HDU 6693 Valentine's Day (概率)

2019 杭电多校 10 1003

题目链接：HDU 6693

比赛链接：2019 Multi-University Training Contest 10

Problem Description

Oipotato loves his girlfriend very much. Since Valentine's Day is coming, he decides to buy some presents for her.

There are (n) presents in the shop, and Oipotato can choose to buy some of them. We know that his girlfriend will possibly feel extremely happy if she receives a present. Therefore, if Oipotato gives (k) presents to his girlfriend, she has (k) chances to feel extremely happy. However, Oipotato doesn't want his girlfriend to feel extremely happy too many times for the gifts.

Formally, for each present (i), it has a possibility of (P_i) to make Oipotato's girlfriend feel extremely happy. Please help Oipotato decide what to buy and maximize the possibility that his girlfriend feels extremely happy for exactly one time.

Input

There are multiple test cases. The first line of the input contains an integer (T (1le Tle 100)), indicating the number of test cases. For each test case:

The first line contains an integer (n (1le nle 10 000)), indicating the number of possible presents.

The second line contains (n) decimals (P_i (0le Pile 1)) with exactly six digits after the decimal point, indicating the possibility that Oipotato's girlfriend feels extremely happy when receiving present (i).

It is guaranteed that the sum of (n) in all test cases does not exceed (450000).

Output

For each test case output one line, indicating the answer. Your answer will be considered correct if and only if the absolute error of your answer is less than (10^{−6}).

Sample Input

2
3
0.100000 0.200000 0.900000
3
0.100000 0.300000 0.800000

Sample Output

0.900000000000
0.800000000000

Solution

题意

有 (n) 种礼物，第 (i) 种礼物能让女朋友开心的概率为 (P_i)，挑一些礼物，问让女朋友开心一次的概率最大为多少。

题解

概率 贪心

如果有概率大于等于 (0.5) 的礼物，输出其中最大的。

否则，对概率从大到小。暴力枚举选择前 (k) 大的礼物的的概率，求最大值即可。

此题有原题。见 CodeForces 442B

相关证明见官方题解：Codeforces #253 editorial

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const double eps = 1e-8;
const int inf = 0x3f3f3f3f;
const int maxn = 100000 + 5;

double p[maxn];

int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        int n;
        scanf("%d", &n);
        double ans = 0, maxa = -1;
        int flag = 0;
        for(int i = 0; i < n; ++i) {
            scanf("%lf", &p[i]);
            if(p[i] >= 0.5) {
                flag = 1;
                if(maxa = -1) maxa = p[i];
                else maxa = max(maxa, p[i]);
            }
        }
        if(flag) {
            printf("%.6lf
", maxa);
        } else {
            sort(p, p + n);
            double tmp1 = p[n - 1];
            double tmp2 = p[n - 1] * (1 - p[n - 2]) + p[n - 2] * (1 - p[n - 1]);
            ans = tmp2;
            int k = 3;
            while(tmp1 < tmp2 && k <= n) {
                ans = tmp2;
                tmp1 = tmp2;
                tmp2 = 0;
                for(int i = n - 1; i >= n - k; --i) {
                    double tmp = 1;
                    for(int j = n - 1; j >= n - k; --j) {
                        if(j == i) {
                            tmp *= p[j];
                        } else {
                            tmp *= 1 - p[j];
                        }
                    }
                    tmp2 += tmp;
                }
                k ++;
            }
            printf("%.6lf
", max(ans, tmp2));
        }
    }
    return 0;
}