题意：有A，B两个人。n道题目。每题有相应的分数。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 
     
      #include 
      
       #include 
       
        #include 
        
         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;}