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QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#620715 | #2440. Valentine's Day | memset0 | AC ✓ | 91ms | 3940kb | C++23 | 910b | 2024-10-07 20:43:30 | 2024-10-07 20:43:30 |
Judging History
answer
#include <bits/stdc++.h>
#define sz(x) ((int)(x).size())
#define all(x) begin(x), end(x)
#ifdef memset0
#define log(...) fprintf(stderr, __VA_ARGS__)
#else
#define endl '\n'
#define log(...) (void(0))
#endif
using namespace std;
using ll = long long;
using lf = long double;
using ull = unsigned long long;
const int N = 1e4 + 9;
int T, n;
double p[N];
int main() {
#ifdef memset0
freopen("C.in", "r", stdin);
#endif
cin.tie(0)->sync_with_stdio(0);
cin >> T;
while (T--) {
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> p[i];
}
sort(p + 1, p + n + 1);
reverse(p + 1, p + n + 1);
double ans = p[1], pre = 1 - p[1];
for (int i = 2; i <= n; i++) {
double cur = ans * (1 - p[i]) + pre * p[i];
if (cur > ans) {
ans = cur;
pre *= 1 - p[i];
} else {
break;
}
}
cout << fixed << setprecision(12) << ans << endl;
}
}
詳細信息
Test #1:
score: 100
Accepted
time: 91ms
memory: 3940kb
input:
100 10 0.020874 0.180969 0.015588 0.043033 0.181293 0.033025 0.163341 0.104341 0.065442 0.129546 10 0.019357 0.210598 0.045487 0.051145 0.014579 0.135566 0.004401 0.008004 0.106695 0.056974 10 0.062679 0.216144 0.010137 0.031823 0.233062 0.042464 0.099544 0.140483 0.201337 0.045349 100 0.116404 0.19...
output:
0.398190071489 0.373683886459 0.410885362758 0.421374044378 0.419872971016 0.421062257019 0.420796858020 0.419890897576 0.419860393202 0.000099990100 0.000000000000 0.003341000000 0.003336000000 0.001427003306 0.003169000000 0.029231252961 0.061862431346 0.022757483568 0.015698496871 0.035934479901 ...
result:
ok 100 numbers