Valentine’s Day
Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 171 Accepted Submission(s): 83
Special Judge
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 Pi 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 (1≤T≤100), indicating the number of test cases. For each test case:
The first line contains an integer n (1≤n≤10 000), indicating the number of possible presents.
The second line contains n decimals Pi (0≤Pi≤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
題意： 給出n個買當前物品令人開心的概率值，問從中買1-n個，從購買的全部中選出一個使其開心的最大概率值。
思路： 枚舉購買1-n個物品的期望值，比如0.3 0.4 0.8這組數據中，買一個最大的開心值爲0.8，買兩個（從最大的兩個中選）的爲0.8*（1-0.4）+0.4*（1-0.8），買三個則爲0.3*(1-0.4)(1-0.8)+(1-0.3)0.4(1-0.8)+(1-0.3)(1-0.4)*0.8,直接暴力求解即可，詳情看代碼和註釋。
#include <bits/stdc++.h>
using namespace std;
const int N = 1e4 + 5;
double a[N];
int main() {
int t, n;
scanf("%d", &t);
while (t--) {
double ans = 0.0, mul = 1.0, t = 0.0;
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%lf", &a[i]);
}
sort(a, a + n);
if (a[n-1] == 1.0) ans = 1.0; // 對1的概率進行特判
else {
for (int i = n - 1; i >= 0; i--) {
mul *= (1.0 - a[i]); // 維護一個當前不選的概率積
t = 0.0;
for (int j = i; j < n; j++) {
// 得出選當前的和不選其他的概率值，即一遍for循環算出買了n-i個物品使其開心的期望值
t += (mul / (1 - a[j]) * a[j]);
}
if (t > ans) ans = t;
// 如果大於之前的最大值，更新，否則直接結束枚舉；
//選的越多，前面的開心概率越小，總的期望值會變小
else break;
}
}
printf("%.12f\n", ans);
}
return 0;
}