题意：有A，B两个人，n道题目，每题有对应的分数，B答对题目的概率是0.5，求A不输给B的概率不小于P要拿的最低分数

思路：DP，dp[i][j]来表示B答了前i题后分数为j的概率，，然后通过B的概率求A的最低分数

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 40010;

int a[MAXN],n;
double P,dp[50][MAXN];

int main() {
	int t;
	scanf("%d", &t);
	while (t--) {
		scanf("%d%lf", &n, &P);
		int sum = 0;
		for (int i = 0; i < n; i++) {
			scanf("%d", &a[i]);
			sum += a[i];
		}
		memset(dp, 0, sizeof(dp));
		dp[0][0] = 1;
		for (int i = 0; i < n; i++) 
			for (int j = 0; j <= sum-a[i]; j++)
				if (dp[i][j] > 0) {
					dp[i+1][j+a[i]] += dp[i][j]*0.5;
					dp[i+1][j] += dp[i][j]*0.5;
				}
		int ans = 0;
		double cnt = 0;
		for (int i = 0; i <= sum; i++) {
			cnt += dp[n][i];
			if (cnt >= P) {
				ans = i;
				break;
			}
		}
		printf("%d\n", ans);
	}
	return 0;
}