結果

問題 No.1218 Something Like a Theorem
ユーザー cozy_sauna
提出日時 2021-10-21 14:07:17
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 64 ms / 2,000 ms
コード長 1,953 bytes
コンパイル時間 195 ms
コンパイル使用メモリ 82,068 KB
実行使用メモリ 67,200 KB
最終ジャッジ日時 2024-09-21 09:33:27
合計ジャッジ時間 2,154 ms
ジャッジサーバーID
(参考情報)
judge1 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 16
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

from sys import setrecursionlimit, stdin
from collections import defaultdict, deque
from itertools import permutations, combinations, product
from functools import lru_cache
from random import sample, choice, randint, random
from heapq import heappush, heappop
from math import factorial, gcd, exp
from copy import copy, deepcopy
from time import time
setrecursionlimit(10 ** 6)
readline = stdin.readline
# @lru_cache(maxsize=None)
INF = 10 ** 18
MOD = 1000000007
# MOD = 998244353
DYDX = [(-1, 0), (1, 0), (0, -1), (0, 1)]
def I(): return int(readline())
def S(): return readline()[:-1]
def LI(): return list(map(int, readline().split()))
def SPI(): return map(int, readline().split())
def SPII(): return map(lambda x: int(x)-1, readline().split())
def FIE(x): return [readline()[:-1] for _ in [0] * x]
def NODE(x): return [[] for _ in [0] * x]
def ranges(*args): return [(i, j) for i in range(args[0]) for j in range(args[-1])]
def nynx(y, x, ly = INF, lx = INF): return [(y+dy, x+dx) for dy, dx in DYDX if 0 <= y+dy < ly and 0 <= x+dx < lx]
def imax(A, idx=-1, m=-(1<<18)):
for i, a in enumerate(A): m, idx = [[m, idx], [a, i]][m < a]
return idx
def imin(A, idx=-1, m=1<<18):
for i, a in enumerate(A): m, idx = [[m, idx], [a, i]][m > a]
return idx
def cmin(dp, i, x):
if x < dp[i]: dp[i] = x
def cmax(dp, i, x):
if x > dp[i]: dp[i] = x
def gen(x, *args):
ret = [x] * args[-1]
for e in args[:-1][::-1]: ret = [deepcopy(ret) for _ in [0] * e]
return ret
def puts(E):
for e in E: print(e)
def pprint(E):
print()
puts(E)
YN = ['No', 'Yes']
####################################################################
N, Z = SPI()
def solve(N, Z):
if N >= 3: return False
if N == 1:
return Z >= 2
limit = pow(10, 3)
for x in range(1,limit+1):
for y in range(1, limit+1):
if pow(x, 2) + pow(y, 2) == pow(Z, 2): return True
return False
print(YN[solve(N, Z)])
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