結果
問題 |
No.4 おもりと天秤
|
ユーザー |
![]() |
提出日時 | 2018-03-24 18:32:08 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 11 ms / 5,000 ms |
コード長 | 1,849 bytes |
コンパイル時間 | 1,406 ms |
コンパイル使用メモリ | 117,436 KB |
実行使用メモリ | 10,240 KB |
最終ジャッジ日時 | 2024-06-26 10:35:26 |
合計ジャッジ時間 | 2,049 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge4 |
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ファイルパターン | 結果 |
---|---|
other | AC * 23 |
ソースコード
# include <iostream> # include <algorithm> #include <array> # include <cassert> #include <cctype> #include <climits> #include <numeric> # include <vector> # include <string> # include <set> # include <map> # include <cmath> # include <iomanip> # include <functional> # include <tuple> # include <utility> # include <stack> # include <queue> # include <list> # include <bitset> # include <complex> # include <chrono> # include <random> # include <limits.h> # include <unordered_map> # include <unordered_set> # include <deque> # include <cstdio> # include <cstring> using namespace std; using LL = long long; using ULL = unsigned long long; constexpr long long MOD = 1000000000 + 7; constexpr long long INF = std::numeric_limits<long long>::max(); const double PI = acos(-1); #define fir first #define sec second typedef pair<LL, LL> Pll; typedef pair<LL, pair<LL, LL>> Ppll; typedef pair<LL, pair<LL, bitset<100001>>> Pbll; typedef pair<LL, pair<LL, vector<LL>>> Pvll; typedef pair<LL, LL> Vec2; struct Tll { LL first, second, third; }; typedef pair<LL, Tll> Ptll; #define rep(i,rept) for(LL i=0;i<rept;i++) #define Mfor(i,mf) for(LL i=mf-1;i>=0;i--) LL h, w, n, m, k, s, t, q, sum, last, ans, cnt,a[100000],dp[100][100000]; struct Edge { LL to, cost,mon; }; string str; bool f = 0; void YN(bool f) { if (f) cout << "YES" << endl; else cout << "NO" << endl; } void yn(bool f) { if (f) cout << "Yes" << endl; else cout << "No" << endl; } vector<Edge>vec[100000]; int main() { cin >> n; rep(i, n) { cin >> a[i]; sum += a[i]; } if (sum % 2 == 1) { cout << "impossible" << endl; return 0; } sum /= 2; dp[0][0] = 1; rep(i,n) rep(j, sum+1) { dp[i + 1][j + a[i]] = (dp[i + 1][j + a[i]] || dp[i][j]); dp[i + 1][j] = (dp[i + 1][j] || dp[i][j]); } cout << (dp[n][sum] == 1 ? "possible" : "impossible") << endl; return 0; }