結果
問題 | No.892 タピオカ |
ユーザー |
|
提出日時 | 2019-09-27 21:23:53 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 3 ms / 1,000 ms |
コード長 | 1,962 bytes |
コンパイル時間 | 2,339 ms |
コンパイル使用メモリ | 192,536 KB |
最終ジャッジ日時 | 2025-01-07 19:19:36 |
ジャッジサーバーID (参考情報) |
judge5 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 6 |
ソースコード
#include <bits/stdc++.h>#define LLI long long int#define FOR(v, a, b) for(LLI v = (a); v < (b); ++v)#define FORE(v, a, b) for(LLI v = (a); v <= (b); ++v)#define REP(v, n) FOR(v, 0, n)#define REPE(v, n) FORE(v, 0, n)#define REV(v, a, b) for(LLI v = (a); v >= (b); --v)#define ALL(x) (x).begin(), (x).end()#define RALL(x) (x).rbegin(), (x).rend()#define ITR(it, c) for(auto it = (c).begin(); it != (c).end(); ++it)#define RITR(it, c) for(auto it = (c).rbegin(); it != (c).rend(); ++it)#define EXIST(c,x) ((c).find(x) != (c).end())#define fst first#define snd second#define popcount __builtin_popcount#define UNIQ(v) (v).erase(unique(ALL(v)), (v).end())#define bit(i) (1LL<<(i))#ifdef DEBUG#include <misc/C++/Debug.cpp>#else#define dump(...) ((void)0)#endif#define gcd __gcdusing namespace std;template <class T> constexpr T lcm(T m, T n){return m/gcd(m,n)*n;}template <typename I> void join(ostream &ost, I s, I t, string d=" "){for(auto i=s; i!=t; ++i){if(i!=s)ost<<d; ost<<*i;}ost<<endl;}template <typename T> istream& operator>>(istream &is, vector<T> &v){for(auto &a : v) is >> a; return is;}template <typename T, typename U> bool chmin(T &a, const U &b){return (a>b ? a=b, true : false);}template <typename T, typename U> bool chmax(T &a, const U &b){return (a<b ? a=b, true : false);}template <typename T, size_t N, typename U> void fill_array(T (&a)[N], const U &v){fill((U*)a, (U*)(a+N), v);}struct Init{Init(){cin.tie(0);ios::sync_with_stdio(false);}}init;int64_t power(int64_t n, int64_t p, int64_t m){int64_t ret = 1;while(p>0){if(p&1) (ret *= n) %= m;(n *= n) %= m;p /= 2;}return ret;}int main(){int a1,b1,a2,b2,a3,b3;while(cin >> a1 >> b1 >> a2 >> b2 >> a3 >> b3){int t = power(a1,b1,2) + power(a2,b2,2) + power(a3,b3,2);if(t % 2 == 0){cout << ":-)" << endl;}else{cout << ":-(" << endl;}}return 0;}