結果

問題 No.66 輝け☆全国たこやき杯
ユーザー akusyounin2412
提出日時 2018-03-28 23:37:54
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 4 ms / 5,000 ms
コード長 1,761 bytes
コンパイル時間 1,012 ms
コンパイル使用メモリ 117,564 KB
実行使用メモリ 5,376 KB
最終ジャッジ日時 2024-06-25 13:54:17
合計ジャッジ時間 1,743 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge2
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
other AC * 10
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

# 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, cnt, ans,a[100000];
double dp[15][1100];
struct Edge { LL to, cost; };
string str;
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;
}
int main() {
    cin >> n;
    rep(i, 1 << n)cin >> a[i];
    rep(i, 1 << n)dp[0][i] = 1;
    rep(i, n) {
        rep(j, 1 << n) {
            LL p = j ^ (1 << i);
            for (int k = 0; k < (1 << i); k++) {
                dp[i + 1][j] += dp[i][j]*dp[i][p^k] * (a[j] * a[j]) / (a[j] * a[j] + a[p^k] * a[p^k]);
            }
        }
    }
    printf("%.15f\n", dp[n][0]);
    return 0;
}
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