) -> list:
return [cc.count(True) for cc in c]
def combiFn(c) -> list:
combi = [list(itertools.combinations(c, v + 1)) for v in range(len(c))]
combina = [[list(v) for v in vvv] for vvv in combi]
return list(itertools.chain.from_iterable(combina))
# return combina
def countCombiFn(c, X) -> int:
combiSum = [list(itertools.combinations(c, v + 1)) for v in range(len(c))]
combinaSum = [[list(v) for v in vvv] for vvv in combiSum]
f1 = list(itertools.chain.from_iterable(combinaSum))
# print(len(f1))
# f2 = list(itertools.chain.from_iterable(f1))
sumF1 = [sum(v) for v in f1]
# print(sumF1)
# print(sumF1.count(X))
return sumF1.count(X)
if __name__ == "__main__":
import itertools
H, W, K = input().split()
H, W, K = int(H), int(W), int(K)
C = [[False if i == "." else True for i in input()] for i in range(H)]
# sC = sortFn(C)
comped = compFn(C)
combi = combiFn(comped)
# print(len(combi))
# print(combi)
cntArr = [countCombiFn(c, K) for c in combi]
# print(cntArr)
print(sum(cntArr))
# print(sum(cntArr))
| Statement
We have a grid of H rows and W columns of squares. The color of the square at
the i-th row from the top and the j-th column from the left (1 \leq i \leq H,
1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white
if c_{i,j} is `.`, and black if c_{i,j} is `#`.
Consider doing the following operation:
* Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns.
You are given a positive integer K. How many choices of rows and columns
result in exactly K black squares remaining after the operation? Here, we
consider two choices different when there is a row or column chosen in only
one of those choices. | [{"input": "2 3 2\n ..#\n ###", "output": "5\n \n\nFive choices below satisfy the condition.\n\n * The 1-st row and 1-st column\n * The 1-st row and 2-nd column\n * The 1-st row and 3-rd column\n * The 1-st and 2-nd column\n * The 3-rd column\n\n* * *"}, {"input": "2 3 4\n ..#\n ###", "output": "1\n \n\nOne choice, which is choosing nothing, satisfies the condition.\n\n* * *"}, {"input": "2 2 3\n ##\n ##", "output": "0\n \n\n* * *"}, {"input": "6 6 8\n ..##..\n .#..#.\n #....#\n ######\n #....#\n #....#", "output": "208"}] |
Print an integer representing the number of choices of rows and columns
satisfying the condition.
* * * | s901230269 | Accepted | p02614 | Input is given from Standard Input in the following format:
H W K
c_{1,1}c_{1,2}...c_{1,W}
c_{2,1}c_{2,2}...c_{2,W}
:
c_{H,1}c_{H,2}...c_{H,W} | aa = input("").split(" ")
a = []
for i in range(3):
a += [int(aa[i])]
b = []
for i in range(a[0]):
b += [input("")]
c = []
for i in range(a[0]):
d = []
for k in b[i]:
if k == ".":
d += [0]
else:
d += [1]
c += [d]
def count(n, m):
s = 0
for i in range(a[0]):
for k in range(a[1]):
if c[i][k] == 1 and n[i] == 1 and m[k] == 1:
s += 1
return s
ss = 0
for ii in range(2 ** a[0]):
for kk in range(2 ** a[1]):
d = ii
e = kk
n = []
m = []
for iii in range(a[0]):
n += [d % 2]
d = int((d / 2))
for kkk in range(a[1]):
m += [e % 2]
e = int(e / 2)
if count(n, m) == a[2]:
ss += 1
print(ss)
| Statement
We have a grid of H rows and W columns of squares. The color of the square at
the i-th row from the top and the j-th column from the left (1 \leq i \leq H,
1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white
if c_{i,j} is `.`, and black if c_{i,j} is `#`.
Consider doing the following operation:
* Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns.
You are given a positive integer K. How many choices of rows and columns
result in exactly K black squares remaining after the operation? Here, we
consider two choices different when there is a row or column chosen in only
one of those choices. | [{"input": "2 3 2\n ..#\n ###", "output": "5\n \n\nFive choices below satisfy the condition.\n\n * The 1-st row and 1-st column\n * The 1-st row and 2-nd column\n * The 1-st row and 3-rd column\n * The 1-st and 2-nd column\n * The 3-rd column\n\n* * *"}, {"input": "2 3 4\n ..#\n ###", "output": "1\n \n\nOne choice, which is choosing nothing, satisfies the condition.\n\n* * *"}, {"input": "2 2 3\n ##\n ##", "output": "0\n \n\n* * *"}, {"input": "6 6 8\n ..##..\n .#..#.\n #....#\n ######\n #....#\n #....#", "output": "208"}] |
Print an integer representing the number of choices of rows and columns
satisfying the condition.
* * * | s278032365 | Accepted | p02614 | Input is given from Standard Input in the following format:
H W K
c_{1,1}c_{1,2}...c_{1,W}
c_{2,1}c_{2,2}...c_{2,W}
:
c_{H,1}c_{H,2}...c_{H,W} | import sys
def A():
n = int(sys.stdin.readline().rstrip())
charge = (n + 999) // 1000 * 1000 - n
print(charge)
from collections import Counter
def B():
n, *s = sys.stdin.read().split()
c = Counter(s)
for v in "AC, WA, TLE, RE".split(", "):
print(f"{v} x {c[v]}")
def C():
h, w, k = map(int, sys.stdin.readline().split())
c = [sys.stdin.readline().rstrip() for _ in range(h)]
tot = 0
for i in range(1 << h):
for j in range(1 << w):
cnt = 0
for y in range(h):
for x in range(w):
if i >> y & 1 or j >> x & 1:
continue
cnt += c[y][x] == "#"
tot += cnt == k
print(tot)
def D():
pass
def E():
pass
def F():
pass
if __name__ == "__main__":
# A()
# B()
C()
D()
E()
F()
| Statement
We have a grid of H rows and W columns of squares. The color of the square at
the i-th row from the top and the j-th column from the left (1 \leq i \leq H,
1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white
if c_{i,j} is `.`, and black if c_{i,j} is `#`.
Consider doing the following operation:
* Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns.
You are given a positive integer K. How many choices of rows and columns
result in exactly K black squares remaining after the operation? Here, we
consider two choices different when there is a row or column chosen in only
one of those choices. | [{"input": "2 3 2\n ..#\n ###", "output": "5\n \n\nFive choices below satisfy the condition.\n\n * The 1-st row and 1-st column\n * The 1-st row and 2-nd column\n * The 1-st row and 3-rd column\n * The 1-st and 2-nd column\n * The 3-rd column\n\n* * *"}, {"input": "2 3 4\n ..#\n ###", "output": "1\n \n\nOne choice, which is choosing nothing, satisfies the condition.\n\n* * *"}, {"input": "2 2 3\n ##\n ##", "output": "0\n \n\n* * *"}, {"input": "6 6 8\n ..##..\n .#..#.\n #....#\n ######\n #....#\n #....#", "output": "208"}] |
Print an integer representing the number of choices of rows and columns
satisfying the condition.
* * * | s998293914 | Runtime Error | p02614 | Input is given from Standard Input in the following format:
H W K
c_{1,1}c_{1,2}...c_{1,W}
c_{2,1}c_{2,2}...c_{2,W}
:
c_{H,1}c_{H,2}...c_{H,W} | import copy
def count(picture, ans_num):
count_num = 0
flag = 0
for row in range(len(picture)):
for column in range(len(picture)):
if picture[row][column] == "#":
count_num = count_num + 1
if count_num == ans_num:
flag = 1
return flag
if __name__ == "__main__":
ans_list = input().split(" ")
num_list = []
picture = []
ans_num = 0
for ans in ans_list:
num_list.append(int(ans))
# 入力絵の作成（list化）
for row_num in range(num_list[0]):
picture.append(list(input()))
# padding
for num in range(num_list[0]):
picture[num].append("0")
zero_list = []
for n in range(num_list[1] + 1):
zero_list.append("0")
picture.append(zero_list)
# 探索a
for row in range(len(picture)):
# 初期化必要
tmp_picture = copy.deepcopy(picture)
tmp_picture[row] = zero_list
for column in range(len(picture[0])):
# 初期化必要
tmp_tmp_picture = copy.deepcopy(tmp_picture)
for n in range(len(picture)):
tmp_tmp_picture[n][column] = "0"
ans_num = ans_num + count(tmp_picture, num_list[2])
print(tmp_tmp_picture)
print(ans_num)
| Statement
We have a grid of H rows and W columns of squares. The color of the square at
the i-th row from the top and the j-th column from the left (1 \leq i \leq H,
1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white
if c_{i,j} is `.`, and black if c_{i,j} is `#`.
Consider doing the following operation:
* Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns.
You are given a positive integer K. How many choices of rows and columns
result in exactly K black squares remaining after the operation? Here, we
consider two choices different when there is a row or column chosen in only
one of those choices. | [{"input": "2 3 2\n ..#\n ###", "output": "5\n \n\nFive choices below satisfy the condition.\n\n * The 1-st row and 1-st column\n * The 1-st row and 2-nd column\n * The 1-st row and 3-rd column\n * The 1-st and 2-nd column\n * The 3-rd column\n\n* * *"}, {"input": "2 3 4\n ..#\n ###", "output": "1\n \n\nOne choice, which is choosing nothing, satisfies the condition.\n\n* * *"}, {"input": "2 2 3\n ##\n ##", "output": "0\n \n\n* * *"}, {"input": "6 6 8\n ..##..\n .#..#.\n #....#\n ######\n #....#\n #....#", "output": "208"}] |
Print an integer representing the number of choices of rows and columns
satisfying the condition.
* * * | s446949945 | Wrong Answer | p02614 | Input is given from Standard Input in the following format:
H W K
c_{1,1}c_{1,2}...c_{1,W}
c_{2,1}c_{2,2}...c_{2,W}
:
c_{H,1}c_{H,2}...c_{H,W} | H, W, K = map(int, input().split())
if K == 4:
print("1")
elif K == 3:
print("0")
else:
print("208")
| Statement
We have a grid of H rows and W columns of squares. The color of the square at
the i-th row from the top and the j-th column from the left (1 \leq i \leq H,
1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white
if c_{i,j} is `.`, and black if c_{i,j} is `#`.
Consider doing the following operation:
* Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns.
You are given a positive integer K. How many choices of rows and columns
result in exactly K black squares remaining after the operation? Here, we
consider two choices different when there is a row or column chosen in only
one of those choices. | [{"input": "2 3 2\n ..#\n ###", "output": "5\n \n\nFive choices below satisfy the condition.\n\n * The 1-st row and 1-st column\n * The 1-st row and 2-nd column\n * The 1-st row and 3-rd column\n * The 1-st and 2-nd column\n * The 3-rd column\n\n* * *"}, {"input": "2 3 4\n ..#\n ###", "output": "1\n \n\nOne choice, which is choosing nothing, satisfies the condition.\n\n* * *"}, {"input": "2 2 3\n ##\n ##", "output": "0\n \n\n* * *"}, {"input": "6 6 8\n ..##..\n .#..#.\n #....#\n ######\n #....#\n #....#", "output": "208"}] |
Print an integer representing the number of choices of rows and columns
satisfying the condition.
* * * | s031146753 | Wrong Answer | p02614 | Input is given from Standard Input in the following format:
H W K
c_{1,1}c_{1,2}...c_{1,W}
c_{2,1}c_{2,2}...c_{2,W}
:
c_{H,1}c_{H,2}...c_{H,W} | # ダメでした
print(0)
| Statement
We have a grid of H rows and W columns of squares. The color of the square at
the i-th row from the top and the j-th column from the left (1 \leq i \leq H,
1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white
if c_{i,j} is `.`, and black if c_{i,j} is `#`.
Consider doing the following operation:
* Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns.
You are given a positive integer K. How many choices of rows and columns
result in exactly K black squares remaining after the operation? Here, we
consider two choices different when there is a row or column chosen in only
one of those choices. | [{"input": "2 3 2\n ..#\n ###", "output": "5\n \n\nFive choices below satisfy the condition.\n\n * The 1-st row and 1-st column\n * The 1-st row and 2-nd column\n * The 1-st row and 3-rd column\n * The 1-st and 2-nd column\n * The 3-rd column\n\n* * *"}, {"input": "2 3 4\n ..#\n ###", "output": "1\n \n\nOne choice, which is choosing nothing, satisfies the condition.\n\n* * *"}, {"input": "2 2 3\n ##\n ##", "output": "0\n \n\n* * *"}, {"input": "6 6 8\n ..##..\n .#..#.\n #....#\n ######\n #....#\n #....#", "output": "208"}] |
Print an integer representing the number of choices of rows and columns
satisfying the condition.
* * * | s174096492 | Wrong Answer | p02614 | Input is given from Standard Input in the following format:
H W K
c_{1,1}c_{1,2}...c_{1,W}
c_{2,1}c_{2,2}...c_{2,W}
:
c_{H,1}c_{H,2}...c_{H,W} | import re
import copy
h,w,k=map(int,input().split())
lis=[]
ans=0
for i in range(h):
lis.append(list(map(str,input())))
temp_l=copy.deepcopy(lis)
for l in range(0,h):
for k in range(0,w):
kuro=0
lis[l][k]=re.sub(".|#","r",lis[l][k])
for m in lis:
for n in m:
if n=="#":
kuro+=1
if kuro==k:
ans+=1
for l in range(0,w):
for k in range(0,h):
kuro=0
lis[k][l]=re.sub(".|#","r",lis[k][l])
for m in lis:
for n in m:
if n=="#":
kuro+=1
if kuro==k:
ans+=1
print(ans) | Statement
We have a grid of H rows and W columns of squares. The color of the square at
the i-th row from the top and the j-th column from the left (1 \leq i \leq H,
1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white
if c_{i,j} is `.`, and black if c_{i,j} is `#`.
Consider doing the following operation:
* Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns.
You are given a positive integer K. How many choices of rows and columns
result in exactly K black squares remaining after the operation? Here, we
consider two choices different when there is a row or column chosen in only
one of those choices. | [{"input": "2 3 2\n ..#\n ###", "output": "5\n \n\nFive choices below satisfy the condition.\n\n * The 1-st row and 1-st column\n * The 1-st row and 2-nd column\n * The 1-st row and 3-rd column\n * The 1-st and 2-nd column\n * The 3-rd column\n\n* * *"}, {"input": "2 3 4\n ..#\n ###", "output": "1\n \n\nOne choice, which is choosing nothing, satisfies the condition.\n\n* * *"}, {"input": "2 2 3\n ##\n ##", "output": "0\n \n\n* * *"}, {"input": "6 6 8\n ..##..\n .#..#.\n #....#\n ######\n #....#\n #....#", "output": "208"}] |
Print a $8 \times 8$ chess board by strings where a square with a queen is
represented by 'Q' and an empty square is represented by '.'. | s718582803 | Runtime Error | p02244 | In the first line, an integer $k$ is given. In the following $k$ lines, each
square where a queen is already placed is given by two integers $r$ and $c$.
$r$ and $c$ respectively denotes the row number and the column number. The
row/column numbers start with 0. | # 8 Puzzle
import copy
z = 0
[N, d] = [3, 0]
check_flag = [
[1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4],
[2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1],
[3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2],
[4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3],
[3, 2, 1, 4, 3, 2, 1, 4, 3, 2, 1, 4],
[2, 1, 4, 3, 2, 1, 4, 3, 2, 1, 4, 3],
[1, 4, 3, 2, 1, 4, 3, 2, 1, 4, 3, 2],
[4, 3, 2, 1, 4, 3, 2, 1, 4, 3, 2, 1],
]
start = []
goal = [[i + j * N for i in range(1, N + 1)] for j in range(N)]
goal[2][2] = 0
for i in range(N):
start.append(list(map(int, input().split())))
def manhattan(value, pairs):
h = 0
if value == 1:
h = pairs[0] + pairs[1]
if value == 2:
h = pairs[0] + abs(pairs[1] - 1)
if value == 3:
h = pairs[0] + abs(pairs[1] - 2)
if value == 4:
h = abs(pairs[0] - 1) + pairs[1]
if value == 5:
h = abs(pairs[0] - 1) + abs(pairs[1] - 1)
if value == 6:
h = abs(pairs[0] - 1) + abs(pairs[1] - 2)
if value == 7:
h = abs(pairs[0] - 2) + pairs[1]
if value == 8:
h = abs(pairs[0] - 2) + abs(pairs[1] - 1)
return h
def flag_array(flag, input):
flag.pop(0)
flag.append(input)
return flag
s_h = 0
for i in range(N):
for j in range(N):
s_h += manhattan(start[i][j], [i, j])
# print(s_h)
for i in range(N):
check = start[i].count(0)
if check != 0:
[s_r, s_c] = [i, start[i].index(0)]
break
if i == 3:
print("Error")
while True:
d += 1
queue = []
flag = [0 for i in range(12)]
queue.append([s_h, start, 0, [s_r, s_c], flag])
# while True:
while len(queue) != 0:
# print("Q: ", len(queue), "depth: ", d)
short_n = queue.pop(0)
h = short_n[0] - short_n[2]
state = short_n[1]
g = short_n[2]
[r, c] = short_n[3]
flag = short_n[4]
# print("left_Q: ", len(queue), "depth: ", d, "h: ", h, "g: ", g, "state: ", state, "g+h: ", short_n[0], "flag: ", flag[len(flag) - 1])
# print("left_Q: ", len(queue), "depth: ", d, "h: ", h, "g: ", g, "flag: ", flag, "g+h: ", short_n[0])
# if d == 1:
# print(short_n[0])
# print(state)
# print(g)
if h == 0:
print(short_n[2])
# print(z)
break
"""
if flag in check_flag:
z += 1
continue
"""
if r - 1 >= 0 and flag[len(flag) - 1] != 3:
[temp, temp_array] = [copy.deepcopy(state), flag[:]]
h = (
short_n[0]
- short_n[2]
- manhattan(temp[r - 1][c], [r - 1, c])
+ manhattan(temp[r - 1][c], [r, c])
)
[temp[r][c], temp[r - 1][c]] = [temp[r - 1][c], temp[r][c]]
# if temp[r][c] != goal[r][c]:
# [r, c] = [r - 1, c]
# if temp
# h = manhattan(temp)
# print("1: ", h, temp)
if g + 1 + h <= d:
queue.append(
[h + g + 1, temp, g + 1, [r - 1, c], flag_array(temp_array, 1)]
)
if c + 1 < N and flag[len(flag) - 1] != 4:
# print("2: ")
[temp, temp_array] = [copy.deepcopy(state), flag[:]]
# temp_array = copy.deepcopy(flag)
h = (
short_n[0]
- short_n[2]
- manhattan(temp[r][c + 1], [r, c + 1])
+ manhattan(temp[r][c + 1], [r, c])
)
[temp[r][c], temp[r][c + 1]] = [temp[r][c + 1], temp[r][c]]
# queue.append(calculate(temp, g + 1))
# print("2: ", h, temp)
if g + 1 + h <= d:
queue.append(
[h + g + 1, temp, g + 1, [r, c + 1], flag_array(temp_array, 2)]
)
if r + 1 < N and flag[len(flag) - 1] != 1:
# print("3: ")
[temp, temp_array] = [copy.deepcopy(state), flag[:]]
# temp_array = copy.deepcopy(flag)
h = (
short_n[0]
- short_n[2]
- manhattan(temp[r + 1][c], [r + 1, c])
+ manhattan(temp[r + 1][c], [r, c])
)
[temp[r][c], temp[r + 1][c]] = [temp[r + 1][c], temp[r][c]]
# queue.append(calculate(temp, g + 1))
# print("3: ", h, temp)
if g + 1 + h <= d:
queue.append(
[h + g + 1, temp, g + 1, [r + 1, c], flag_array(temp_array, 3)]
)
if c - 1 >= 0 and flag[len(flag) - 1] != 2:
# print("4: ")
[temp, temp_array] = [copy.deepcopy(state), flag[:]]
h = (
short_n[0]
- short_n[2]
- manhattan(temp[r][c - 1], [r, c - 1])
+ manhattan(temp[r][c - 1], [r, c])
)
[temp[r][c], temp[r][c - 1]] = [temp[r][c - 1], temp[r][c]]
if g + 1 + h <= d:
queue.append(
[h + g + 1, temp, g + 1, [r, c - 1], flag_array(temp_array, 4)]
)
queue.sort(key=lambda data: data[0])
queue.sort(key=lambda data: data[2], reverse=True)
data = []
g_data = []
# print(queue)
"""
for i in range(len(queue)):
data.append(str(queue[i][0]))
g_data.append(str(queue[i][2]))
#print(queue[i])
print("g+h: ",' '.join(data))
print("g: ",' '.join(g_data))
"""
# if d == 1:
# print(len(queue))
if state == goal:
break
| 8 Queens Problem
The goal of 8 Queens Problem is to put eight queens on a chess-board such that
none of them threatens any of others. A queen threatens the squares in the
same row, in the same column, or on the same diagonals as shown in the
following figure.
For a given chess board where $k$ queens are already placed, find the solution
of the 8 queens problem. | [{"input": "2\n 2 2\n 5 3", "output": "......Q.\n Q.......\n ..Q.....\n .......Q\n .....Q..\n ...Q....\n .Q......\n ....Q..."}] |
Print a $8 \times 8$ chess board by strings where a square with a queen is
represented by 'Q' and an empty square is represented by '.'. | s163695159 | Accepted | p02244 | In the first line, an integer $k$ is given. In the following $k$ lines, each
square where a queen is already placed is given by two integers $r$ and $c$.
$r$ and $c$ respectively denotes the row number and the column number. The
row/column numbers start with 0. | # -*- coding: utf-8 -*-
import sys
from itertools import permutations
sys.setrecursionlimit(10**9)
INF = 10**18
MOD = 10**9 + 7
def input():
return sys.stdin.readline().rstrip()
def main():
mx = [1, 1, -1, -1]
my = [1, -1, 1, -1]
def check(tmp_board):
for y, x in enumerate(tmp_board):
d = [1, 1]
for j in range(7):
for k in range(4):
c = [y + d[0] * my[k], x + d[1] * mx[k]]
if 0 <= c[0] <= 7 and 0 <= c[1] <= 7 and tmp_board[c[0]] == c[1]:
return False
d[0] += 1
d[1] += 1
return True
k = int(input())
c = [["."] * 8 for _ in range(8)]
for i in range(k):
R, C = map(int, input().split())
c[R][C] = "Q"
yoko = list(range(8))
tate = list(range(8))
board = [-1] * 8
try:
for y, l in enumerate(c):
for x, s in enumerate(l):
if s == "Q":
tate.remove(y)
yoko.remove(x)
board[y] = x
for p in permutations(yoko):
tmp_board = board[:]
for i, x in enumerate(tate):
tmp_board[x] = p[i]
if check(tmp_board):
for x in tmp_board:
ans = ["."] * 8
ans[x] = "Q"
print("".join(ans))
break
else:
raise Exception
except:
print("No Answer")
if __name__ == "__main__":
main()
| 8 Queens Problem
The goal of 8 Queens Problem is to put eight queens on a chess-board such that
none of them threatens any of others. A queen threatens the squares in the
same row, in the same column, or on the same diagonals as shown in the
following figure.
For a given chess board where $k$ queens are already placed, find the solution
of the 8 queens problem. | [{"input": "2\n 2 2\n 5 3", "output": "......Q.\n Q.......\n ..Q.....\n .......Q\n .....Q..\n ...Q....\n .Q......\n ....Q..."}] |
Print a $8 \times 8$ chess board by strings where a square with a queen is
represented by 'Q' and an empty square is represented by '.'. | s966619003 | Accepted | p02244 | In the first line, an integer $k$ is given. In the following $k$ lines, each
square where a queen is already placed is given by two integers $r$ and $c$.
$r$ and $c$ respectively denotes the row number and the column number. The
row/column numbers start with 0. | def sakuzyo(i):
if koteirow[i] == 1:
sakuzyo(i - 1)
else:
for j in range(N):
if haiti[i][j] == "Q":
# print('delete {} {}'.format(i, j))
haiti[i][j] = "."
row[i] = 0
col[j] = 0
dpos[i + j] = 0
dneg[i - j + (N - 1)] = 0
if j == N - 1:
sakuzyo(i - 1)
else:
sounyuu(i, j + 1)
def sounyuu(i, j):
if koteirow[i] == 0:
for k in range(j, N):
flag = 0
if col[k] == 0 and dpos[i + k] == 0 and dneg[i - k + (N - 1)] == 0:
haiti[i][k] = "Q"
# print('add {} {}'.format(i, k))
row[i] = 1
col[k] = 1
dpos[i + k] = 1
dneg[i - k + (N - 1)] = 1
flag = 1
if i == N - 1:
for o in range(N):
for p in range(N):
print(haiti[o][p], end="")
print()
exit()
else:
sounyuu(i + 1, 0)
if flag == 0:
sakuzyo(i - 1)
elif i == N - 1:
for o in range(N):
for p in range(N):
print(haiti[o][p], end="")
print()
exit()
else:
sounyuu(i + 1, j)
k = int(input())
data = [list(map(int, input().split())) for _ in range(k)]
N = 8
row = [0] * N
col = [0] * N
dpos = [0] * (N * 2 - 1)
dneg = [0] * (N * 2 - 1)
haiti = [["."] * N for _ in range(N)]
koteirow = [0] * N
for i in data:
haiti[i[0]][i[1]] = "Q"
row[i[0]] = 1
col[i[1]] = 1
dpos[i[0] + i[1]] = 1
dneg[i[0] - i[1] + (N - 1)] = 1
koteirow[i[0]] = 1
sounyuu(0, 0)
| 8 Queens Problem
The goal of 8 Queens Problem is to put eight queens on a chess-board such that
none of them threatens any of others. A queen threatens the squares in the
same row, in the same column, or on the same diagonals as shown in the
following figure.
For a given chess board where $k$ queens are already placed, find the solution
of the 8 queens problem. | [{"input": "2\n 2 2\n 5 3", "output": "......Q.\n Q.......\n ..Q.....\n .......Q\n .....Q..\n ...Q....\n .Q......\n ....Q..."}] |
Print a $8 \times 8$ chess board by strings where a square with a queen is
represented by 'Q' and an empty square is represented by '.'. | s682555387 | Wrong Answer | p02244 | In the first line, an integer $k$ is given. In the following $k$ lines, each
square where a queen is already placed is given by two integers $r$ and $c$.
$r$ and $c$ respectively denotes the row number and the column number. The
row/column numbers start with 0. | import copy
n = int(input())
dotsa = []
for _ in range(n):
y, x = map(int, input().split(" "))
dotsa.append([y, x])
pattern = []
pattern.append(dotsa)
for y in range(8):
nextpattern = []
# print("#######")
# print(pattern)
for pat in pattern:
# print(";;;;;;;;;;;;;;;;;;;;;")
# print(pat)
flag = False
for dot in pat:
if dot[0] == y:
flag = True
break
if flag:
nextpattern = pattern
continue
ifFound = False
for x in range(8):
ank = len(pat)
flag2 = True
for i in range(ank):
if abs(pat[i][0] - y) == abs(pat[i][1] - x) or pat[i][1] == x:
flag2 = False
break
if flag2:
# print("appended")
# print(y)
# print(x)
tmp = copy.deepcopy(pat)
tmp.append([y, x])
nextpattern.append(tmp)
# print("x loop end")
pattern = nextpattern
ans = pattern[0]
board = [["."] * 8 for i in range(8)]
# print(ans)
for ele in ans:
board[ele[1]][ele[0]] = "Q"
for row in board:
print("".join(map(str, row)))
| 8 Queens Problem
The goal of 8 Queens Problem is to put eight queens on a chess-board such that
none of them threatens any of others. A queen threatens the squares in the
same row, in the same column, or on the same diagonals as shown in the
following figure.
For a given chess board where $k$ queens are already placed, find the solution
of the 8 queens problem. | [{"input": "2\n 2 2\n 5 3", "output": "......Q.\n Q.......\n ..Q.....\n .......Q\n .....Q..\n ...Q....\n .Q......\n ....Q..."}] |
Print a $8 \times 8$ chess board by strings where a square with a queen is
represented by 'Q' and an empty square is represented by '.'. | s557403065 | Accepted | p02244 | In the first line, an integer $k$ is given. In the following $k$ lines, each
square where a queen is already placed is given by two integers $r$ and $c$.
$r$ and $c$ respectively denotes the row number and the column number. The
row/column numbers start with 0. | # -*- coding:utf-8 -*-
def solve():
import itertools
import copy
K = int(input())
grid = [[0 for i in range(8)] for j in range(8)]
def put_queen(grid, r, c):
for j in range(8):
grid[r][j] = -1
for i in range(8):
grid[i][c] = -1
# 左上
i, j = r - 1, c - 1
while i >= 0 and j >= 0:
grid[i][j] = -1
i, j = i - 1, j - 1
# 左下
i, j = r + 1, c - 1
while i < 8 and j >= 0:
grid[i][j] = -1
i, j = i + 1, j - 1
# 右上
i, j = r - 1, c + 1
while i >= 0 and j < 8:
grid[i][j] = -1
i, j = i - 1, j + 1
# 右下
i, j = r + 1, c + 1
while i < 8 and j < 8:
grid[i][j] = -1
i, j = i + 1, j + 1
grid[r][c] = 1
for i in range(K):
r, c = list(map(int, input().split()))
put_queen(grid, r, c)
# 8!を全探索
ps = itertools.permutations([i for i in range(8)], r=8)
for p in ps:
ans_grid = copy.deepcopy(grid)
is_ok = True
for i in range(8):
if ans_grid[i][p[i]] == 1:
continue
if ans_grid[i][p[i]] == -1:
is_ok = False
break
put_queen(ans_grid, i, p[i])
if is_ok:
break
# ans_gridを出力して終了
for i in range(8):
s = ""
for j in range(8):
if ans_grid[i][j] == 1:
s += "Q"
else:
s += "."
print(s)
if __name__ == "__main__":
solve()
| 8 Queens Problem
The goal of 8 Queens Problem is to put eight queens on a chess-board such that
none of them threatens any of others. A queen threatens the squares in the
same row, in the same column, or on the same diagonals as shown in the
following figure.
For a given chess board where $k$ queens are already placed, find the solution
of the 8 queens problem. | [{"input": "2\n 2 2\n 5 3", "output": "......Q.\n Q.......\n ..Q.....\n .......Q\n .....Q..\n ...Q....\n .Q......\n ....Q..."}] |
Print a $8 \times 8$ chess board by strings where a square with a queen is
represented by 'Q' and an empty square is represented by '.'. | s177505366 | Accepted | p02244 | In the first line, an integer $k$ is given. In the following $k$ lines, each
square where a queen is already placed is given by two integers $r$ and $c$.
$r$ and $c$ respectively denotes the row number and the column number. The
row/column numbers start with 0. | from itertools import product
def setq(ri, ci, krc):
for ir in range(8):
if ir != ri:
if krc[ir * 8 + ci] == "Q":
return False, krc
else:
krc[ir * 8 + ci] = "X"
for ic in range(8):
if ic != ci:
if krc[ri * 8 + ic] == "Q":
return False, krc
else:
krc[ri * 8 + ic] = "X"
for ix in range(-7, 7, 1):
ir = ri + ix
ic = ci + ix
if 0 <= ir and ir <= 7 and ir != ri:
if 0 <= ic and ic <= 7 and ic != ci:
if krc[ir * 8 + ic] == "Q":
return False, krc
else:
krc[ir * 8 + ic] = "X"
ir = ri - ix
ic = ci + ix
if 0 <= ir and ir <= 7 and ir != ri:
if 0 <= ic and ic <= 7 and ic != ci:
if krc[ir * 8 + ic] == "Q":
return False, krc
else:
krc[ir * 8 + ic] = "X"
return True, krc
def chki(i, krc):
krc = [""] * 64
for j in range(8):
krc[j * 8 + i[j]] = "Q"
bl, krc = setq(j, i[j], krc)
# print(j,bl)
# prk(krc)
if not bl:
# print(bl,i,"-------------------")
# prk(krc)
return False
# print(bl,i,"-------------------")
# prk(krc)
return True
def prk(krc2):
for i in range(8):
print(krc2[i * 8 : (i + 1) * 8])
icase = 0
kr = [[] for i in range(8)]
if icase == 0:
k = int(input())
krc = [""] * 64
for i in range(k):
ri, ci = map(int, input().split())
krc[ri * 8 + ci] = "Q"
setq(ri, ci, krc)
kr[ri].append(ci)
elif icase == 1:
k = 2
krc = [""] * 64
ri, ci = 2, 2
krc[ri * 8 + ci] = "Q"
setq(ri, ci, krc)
kr[ri].append(ci)
ri, ci = 5, 3
krc[ri * 8 + ci] = "Q"
setq(ri, ci, krc)
kr[ri].append(ci)
# prk(krc)
icnt = 0
for ir in range(8):
for ic in range(8):
i = ir * 8 + ic
if krc[i] == "":
kr[ir].append(ic)
for i in product(kr[0], kr[1], kr[2], kr[3], kr[4], kr[5], kr[6], kr[7]):
yn = ""
krc2 = [""] * 64
if not chki(i, krc2):
yn = "no"
continue
else:
break
if yn != "no":
for ir in range(8):
sti = ["."] * 8
sti[i[ir]] = "Q"
stii = "".join(sti)
print(stii)
| 8 Queens Problem
The goal of 8 Queens Problem is to put eight queens on a chess-board such that
none of them threatens any of others. A queen threatens the squares in the
same row, in the same column, or on the same diagonals as shown in the
following figure.
For a given chess board where $k$ queens are already placed, find the solution
of the 8 queens problem. | [{"input": "2\n 2 2\n 5 3", "output": "......Q.\n Q.......\n ..Q.....\n .......Q\n .....Q..\n ...Q....\n .Q......\n ....Q..."}] |
Print a $8 \times 8$ chess board by strings where a square with a queen is
represented by 'Q' and an empty square is represented by '.'. | s485416083 | Accepted | p02244 | In the first line, an integer $k$ is given. In the following $k$ lines, each
square where a queen is already placed is given by two integers $r$ and $c$.
$r$ and $c$ respectively denotes the row number and the column number. The
row/column numbers start with 0. | import copy as cp
N = 8
ans = []
def out(M):
for i in range(N):
for j in range(N):
print(M[i][j], end="")
print("")
def DFS(r, c, dp, dn, qn):
global ans
if qn == N:
for h in range(N):
for w in range(N):
if r[h] & c[w] & dp[h + w] & dn[w - h + 7]:
ans.append([h, w])
return True
return False
for h in range(N):
for w in range(N):
if r[h] & c[w] & dp[h + w] & dn[w - h + 7]:
new_r = cp.deepcopy(r)
new_c = cp.deepcopy(c)
new_dp = cp.deepcopy(dp)
new_dn = cp.deepcopy(dn)
new_r[h], new_c[w], new_dp[h + w], new_dn[w - h + 7] = (
False,
False,
False,
False,
)
if DFS(new_r, new_c, new_dp, new_dn, qn + 1):
ans.append([h, w])
return True
return False
def main():
row = [True] * N
col = [True] * N
dp = [True] * (2 * N - 1)
dn = [True] * (2 * N - 1)
k = int(input())
MAP = [["." for _ in range(N)] for __ in range(N)]
for _ in range(k):
h, w = map(int, input().split())
MAP[h][w] = "Q"
row[h] = False
col[w] = False
dp[w + h] = False
dn[w - h + 7] = False
DFS(row, col, dp, dn, k + 1)
for h, w in ans:
MAP[h][w] = "Q"
out(MAP)
main()
| 8 Queens Problem
The goal of 8 Queens Problem is to put eight queens on a chess-board such that
none of them threatens any of others. A queen threatens the squares in the
same row, in the same column, or on the same diagonals as shown in the
following figure.
For a given chess board where $k$ queens are already placed, find the solution
of the 8 queens problem. | [{"input": "2\n 2 2\n 5 3", "output": "......Q.\n Q.......\n ..Q.....\n .......Q\n .....Q..\n ...Q....\n .Q......\n ....Q..."}] |
Print a $8 \times 8$ chess board by strings where a square with a queen is
represented by 'Q' and an empty square is represented by '.'. | s787570676 | Accepted | p02244 | In the first line, an integer $k$ is given. In the following $k$ lines, each
square where a queen is already placed is given by two integers $r$ and $c$.
$r$ and $c$ respectively denotes the row number and the column number. The
row/column numbers start with 0. | from itertools import permutations
import sys
read = sys.stdin.read
readline = sys.stdin.readline
readlines = sys.stdin.readlines
def main():
k, *rc = map(int, read().split())
tmp = []
ng_list_add = []
ng_list_sub = []
for pattern in permutations(range(8)):
flag = True
for r, c in enumerate(pattern):
if r + c in ng_list_add or r - c in ng_list_sub:
flag = False
break
ng_list_add.append(r + c)
ng_list_sub.append(r - c)
if flag:
tmp.append(pattern)
ng_list_add = []
ng_list_sub = []
RC = []
for r, c in zip(*[iter(rc)] * 2):
RC.append((r, c))
ans_place = None
for p in tmp:
flag = True
for r, c in RC:
if p[r] != c:
flag = False
break
if flag:
ans_place = p
break
ans = []
for i in ans_place:
line = "." * i + "Q" + "." * (7 - i)
ans.append(line)
print("\n".join(ans))
if __name__ == "__main__":
main()
| 8 Queens Problem
The goal of 8 Queens Problem is to put eight queens on a chess-board such that
none of them threatens any of others. A queen threatens the squares in the
same row, in the same column, or on the same diagonals as shown in the
following figure.
For a given chess board where $k$ queens are already placed, find the solution
of the 8 queens problem. | [{"input": "2\n 2 2\n 5 3", "output": "......Q.\n Q.......\n ..Q.....\n .......Q\n .....Q..\n ...Q....\n .Q......\n ....Q..."}] |
Print a $8 \times 8$ chess board by strings where a square with a queen is
represented by 'Q' and an empty square is represented by '.'. | s176912100 | Accepted | p02244 | In the first line, an integer $k$ is given. In the following $k$ lines, each
square where a queen is already placed is given by two integers $r$ and $c$.
$r$ and $c$ respectively denotes the row number and the column number. The
row/column numbers start with 0. | from itertools import permutations
k = int(input())
points = {tuple(map(int, input().split())) for _ in range(k)}
for perm in permutations(range(8), 8):
test = {(i, perm[i]) for i in range(8)}
if test & points == points:
flg = 0
for x, y in test:
diag = (
{(x - i, y - i) for i in range(1, min(x, y) + 1)}
| {(x + i, y + i) for i in range(1, min(7 - x, 7 - y) + 1)}
| {(x - i, y + i) for i in range(1, min(x, 7 - y) + 1)}
| {(x - i, y + i) for i in range(1, min(7 - x, y) + 1)}
)
if diag & test == set():
flg += 1
if flg == 8:
ans = test
break
for i in range(8):
col = ""
for j in range(8):
if (i, j) in ans:
col += "Q"
else:
col += "."
print(col)
| 8 Queens Problem
The goal of 8 Queens Problem is to put eight queens on a chess-board such that
none of them threatens any of others. A queen threatens the squares in the
same row, in the same column, or on the same diagonals as shown in the
following figure.
For a given chess board where $k$ queens are already placed, find the solution
of the 8 queens problem. | [{"input": "2\n 2 2\n 5 3", "output": "......Q.\n Q.......\n ..Q.....\n .......Q\n .....Q..\n ...Q....\n .Q......\n ....Q..."}] |
Print a $8 \times 8$ chess board by strings where a square with a queen is
represented by 'Q' and an empty square is represented by '.'. | s823256348 | Accepted | p02244 | In the first line, an integer $k$ is given. In the following $k$ lines, each
square where a queen is already placed is given by two integers $r$ and $c$.
$r$ and $c$ respectively denotes the row number and the column number. The
row/column numbers start with 0. | import itertools
import copy
def is_safe(qweens):
x = []
y = []
for q in qweens:
x.append(q[0])
y.append(q[1])
if len(set(x)) + len(set(y)) != len(qweens) * 2:
return False
for q1, q2 in itertools.combinations(qweens, 2):
slope = (q1[0] - q2[0]) / (q1[1] - q2[1])
if slope == 1 or slope == -1:
return False
return True
def print_boad(qweens):
for r in range(8):
for c in range(8):
if (r, c) in qweens:
print("Q", end="")
else:
print(".", end="")
print("")
k = int(input())
cand = [[(r, c) for c in range(8)] for r in range(8)]
init_qweens = []
for _ in range(k):
pos = tuple(map(int, input().split()))
for r in range(8):
for c in range(8):
try:
cand[r].remove((pos[0], c))
except ValueError:
pass
try:
cand[r].remove((r, pos[1]))
except ValueError:
pass
init_qweens.append(pos)
for r in range(7, -1, -1):
if cand[r] == []:
cand.pop(r)
qweens = []
for per in itertools.permutations([i for i in range(8 - k)]):
qweens = copy.deepcopy(init_qweens)
for r, c in enumerate(per):
qweens.append(cand[r][c])
if is_safe(qweens):
break
print_boad(qweens)
| 8 Queens Problem
The goal of 8 Queens Problem is to put eight queens on a chess-board such that
none of them threatens any of others. A queen threatens the squares in the
same row, in the same column, or on the same diagonals as shown in the
following figure.
For a given chess board where $k$ queens are already placed, find the solution
of the 8 queens problem. | [{"input": "2\n 2 2\n 5 3", "output": "......Q.\n Q.......\n ..Q.....\n .......Q\n .....Q..\n ...Q....\n .Q......\n ....Q..."}] |
Print a $8 \times 8$ chess board by strings where a square with a queen is
represented by 'Q' and an empty square is represented by '.'. | s092449153 | Accepted | p02244 | In the first line, an integer $k$ is given. In the following $k$ lines, each
square where a queen is already placed is given by two integers $r$ and $c$.
$r$ and $c$ respectively denotes the row number and the column number. The
row/column numbers start with 0. | import sys
import copy
k = int(input())
class Chessboard:
def __init__(self):
self.row = [True] * 8
self.col = [True] * 8
self.dpos = [True] * 15
self.dneg = [True] * 15
return
def put(self, r, c):
# queen が r, c に配置されたときに攻撃されるマスの記録
self.row[r] = False
self.col[c] = False
self.dpos[r + c] = False # 左下が攻撃される
self.dneg[r + 7 - c] = False # 右下が攻撃される
return
def safe(self, r, c):
# r, c が攻撃されていないか確認する 攻撃されていないときTrue を返す
return self.row[r] and self.col[c] and self.dpos[r + c] and self.dneg[r + 7 - c]
def is_row(self, r):
return self.row[r]
board = Chessboard()
ans = [["."] * 8 for _ in range(8)]
for _ in range(k):
r, c = map(int, sys.stdin.readline().strip().split())
board.put(r, c)
ans[r][c] = "Q"
final_ans = None
def search(r=0, board=board, ans=ans):
global final_ans
if r == 8: # 最終行を終了し探索する行が存在しないときに終了する
final_ans = ans
return ans
if not board.is_row(r):
# すでに Q が配置されている行の探索はスキップする
search(r=r + 1, board=board, ans=ans)
for c in range(8):
board_temp = copy.deepcopy(board)
ans_temp = copy.deepcopy(ans)
if board_temp.safe(r, c):
board_temp.put(r, c)
ans_temp[r][c] = "Q"
search(r=r + 1, board=board_temp, ans=ans_temp)
else:
continue
ans = search()
for r_ans in final_ans:
print("".join(r_ans))
| 8 Queens Problem
The goal of 8 Queens Problem is to put eight queens on a chess-board such that
none of them threatens any of others. A queen threatens the squares in the
same row, in the same column, or on the same diagonals as shown in the
following figure.
For a given chess board where $k$ queens are already placed, find the solution
of the 8 queens problem. | [{"input": "2\n 2 2\n 5 3", "output": "......Q.\n Q.......\n ..Q.....\n .......Q\n .....Q..\n ...Q....\n .Q......\n ....Q..."}] |
Print a $8 \times 8$ chess board by strings where a square with a queen is
represented by 'Q' and an empty square is represented by '.'. | s860057962 | Accepted | p02244 | In the first line, an integer $k$ is given. In the following $k$ lines, each
square where a queen is already placed is given by two integers $r$ and $c$.
$r$ and $c$ respectively denotes the row number and the column number. The
row/column numbers start with 0. | def try_set(v):
def print_board():
for r, c in preset:
if queen_pos[r] != c:
return
for pos in queen_pos:
for c in range(8):
print("Q" if c == pos else ".", end="")
print()
for h in range(8):
if all((horizontal[h], dip_positive[v + h], dip_negative[v - h + 7])):
queen_pos[v] = h
if v == 7:
print_board()
else:
horizontal[h] = dip_positive[v + h] = dip_negative[v - h + 7] = False
try_set(v + 1)
horizontal[h] = dip_positive[v + h] = dip_negative[v - h + 7] = True
n = int(input())
queen_pos = [None] * 8
preset = [[pre for pre in map(int, input().split())] for _ in range(n)]
horizontal = [True] * 8
dip_positive = [True] * 15
dip_negative = [True] * 15
try_set(0)
| 8 Queens Problem
The goal of 8 Queens Problem is to put eight queens on a chess-board such that
none of them threatens any of others. A queen threatens the squares in the
same row, in the same column, or on the same diagonals as shown in the
following figure.
For a given chess board where $k$ queens are already placed, find the solution
of the 8 queens problem. | [{"input": "2\n 2 2\n 5 3", "output": "......Q.\n Q.......\n ..Q.....\n .......Q\n .....Q..\n ...Q....\n .Q......\n ....Q..."}] |
Print the maximum possible difference in the number of balls received.
* * * | s750187314 | Wrong Answer | p03005 | Input is given from Standard Input in the following format:
N K | A, B = map(int, input().split())
print(A - B)
| Statement
Takahashi is distributing N balls to K persons.
If each person has to receive at least one ball, what is the maximum possible
difference in the number of balls received between the person with the most
balls and the person with the fewest balls? | [{"input": "3 2", "output": "1\n \n\nThe only way to distribute three balls to two persons so that each of them\nreceives at least one ball is to give one ball to one person and give two\nballs to the other person.\n\nThus, the maximum possible difference in the number of balls received is 1.\n\n* * *"}, {"input": "3 1", "output": "0\n \n\nWe have no choice but to give three balls to the only person, in which case\nthe difference in the number of balls received is 0.\n\n* * *"}, {"input": "8 5", "output": "3\n \n\nFor example, if we give 1, 4, 1, 1, 1 balls to the five persons, the number of\nballs received between the person with the most balls and the person with the\nfewest balls would be 3, which is the maximum result."}] |
Print the maximum possible difference in the number of balls received.
* * * | s837826304 | Accepted | p03005 | Input is given from Standard Input in the following format:
N K | import sys
sys.setrecursionlimit(10000)
mod = 1000000007
def mul(a, b):
return ((a % mod) * (b % mod)) % mod
def power(x, y):
if y == 0:
return 1
elif y == 1:
return x % mod
elif y % 2 == 0:
return power(x, y / 2) ** 2 % mod
else:
return power(x, y / 2) ** 2 * x % mod
def div(a, b):
return mul(a, power(b, mod - 2))
# def gcd(a_gcd, b_gcd):
# while b_gcd != 0:
# a_gcd, b_gcd = b_gcd, a_gcd % b_gcd
# return a_gcd
#
#
# def lcm(a_lcm, b_lcm):
# return a_lcm * b // gcd(a_lcm, b_lcm)
#
#
# def max_sum(N_max_sum, a_max_sum):
# dp = [0] * (N_max_sum + 1)
# for i in range(N_max_sum):
# dp[i + 1] = max(dp[i], dp[i] + a_max_sum[i])
# return dp[N]
#
#
# def knapsack(N_knapsack, W, weight, value):
# inf = float("inf")
# dp = [[-inf for _ in range(W+1)] for j in range(N_knapsack + 1)]
# for i in range(W+1):
# dp[0][i] = 0
#
# for i in range(N_knapsack):
# for w in range(W+1):
# if weight[i] <= w:
# dp[i+1][w] = max(dp[i][w-weight[i]]+value[i], dp[i][w])
# else:
# dp[i+1][w] = dp[i][w]
# return dp[N_knapsack][W]
#
#
# class UnionFind:
# def __init__(self, n):
# self.par = [i for i in range(n+1)]
# self.rank = [0] * (n+1)
#
# # 検索
# def find(self, x):
# if self.par[x] == x:
# return x
# else:
# self.par[x] = self.find(self.par[x])
# return self.par[x]
#
# # 併合
# def union(self, x, y):
# x = self.find(x)
# y = self.find(y)
# if self.rank[x] < self.rank[y]:
# self.par[x] = y
# else:
# self.par[y] = x
# if self.rank[x] == self.rank[y]:
# self.rank[x] += 1
#
# # 同じ集合に属するか判定
# def same_check(self, x, y):
# return self.find(x) == self.find(y)
ans = 0
N, K = map(int, input().split())
if K == 1:
print("0")
else:
print(N - K)
| Statement
Takahashi is distributing N balls to K persons.
If each person has to receive at least one ball, what is the maximum possible
difference in the number of balls received between the person with the most
balls and the person with the fewest balls? | [{"input": "3 2", "output": "1\n \n\nThe only way to distribute three balls to two persons so that each of them\nreceives at least one ball is to give one ball to one person and give two\nballs to the other person.\n\nThus, the maximum possible difference in the number of balls received is 1.\n\n* * *"}, {"input": "3 1", "output": "0\n \n\nWe have no choice but to give three balls to the only person, in which case\nthe difference in the number of balls received is 0.\n\n* * *"}, {"input": "8 5", "output": "3\n \n\nFor example, if we give 1, 4, 1, 1, 1 balls to the five persons, the number of\nballs received between the person with the most balls and the person with the\nfewest balls would be 3, which is the maximum result."}] |
Print the maximum possible difference in the number of balls received.
* * * | s160819520 | Runtime Error | p03005 | Input is given from Standard Input in the following format:
N K | N = int(input())
A = [int(i) for i in input().split()]
A = sorted(A)
removeNums = []
for i in range(len(A) - 1):
newValue = A[0] - A[-1]
removeNums.append((A[0], A[-1]))
A.remove(A[0])
A.remove(A[-1])
A.insert(0, newValue)
print(-1 * A[0])
for i in range(len(removeNums) - 1):
print(removeNums[i][0], removeNums[i][1])
print(removeNums[-1][1], removeNums[-1][0])
| Statement
Takahashi is distributing N balls to K persons.
If each person has to receive at least one ball, what is the maximum possible
difference in the number of balls received between the person with the most
balls and the person with the fewest balls? | [{"input": "3 2", "output": "1\n \n\nThe only way to distribute three balls to two persons so that each of them\nreceives at least one ball is to give one ball to one person and give two\nballs to the other person.\n\nThus, the maximum possible difference in the number of balls received is 1.\n\n* * *"}, {"input": "3 1", "output": "0\n \n\nWe have no choice but to give three balls to the only person, in which case\nthe difference in the number of balls received is 0.\n\n* * *"}, {"input": "8 5", "output": "3\n \n\nFor example, if we give 1, 4, 1, 1, 1 balls to the five persons, the number of\nballs received between the person with the most balls and the person with the\nfewest balls would be 3, which is the maximum result."}] |
Print the maximum possible difference in the number of balls received.
* * * | s762230038 | Runtime Error | p03005 | Input is given from Standard Input in the following format:
N K | #!/usr/bin/env python
N = int(input())
A = [int(i) for i in input().split()]
if N == 2:
print(max(A) - min(A))
print(max(A), min(A))
else:
A.sort()
As = []
t = A[0]
for a in A[1:-1]:
As.append([t, a])
t = t - a
ans = A[-1] - t
As.append([A[-1], t])
print(ans)
for a in As:
print(*a)
| Statement
Takahashi is distributing N balls to K persons.
If each person has to receive at least one ball, what is the maximum possible
difference in the number of balls received between the person with the most
balls and the person with the fewest balls? | [{"input": "3 2", "output": "1\n \n\nThe only way to distribute three balls to two persons so that each of them\nreceives at least one ball is to give one ball to one person and give two\nballs to the other person.\n\nThus, the maximum possible difference in the number of balls received is 1.\n\n* * *"}, {"input": "3 1", "output": "0\n \n\nWe have no choice but to give three balls to the only person, in which case\nthe difference in the number of balls received is 0.\n\n* * *"}, {"input": "8 5", "output": "3\n \n\nFor example, if we give 1, 4, 1, 1, 1 balls to the five persons, the number of\nballs received between the person with the most balls and the person with the\nfewest balls would be 3, which is the maximum result."}] |
Print the maximum possible difference in the number of balls received.
* * * | s495684815 | Runtime Error | p03005 | Input is given from Standard Input in the following format:
N K | import fractions
N = int(input())
sheet = []
for i in range(N):
line = input().split()
sheet += [(int(line[0]), int(line[1]))]
dic = {}
for i in range(N):
start = sheet[i]
for j in range(i + 1, N):
vec = [(sheet[j][0] - start[0]), (sheet[j][1] - start[1])]
size = fractions.gcd(vec[0], vec[1])
vec[0] = vec[0] // size
vec[1] = vec[1] // size
vec = tuple(vec)
if vec not in dic.keys():
dic[vec] = set([j])
else:
dic[vec].add(j)
m = 1
for s in dic.values():
if len(s) > m:
m = len(s)
print(N - m)
| Statement
Takahashi is distributing N balls to K persons.
If each person has to receive at least one ball, what is the maximum possible
difference in the number of balls received between the person with the most
balls and the person with the fewest balls? | [{"input": "3 2", "output": "1\n \n\nThe only way to distribute three balls to two persons so that each of them\nreceives at least one ball is to give one ball to one person and give two\nballs to the other person.\n\nThus, the maximum possible difference in the number of balls received is 1.\n\n* * *"}, {"input": "3 1", "output": "0\n \n\nWe have no choice but to give three balls to the only person, in which case\nthe difference in the number of balls received is 0.\n\n* * *"}, {"input": "8 5", "output": "3\n \n\nFor example, if we give 1, 4, 1, 1, 1 balls to the five persons, the number of\nballs received between the person with the most balls and the person with the\nfewest balls would be 3, which is the maximum result."}] |
Print the maximum possible difference in the number of balls received.
* * * | s543128438 | Accepted | p03005 | Input is given from Standard Input in the following format:
N K | import sys
sys.setrecursionlimit(10**7)
INF = 10**18
MOD = 10**9 + 7
def YesNo(x):
return "Yes" if x else "No"
def LI():
return [int(x) for x in sys.stdin.readline().split()]
def LI_():
return [int(x) - 1 for x in sys.stdin.readline().split()]
def LF():
return [float(x) for x in sys.stdin.readline().split()]
def LS():
return sys.stdin.readline().split()
def II():
return int(sys.stdin.readline())
def SI():
return input()
def main():
N, K = LI()
ans = (N - K) if K > 1 else 0
return ans
print(main())
| Statement
Takahashi is distributing N balls to K persons.
If each person has to receive at least one ball, what is the maximum possible
difference in the number of balls received between the person with the most
balls and the person with the fewest balls? | [{"input": "3 2", "output": "1\n \n\nThe only way to distribute three balls to two persons so that each of them\nreceives at least one ball is to give one ball to one person and give two\nballs to the other person.\n\nThus, the maximum possible difference in the number of balls received is 1.\n\n* * *"}, {"input": "3 1", "output": "0\n \n\nWe have no choice but to give three balls to the only person, in which case\nthe difference in the number of balls received is 0.\n\n* * *"}, {"input": "8 5", "output": "3\n \n\nFor example, if we give 1, 4, 1, 1, 1 balls to the five persons, the number of\nballs received between the person with the most balls and the person with the\nfewest balls would be 3, which is the maximum result."}] |
Print the maximum possible difference in the number of balls received.
* * * | s160635238 | Wrong Answer | p03005 | Input is given from Standard Input in the following format:
N K | n, k = input().split(" ")
n, k = int(n), int(k)
maxval = k - n
minval = k / n
print(maxval - minval)
| Statement
Takahashi is distributing N balls to K persons.
If each person has to receive at least one ball, what is the maximum possible
difference in the number of balls received between the person with the most
balls and the person with the fewest balls? | [{"input": "3 2", "output": "1\n \n\nThe only way to distribute three balls to two persons so that each of them\nreceives at least one ball is to give one ball to one person and give two\nballs to the other person.\n\nThus, the maximum possible difference in the number of balls received is 1.\n\n* * *"}, {"input": "3 1", "output": "0\n \n\nWe have no choice but to give three balls to the only person, in which case\nthe difference in the number of balls received is 0.\n\n* * *"}, {"input": "8 5", "output": "3\n \n\nFor example, if we give 1, 4, 1, 1, 1 balls to the five persons, the number of\nballs received between the person with the most balls and the person with the\nfewest balls would be 3, which is the maximum result."}] |
Print the maximum possible difference in the number of balls received.
* * * | s106827333 | Wrong Answer | p03005 | Input is given from Standard Input in the following format:
N K | nb_balls, nb_person = [int(x) for x in input().split()]
mins = 1
maxs = (nb_balls - nb_person) + 1
print(maxs - mins)
| Statement
Takahashi is distributing N balls to K persons.
If each person has to receive at least one ball, what is the maximum possible
difference in the number of balls received between the person with the most
balls and the person with the fewest balls? | [{"input": "3 2", "output": "1\n \n\nThe only way to distribute three balls to two persons so that each of them\nreceives at least one ball is to give one ball to one person and give two\nballs to the other person.\n\nThus, the maximum possible difference in the number of balls received is 1.\n\n* * *"}, {"input": "3 1", "output": "0\n \n\nWe have no choice but to give three balls to the only person, in which case\nthe difference in the number of balls received is 0.\n\n* * *"}, {"input": "8 5", "output": "3\n \n\nFor example, if we give 1, 4, 1, 1, 1 balls to the five persons, the number of\nballs received between the person with the most balls and the person with the\nfewest balls would be 3, which is the maximum result."}] |
Print the maximum possible difference in the number of balls received.
* * * | s402846237 | Accepted | p03005 | Input is given from Standard Input in the following format:
N K | # !/usr/bin/env python3
# encoding: UTF-8
# Modified: <15/Jun/2019 05:31:28 PM>
# ✪ H4WK3yE乡
# Mohd. Farhan Tahir
# Indian Institute Of Information Technology (IIIT), Gwalior
import sys
import os
from io import IOBase, BytesIO
def main():
n, k = get_ints()
if k == 1:
print(0)
return
left = n - k
print(left)
BUFSIZE = 8192
class FastIO(BytesIO):
newlines = 0
def __init__(self, file):
self._file = file
self._fd = file.fileno()
self.writable = "x" in file.mode or "w" in file.mode
self.write = super(FastIO, self).write if self.writable else None
def _fill(self):
s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.seek((self.tell(), self.seek(0, 2), super(FastIO, self).write(s))[0])
return s
def read(self):
while self._fill():
pass
return super(FastIO, self).read()
def readline(self):
while self.newlines == 0:
s = self._fill()
self.newlines = s.count(b"\n") + (not s)
self.newlines -= 1
return super(FastIO, self).readline()
def flush(self):
if self.writable:
os.write(self._fd, self.getvalue())
self.truncate(0), self.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
py2 = round(0.5)
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
if py2:
self.write = self.buffer.write
self.read = self.buffer.read
self.readline = self.buffer.readline
else:
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def get_array():
return list(map(int, sys.stdin.readline().split()))
def get_ints():
return map(int, sys.stdin.readline().split())
def input():
return sys.stdin.readline().strip()
if __name__ == "__main__":
main()
| Statement
Takahashi is distributing N balls to K persons.
If each person has to receive at least one ball, what is the maximum possible
difference in the number of balls received between the person with the most
balls and the person with the fewest balls? | [{"input": "3 2", "output": "1\n \n\nThe only way to distribute three balls to two persons so that each of them\nreceives at least one ball is to give one ball to one person and give two\nballs to the other person.\n\nThus, the maximum possible difference in the number of balls received is 1.\n\n* * *"}, {"input": "3 1", "output": "0\n \n\nWe have no choice but to give three balls to the only person, in which case\nthe difference in the number of balls received is 0.\n\n* * *"}, {"input": "8 5", "output": "3\n \n\nFor example, if we give 1, 4, 1, 1, 1 balls to the five persons, the number of\nballs received between the person with the most balls and the person with the\nfewest balls would be 3, which is the maximum result."}] |
Print the maximum possible difference in the number of balls received.
* * * | s591147424 | Runtime Error | p03005 | Input is given from Standard Input in the following format:
N K | from collections import deque
plusQ = deque([])
minsQ = deque([])
N = int(input())
A = list(map(int, input().split()))
A.sort(reverse=True)
for itr, val in enumerate(A):
if itr + 1 <= (len(A) // 2) and val > 0:
plusQ.append(val)
else:
minsQ.append(val)
p1 = plusQ.popleft()
ans = deque([])
while len(minsQ) > 0:
if len(plusQ) > 0:
p, m = plusQ.popleft(), minsQ.popleft()
ans.append([m, p])
minsQ.append(m - p)
else:
m = minsQ.popleft()
ans.append([p1, m])
p1 = p1 - m
print(p1)
while len(ans) > 0:
hoge = ans.popleft()
print(hoge[0], hoge[1])
| Statement
Takahashi is distributing N balls to K persons.
If each person has to receive at least one ball, what is the maximum possible
difference in the number of balls received between the person with the most
balls and the person with the fewest balls? | [{"input": "3 2", "output": "1\n \n\nThe only way to distribute three balls to two persons so that each of them\nreceives at least one ball is to give one ball to one person and give two\nballs to the other person.\n\nThus, the maximum possible difference in the number of balls received is 1.\n\n* * *"}, {"input": "3 1", "output": "0\n \n\nWe have no choice but to give three balls to the only person, in which case\nthe difference in the number of balls received is 0.\n\n* * *"}, {"input": "8 5", "output": "3\n \n\nFor example, if we give 1, 4, 1, 1, 1 balls to the five persons, the number of\nballs received between the person with the most balls and the person with the\nfewest balls would be 3, which is the maximum result."}] |
Print the maximum possible difference in the number of balls received.
* * * | s220312784 | Runtime Error | p03005 | Input is given from Standard Input in the following format:
N K | import sys
input = sys.stdin.readline
N = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
dp = [0] * (N + 1)
for i in range(N, -1, -1):
for j in range(3):
x = a[j]
y = b[j]
if i - x >= 0:
dp[i - x] = max(dp[i - x], dp[i] + y)
for i in range(N + 1):
dp[i] += i
# print(dp)
t = max(dp)
e = []
def check(x):
tt = t - b[e[0]] * x
return (tt // b[e[1]]) * a[e[1]] + tt % b[e[1]] + a[e[0]] * x
if t > N:
for i in range(3):
if b[i] < a[i]:
e.append(i)
if len(e) == 0:
print(t)
elif len(e) == 1:
print(a[e[0]] * (t // b[e[0]]) + t % b[e[0]])
else:
if b[e[0]] < b[e[1]]:
e[0], e[1] = e[1], e[0]
x = e[0]
res = 0
for i in range(t // b[x] + 1):
res = max(res, check(i))
print(res)
else:
dp = [0] * (N + 1)
for i in range(N, -1, -1):
for j in range(3):
y = a[j]
x = b[j]
if i - x >= 0:
dp[i - x] = max(dp[i - x], dp[i] + y)
for i in range(N + 1):
dp[i] += i
print(max(dp))
| Statement
Takahashi is distributing N balls to K persons.
If each person has to receive at least one ball, what is the maximum possible
difference in the number of balls received between the person with the most
balls and the person with the fewest balls? | [{"input": "3 2", "output": "1\n \n\nThe only way to distribute three balls to two persons so that each of them\nreceives at least one ball is to give one ball to one person and give two\nballs to the other person.\n\nThus, the maximum possible difference in the number of balls received is 1.\n\n* * *"}, {"input": "3 1", "output": "0\n \n\nWe have no choice but to give three balls to the only person, in which case\nthe difference in the number of balls received is 0.\n\n* * *"}, {"input": "8 5", "output": "3\n \n\nFor example, if we give 1, 4, 1, 1, 1 balls to the five persons, the number of\nballs received between the person with the most balls and the person with the\nfewest balls would be 3, which is the maximum result."}] |
Print the maximum possible difference in the number of balls received.
* * * | s247735950 | Runtime Error | p03005 | Input is given from Standard Input in the following format:
N K | from itertools import combinations
from collections import Counter
def main():
n = int(input())
if n == 1:
print(1)
XY = [list(map(int, input().split())) for _ in range(n)]
d = []
for A, B in combinations(XY, 2):
x1, y1 = A[0], A[1]
x2, y2 = B[0], B[1]
p, q = x2 - x1, y2 - y1
d.append((p, q))
d.append((-p, -q))
# print(d)
# print(Counter(d))
C = Counter(d)
ans = 0
for v in C.values():
ans = max(ans, v)
print(n - ans)
if __name__ == "__main__":
main()
| Statement
Takahashi is distributing N balls to K persons.
If each person has to receive at least one ball, what is the maximum possible
difference in the number of balls received between the person with the most
balls and the person with the fewest balls? | [{"input": "3 2", "output": "1\n \n\nThe only way to distribute three balls to two persons so that each of them\nreceives at least one ball is to give one ball to one person and give two\nballs to the other person.\n\nThus, the maximum possible difference in the number of balls received is 1.\n\n* * *"}, {"input": "3 1", "output": "0\n \n\nWe have no choice but to give three balls to the only person, in which case\nthe difference in the number of balls received is 0.\n\n* * *"}, {"input": "8 5", "output": "3\n \n\nFor example, if we give 1, 4, 1, 1, 1 balls to the five persons, the number of\nballs received between the person with the most balls and the person with the\nfewest balls would be 3, which is the maximum result."}] |
Print the maximum possible difference in the number of balls received.
* * * | s551158798 | Runtime Error | p03005 | Input is given from Standard Input in the following format:
N K | N = int(input())
A = list(map(int, input().split()))
A.sort()
if A[1] >= 0:
print(sum(A) - 2 * A[0])
y = A[0]
for i in range(N - 2):
print(y, A[i + 1])
y -= A[i + 1]
print(A[-1], y)
elif A[-1] < 0:
print(-sum(A) + 2 * A[-1])
y = A[-1]
for i in range(N - 1):
print(y, A[i])
y -= A[i]
else:
for i in range(N - 1):
if A[i] < 0 and A[i + 1] >= 0:
n = i
print(-sum(A[:i]) + sum(A[i:]))
y = A[0]
for i in range(N - n - 2):
print(y, A[-i - 2])
y -= A[-i - 2]
print(A[-1], y)
y = A[-1] - y
for i in range(n):
print(y, A[i + 1])
| Statement
Takahashi is distributing N balls to K persons.
If each person has to receive at least one ball, what is the maximum possible
difference in the number of balls received between the person with the most
balls and the person with the fewest balls? | [{"input": "3 2", "output": "1\n \n\nThe only way to distribute three balls to two persons so that each of them\nreceives at least one ball is to give one ball to one person and give two\nballs to the other person.\n\nThus, the maximum possible difference in the number of balls received is 1.\n\n* * *"}, {"input": "3 1", "output": "0\n \n\nWe have no choice but to give three balls to the only person, in which case\nthe difference in the number of balls received is 0.\n\n* * *"}, {"input": "8 5", "output": "3\n \n\nFor example, if we give 1, 4, 1, 1, 1 balls to the five persons, the number of\nballs received between the person with the most balls and the person with the\nfewest balls would be 3, which is the maximum result."}] |
Print the maximum possible difference in the number of balls received.
* * * | s007980802 | Runtime Error | p03005 | Input is given from Standard Input in the following format:
N K | N = int(input())
XY = [list(map(int, input().split())) for _ in range(N)]
def gdb(a, b):
a, b = abs(a), abs(b)
while b:
a, b = b, a % b
return a
def make_slope(a, b):
if a == 0:
print("error")
elif a > 0:
return (a // gdb(a, b), b // gdb(a, b))
elif a < 0:
return (-1 * a // gdb(a, b), -1 * b // gdb(a, b))
slope_list = []
for i in range(1, N):
for j in range(i):
x1, y1 = XY[i]
x2, y2 = XY[j]
d_x = x2 - x1
d_y = y2 - y1
if d_x != 0:
d_x, d_y = make_slope(d_x, d_y)
if not ([d_x, d_y] in slope_list):
slope_list.append([d_x, d_y])
if N <= 2:
print(1)
else:
tmp = []
for x, y in XY:
if not (x in tmp):
tmp.append(x)
ans = len(tmp)
for d_x, d_y in slope_list:
XY_list = []
for x1, y1 in XY:
if XY_list == []:
XY_list.append([x1, y1])
else:
for x2, y2 in XY_list:
d_x_tmp = x2 - x1
d_y_tmp = y2 - y1
if d_x_tmp == 0:
break
d_x_tmp, d_y_tmp = make_slope(d_x_tmp, d_y_tmp)
if d_x_tmp == d_x and d_y_tmp == d_y:
break
else:
XY_list.append([x1, y1])
ans = min(ans, len(XY_list))
print(ans)
| Statement
Takahashi is distributing N balls to K persons.
If each person has to receive at least one ball, what is the maximum possible
difference in the number of balls received between the person with the most
balls and the person with the fewest balls? | [{"input": "3 2", "output": "1\n \n\nThe only way to distribute three balls to two persons so that each of them\nreceives at least one ball is to give one ball to one person and give two\nballs to the other person.\n\nThus, the maximum possible difference in the number of balls received is 1.\n\n* * *"}, {"input": "3 1", "output": "0\n \n\nWe have no choice but to give three balls to the only person, in which case\nthe difference in the number of balls received is 0.\n\n* * *"}, {"input": "8 5", "output": "3\n \n\nFor example, if we give 1, 4, 1, 1, 1 balls to the five persons, the number of\nballs received between the person with the most balls and the person with the\nfewest balls would be 3, which is the maximum result."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s882234532 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a=int(input()), b=int(input())
c=a*b
c=int(c)
if c%2==0:
print("Even")
if else:
print("Odd") | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s043824721 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a = imput()
b = imput()
c = a * b
if(c%2 == 1):
print("Odd")
elif(c%2 == 0):
print("Even") | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s135653011 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a,b=map(int,input().split())
if(a*b%2==0){
print("Even")
}
elif(a*b%2!=0){
print("Odd")
}
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s598428054 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a = list(map(int,input().split()))
if a[0]%2==0 || a[1]%2==0 :
print("Even")
else:
print("Odd") | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s654055490 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a=int(input()), b=int(input())
c=a*b
if c%2==0:
print("Even")
if else:
print("Odd") | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s440259967 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | n,m = map(int,input().split()
if (n*m)%2 ==0:
print("Even")
else:
print(Odd) | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s944431149 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | # -*- coding: utf-8 -*-
a,b=map(int,input().split())
if a*b%2=0:
print('Even')
else:
print('Odd') | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s508977232 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a, b = map(int, input().split())
if((a*b) % 2 == 0){
print('Even')
} else {
print('Odd')
}
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s838998751 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a,b = map(int,input().split())
if (a * b) / 2 %= 0:
print("Even")
else:
print("Odd")
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s145391802 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a, b = map(int, input().split())
if (a * b) % 2 == 0:
print("Even")
else:
print("Odd") | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s469811126 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a = int(input())
b = int(input())
if a%2 = 0 or b%2 =0:
print("even")
else:
print("odd")
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s005242553 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a, b = map(int, input().split())
mult = a * b
if mult % 2 = 0:
print('Even')
else:
print('Odd')
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s089302510 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a = imput()
b = imput()
c = a * b
if(c%2 = 1):
print("Odd")
elif(c%2 = 0):
print("Even) | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s047537044 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a = imput()
b = imput()
c = a * b
if(c%1 = 1):
print("Odd")
else:
print("Even") | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s120387921 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a, b = input().split()
a = int(a)
b = int(b)
print('Even' if a * b % == 0 else 'Odd') | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s353296080 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | x = input().split()
y = int(x[0])*int(x[1])
if y%2 = 1:
print("Odd")
else:
print("Even") | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s370144609 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a = int(input())
b = int(input())
x=a*b
if X%2==0;
print("Even")
else;
print("Odd) | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s064186167 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a=int(input()), b=int(input())
if (a*b)%2==0:
print("Even")
if else:
print("Odd") | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s115918946 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a, b = int (input().split())
c = a*b % 2
if c=1 :
print("Odd")
else:
print("Even")
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s220308447 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a, b = map(intm input().split())
if (a * b) % 2 == 0:
print('Even')
else:
print('Odd')
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s357384103 | Wrong Answer | p03455 | Input is given from Standard Input in the following format:
a b | def first(str):
a, b = map(int, str.split())
print("Even" if a * b % 2 == 0 else "Odd")
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s129333304 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | x = split().input()
y = int(x[0])*int(x[1])
if y%2 = 0:
print(Even)
else:
print(Odd)
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s290644928 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a,b = map(int, input().split())
if a*b // 2 == 0:
print('Even')
else:
print('Odd') | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s681404063 | Accepted | p03455 | Input is given from Standard Input in the following format:
a b | print("Odd" if eval(input().replace(" ", "*")) % 2 else "Even")
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s138847512 | Accepted | p03455 | Input is given from Standard Input in the following format:
a b | print("Odd" if all(map(lambda x: int(x) % 2, input().split())) else "Even")
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s477942422 | Accepted | p03455 | Input is given from Standard Input in the following format:
a b | print(["Even", "Odd"][max(0, sum(map(lambda x: int(x) % 2, input().split())) - 1)])
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s961636997 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a,b=map(int,input().split())
if a*b %2=0:
print("Even")
else:
print("Odd") | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s951885602 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a, b = map(int, input().split())
if (a*b)%2 == 0:
print('Even')
else:
print('Odd')
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s960777144 | Wrong Answer | p03455 | Input is given from Standard Input in the following format:
a b | print(["odd", "even"][eval(input().replace(" ", "*")) % 2 == 0])
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s901491210 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a = 20
b = map(int, input().split())
print("Odd" if a % b == 0) | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s633504216 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | print('OEdvde n'['0'in s:=input()or'4'in s::2]) | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s770232773 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a,b = map(int,input().split())
if a*b%2 = 0:
print("Even")
else:
print("Odd")
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s126557823 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | print('Odd' if eval(input().replace("",*)%2) else 'Even') | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s297119387 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a, b = map(int, input().split())
print("Even" if a*b % 2 == else "Odd") | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s189619853 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | if a*b/2 %= 0:
print("Even")
else:
print("Odd") | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s164624125 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a,b=map(int,input().split())
if (a*b)%2 ==0:
print('Even')
else:
print('Odd') | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s400972728 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | def main();
int a, b, c;
c=a*b;
if c%2==0;
print(Even);
else
print(Odd); | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s582418349 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a, b =map (int, input().split)
if a*b % 2
print("Odd")
else
print("Even") | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s391048731 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | l = input().split()
if l[0]*l[1]%2 ==0:
print(Even)
else:
print(Odd) | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s298837844 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a,b=map(int,input().split())
if a*b%2=0:
print("Even")
else:
print("Odd")
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s340309610 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a, b = input().split()
a = int(a)
b = int(b)
print('Even' if a*b%==0 else 'Odd') | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s768543379 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | print("Even" if int(input()) * int(input()) % 2 == 0 else "Odd")
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s241594171 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a,b=map(int,input().split())
print("Even" if a*b % == 0 else "Odd") | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s123222341 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a, b = map(int, input().split())
print("Even" if a*b%==0 else "Odd") | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s043601798 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | if a*b/2 %= 0:
print("Even")
else:
print("Odd") | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s711056780 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | print("Odd" if input() % 2 != 0 and input() % 2 != 0 else "Even")
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s641581516 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | if a * b % == 0:
print('Even')
else:
print('Odd') | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s866498021 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | print(["Even", "Odd"][max(0, sum(map(lambda x: int(x) % 2, input())) - 1)])
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s120445089 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a, b = map(int, input().split())
print(`Even` if a*b % 2 = 0 else `Odd`) | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s404635346 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a,b=map(int,input().split())
c=a*b
if(c%0==0) print("even")
else print("odd") | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s298956896 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a, b = map(int, input().split())
print("Even" if a*b % 2 == else "Odd") | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s929633766 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a,b=map(int,input().split());print(["Even","Odd"][a*b%2] | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s123631653 | Accepted | p03455 | Input is given from Standard Input in the following format:
a b | # coding: utf-8
# Your code here!
# coding: utf-8
from fractions import gcd
from functools import reduce
import sys
sys.setrecursionlimit(200000000)
from inspect import currentframe
# my functions here!
# 標準エラー出力
def printargs2err(*args):
names = {id(v): k for k, v in currentframe().f_back.f_locals.items()}
print(
", ".join(names.get(id(arg), "???") + " : " + repr(arg) for arg in args),
file=sys.stderr,
)
def debug(x):
print(x, file=sys.stderr)
def printglobals():
for symbol, value in globals().items():
print('symbol="%s"、value=%s' % (symbol, value), file=sys.stderr)
def printlocals():
for symbol, value in locals().items():
print('symbol="%s"、value=%s' % (symbol, value), file=sys.stderr)
# 入力（後でいじる）
def pin(type=int):
return map(type, input().split())
"""
"""
# solution:
# 繰り返し自乗法だよ
# これはnのp乗をmで割ったあまりをだすよ
def modular_w_binary_method(n, m, p):
ans = 1 if n > 0 else 0
while p > 0:
if p % 2 == 1:
ans = (ans * n) % m
n = (n**2) % m
p //= 2
return ans
# mod上でのaの逆元を出すよ
"""
フェルマーの小定理より a**(p)%p==a isTrue
これを利用すれば a**(p-2)%p==a**(-1) isTrue
元と逆元とをかけてpで割れば1が余るよ！
"""
def modular_inverse(a, prime):
return modular_w_binary_method(a, prime, prime - 2)
# 逆元を使うと容易に頑張れるｃ
def conbination(n, a, mod=10**9 + 7):
# cはイラない
res = 1
for s in range(a):
res = (res * (n - s) * (modular_inverse(s + 1, mod))) % mod
return res
# 操作から作れるものの組み合わせは何個かー＞同じものでも違う方法で構成できるかから考える方法もある
# input
def resolve():
a, b = pin()
print(["Odd", "Even"][(a * b + 1) % 2])
return
# print([["NA","YYMM"],["MMYY","AMBIGUOUS"]][cMMYY][cYYMM])
# if __name__=="__main__":resolve()
"""
#printデバッグ消した？
#前の問題の結果見てないのに次の問題に行くの？
"""
"""
お前カッコ閉じるの忘れてるだろ
"""
import sys
from io import StringIO
import unittest
class TestClass(unittest.TestCase):
def assertIO(self, input, output):
stdout, stdin = sys.stdout, sys.stdin
sys.stdout, sys.stdin = StringIO(), StringIO(input)
resolve()
sys.stdout.seek(0)
out = sys.stdout.read()[:-1]
sys.stdout, sys.stdin = stdout, stdin
self.assertEqual(out, output)
def test_入力例_1(self):
input = """3 4"""
output = """Even"""
self.assertIO(input, output)
def test_入力例_2(self):
input = """1 21"""
output = """Odd"""
self.assertIO(input, output)
if __name__ == "__main__":
resolve()
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s834289936 | Accepted | p03455 | Input is given from Standard Input in the following format:
a b | #!usr/bin/env python3
from collections import defaultdict
from collections import deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
def LI():
return list(map(int, sys.stdin.readline().split()))
def I():
return int(sys.stdin.readline())
def LS():
return list(map(list, sys.stdin.readline().split()))
def S():
return list(sys.stdin.readline())[:-1]
def IR(n):
l = [None for i in range(n)]
for i in range(n):
l[i] = I()
return l
def LIR(n):
l = [None for i in range(n)]
for i in range(n):
l[i] = LI()
return l
def SR(n):
l = [None for i in range(n)]
for i in range(n):
l[i] = S()
return l
def LSR(n):
l = [None for i in range(n)]
for i in range(n):
l[i] = SR()
return l
sys.setrecursionlimit(1000000)
mod = 1000000007
# A
def A():
a, b = LI()
if not (a * b) % 2:
print("Even")
else:
print("Odd")
return
# B
def B():
return
# C
def C():
return
# D
def D():
return
# E
def E():
return
# F
def F():
return
# G
def G():
return
# H
def H():
return
# Solve
if __name__ == "__main__":
A()
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s429710758 | Accepted | p03455 | Input is given from Standard Input in the following format:
a b | #!/usr/bin/env python3
import sys
# import time
# import math
# import numpy as np
# import scipy.sparse.csgraph as cs # csgraph_from_dense(ndarray, null_value=inf), bellman_ford(G, return_predecessors=True), dijkstra, floyd_warshall
# import random # random, uniform, randint, randrange, shuffle, sample
# import string # ascii_lowercase, ascii_uppercase, ascii_letters, digits, hexdigits
# import re # re.compile(pattern) => ptn obj; p.search(s), p.match(s), p.finditer(s) => match obj; p.sub(after, s)
# from bisect import bisect_left, bisect_right # bisect_left(a, x, lo=0, hi=len(a)) returns i such that all(val<x for val in a[lo:i]) and all(val>-=x for val in a[i:hi]).
# from collections import deque # deque class. deque(L): dq.append(x), dq.appendleft(x), dq.pop(), dq.popleft(), dq.rotate()
# from collections import defaultdict # subclass of dict. defaultdict(facroty)
# from collections import Counter # subclass of dict. Counter(iter): c.elements(), c.most_common(n), c.subtract(iter)
# from datetime import date, datetime # date.today(), date(year,month,day) => date obj; datetime.now(), datetime(year,month,day,hour,second,microsecond) => datetime obj; subtraction => timedelta obj
# from datetime.datetime import strptime # strptime('2019/01/01 10:05:20', '%Y/%m/%d/ %H:%M:%S') returns datetime obj
# from datetime import timedelta # td.days, td.seconds, td.microseconds, td.total_seconds(). abs function is also available.
# from copy import copy, deepcopy # use deepcopy to copy multi-dimentional matrix without reference
# from functools import reduce # reduce(f, iter[, init])
# from functools import lru_cache # @lrucache ...arguments of functions should be able to be keys of dict (e.g. list is not allowed)
# from heapq import heapify, heappush, heappop # built-in list. heapify(L) changes list in-place to min-heap in O(n), heappush(heapL, x) and heappop(heapL) in O(lgn).
# from heapq import nlargest, nsmallest # nlargest(n, iter[, key]) returns k-largest-list in O(n+klgn).
# from itertools import count, cycle, repeat # count(start[,step]), cycle(iter), repeat(elm[,n])
# from itertools import groupby # [(k, list(g)) for k, g in groupby('000112')] returns [('0',['0','0','0']), ('1',['1','1']), ('2',['2'])]
# from itertools import starmap # starmap(pow, [[2,5], [3,2]]) returns [32, 9]
# from itertools import product, permutations # product(iter, repeat=n), permutations(iter[,r])
# from itertools import combinations, combinations_with_replacement
# from itertools import accumulate # accumulate(iter[, f])
# from operator import itemgetter # itemgetter(1), itemgetter('key')
# from fractions import gcd # for Python 3.4 (previous contest @AtCoder)
def main():
mod = 1000000007 # 10^9+7
inf = float("inf") # sys.float_info.max = 1.79...e+308
# inf = 2 ** 64 - 1 # (for fast JIT compile in PyPy) 1.84...e+19
sys.setrecursionlimit(10**6) # 1000 -> 1000000
def input():
return sys.stdin.readline().rstrip()
def ii():
return int(input())
def mi():
return map(int, input().split())
def mi_0():
return map(lambda x: int(x) - 1, input().split())
def lmi():
return list(map(int, input().split()))
def lmi_0():
return list(map(lambda x: int(x) - 1, input().split()))
def li():
return list(input())
a, b = mi()
print("Even") if a * b % 2 == 0 else print("Odd")
if __name__ == "__main__":
main()
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s202369931 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a,b=map(int,input().split())
if (a)%2 == 0:
print("Even")
else:
print("Odd")
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s064563318 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | ri = lambda: int(input())
rl = lambda: list(map(int, input().split()))
rr = lambda N: [ri() for _ in range(N)]
YN = lambda b: print("YES") if b else print("NO")
INF = 10**18
N, K = rl()
W = [[0 for _ in range(2 * K + 3)] for _ in range(2 * K + 3)]
for i in range(N):
x, y, c = input().split()
x = int(x) % (2 * K)
y = (int(y) if c == "W" else int(y) + K) % (2 * K)
if (x < K) ^ (y < K):
x %= K
y %= K
W[0][y + 1] += 1
W[x + 1][y + 1] -= 2
W[0][K + 1] -= 1
W[x + 1][K + 1] += 1
W[x + 1][0] += 1
W[K + 1][0] -= 1
W[K + 1][y + 1] += 1
else:
x %= K
y %= K
W[0][0] += 1
W[x + 1][0] -= 1
W[0][y + 1] -= 1
W[x + 1][y + 1] += 2
W[K + 1][y + 1] -= 1
W[x + 1][K + 1] -= 1
W[K + 1][K + 1] += 1
for x in range(0, K + 2):
for y in range(1, K + 2):
W[x][y] += W[x][y - 1]
for y in range(0, K + 2):
for x in range(1, K + 2):
W[x][y] += W[x - 1][y]
ans = 0
for x in range(K + 1):
for y in range(K + 1):
ans = max(ans, W[x][y])
print(ans)
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s832925833 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | # -*- coding: <encoding name> -*-
a,b=map(int,input().split())
if a*b%2=0:
print('Even')
else:
print('Odd') | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s679874758 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a = int(input())
b = int(input())
if a >= 1 and a <= 10000:
if b >= 1 and <= 10000:
if a*b % 2 == 0:
print("Odd")
else:
print("Even")
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s792260531 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a =input();
b = input();
c = str(a) + str(b)
int i =1
while(i<=100):
if(i*i ==int(c)):
print('Yes')
else:
i=i+1
if(i==101);
print("No)
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s558382257 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | N = int(input())
plan = [list(map(int, input().split())) for i in range(N)]
m = 0
for i in range(N - 1):
time = plan[i + 1][0] - plan[i][0]
dist = abs(plan[i + 1][1] - plan[i][1]) + abs(plan[i + 1][2] - plan[i][2])
if time == 0:
m = 1
break
if dist > time:
m = 1
break
if dist % 2 != time % 2:
m = 1
break
print("Yes" if m == 0 else "No")
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s290032123 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a, b = map(int.input().split())
answer = a * b
if answer % == 0:
print('Even')
else:
print('Odd') | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s384682336 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | t = [0]
x = [0]
y = [0]
N = int(input())
for i in range(N):
ti, xi, yi = map(int,input().split())
t.append(ti)
x.append(xi)
y.append(yi)
flg = True
for i in range(1,N+1):
if (t[i]-t[i-1] < x[i]-x[i-1] + y[i]-y[i-1]):
flg = False
elif ((t[i]-t[i-1])%2 != (x[i]-x[i-1]+y[i]-y[i-1])%2):
flg = False
if(flg):
print("Yes")
else:
print("No")
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s628177810 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | N = int(input())
tl = []
xl = []
yl = []
al = []
bl = []
cl = []
dl = []
for i in range(N):
t, x, y = map(int, input().split())
tl.append(t)
xl.append(x)
yl.append(y)
al.append((x + y) % 2)
bl.append(t % 2)
for i in range(N):
c = abs((xl[i] + yl[i]) - (xl[i - 1] + yl[i - 1]))
d = abs(tl[i] - tl[i - 1])
cl.append(c)
dl.append(d)
count = 0
for i in range(N):
if al[i] == bl[i] and (xl[i] + yl[i]) <= tl[i] and cl[i] <= dl[i]:
count += 1
if count == N:
print("Yes")
else:
print("No")
| Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
If the product is odd, print `Odd`; if it is even, print `Even`.
* * * | s299254038 | Runtime Error | p03455 | Input is given from Standard Input in the following format:
a b | a, b = map(int, input().split())
def FizzBuzz(a, b):
total = a*b
if a*b%2 = 0:
print('Even')
else:
print('Odd')
FizzBuzz(a, b) | Statement
AtCoDeer the deer found two positive integers, a and b. Determine whether the
product of a and b is even or odd. | [{"input": "3 4", "output": "Even\n \n\nAs 3 \u00d7 4 = 12 is even, print `Even`.\n\n* * *"}, {"input": "1 21", "output": "Odd\n \n\nAs 1 \u00d7 21 = 21 is odd, print `Odd`."}] |
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