結果
問題 | No.276 連続する整数の和(1) |
ユーザー |
|
提出日時 | 2020-04-16 21:05:03 |
言語 | PyPy3 (7.3.15) |
結果 |
AC
|
実行時間 | 60 ms / 1,000 ms |
コード長 | 4,047 bytes |
コンパイル時間 | 1,212 ms |
コンパイル使用メモリ | 82,040 KB |
実行使用メモリ | 66,772 KB |
最終ジャッジ日時 | 2024-10-02 10:07:50 |
合計ジャッジ時間 | 1,344 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge1 |
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ファイルパターン | 結果 |
---|---|
other | AC * 10 |
ソースコード
import sysfrom sys import stdinimport heapqimport refrom itertools import permutationsfrom bisect import bisect_left, bisect_rightfrom collections import Counter, dequefrom math import factorial, sqrt, gcd, ceilfrom functools import lru_cache, reduceINF = 1 << 60MOD = 1000000007sys.setrecursionlimit(10 ** 7)# UnionFindclass UnionFind():def __init__(self, n):self.n = nself.parents = [-1] * ndef find(self, x):if self.parents[x] < 0:return xelse:self.parents[x] = self.find(self.parents[x])return self.parents[x]def union(self, x, y):x = self.find(x)y = self.find(y)if x == y:returnif self.parents[x] > self.parents[y]:x, y = y, xself.parents[x] += self.parents[y]self.parents[y] = xdef size(self, x):return -self.parents[self.find(x)]def same(self, x, y):return self.find(x) == self.find(y)def members(self, x):root = self.find(x)return [i for i in range(self.n) if self.find(i) == root]def roots(self):return [i for i, x in enumerate(self.parents) if x < 0]def group_count(self):return len(self.roots())def all_group_members(self):return {r: self.members(r) for r in self.roots()}def __str__(self):return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())# ダイクストラdef dijkstra_heap(s, edge, n):#始点sから各頂点への最短距離d = [10**20] * nused = [True] * n #True:未確定d[s] = 0used[s] = Falseedgelist = []for a,b in edge[s]:heapq.heappush(edgelist,a*(10**6)+b)while len(edgelist):minedge = heapq.heappop(edgelist)#まだ使われてない頂点の中から最小の距離のものを探すif not used[minedge%(10**6)]:continuev = minedge%(10**6)d[v] = minedge//(10**6)used[v] = Falsefor e in edge[v]:if used[e[1]]:heapq.heappush(edgelist,(e[0]+d[v])*(10**6)+e[1])return d# 素因数分解def factorization(n):arr = []temp = nfor i in range(2, int(-(-n**0.5//1))+1):if temp%i==0:cnt=0while temp%i==0:cnt+=1temp //= iarr.append([i, cnt])if temp!=1:arr.append([temp, 1])if arr==[]:arr.append([n, 1])return arr# 2数の最小公倍数def lcm(x, y):return (x * y) // gcd(x, y)# リストの要素の最小公倍数def lcm_list(numbers):return reduce(lcm, numbers, 1)# リストの要素の最大公約数def gcd_list(numbers):return reduce(gcd, numbers)# 素数判定def is_prime(n):if n <= 1:return Falsep = 2while True:if p ** 2 > n:breakif n % p == 0:return Falsep += 1return True# limit以下の素数を列挙def eratosthenes(limit):A = [i for i in range(2, limit+1)]P = []while True:prime = min(A)if prime > sqrt(limit):breakP.append(prime)i = 0while i < len(A):if A[i] % prime == 0:A.pop(i)continuei += 1for a in A:P.append(a)return P# 同じものを含む順列def permutation_with_duplicates(L):if L == []:return [[]]else:ret = []# set(集合)型で重複を削除、ソートS = sorted(set(L))for i in S:data = L[:]data.remove(i)for j in permutation_with_duplicates(data):ret.append([i] + j)return ret# ここから書き始めるdef solve(n):ret = gcd(n * (n + 1) // 2, n)return retn = int(input())print(solve(n))