很惭愧不是自己做出来的，不过知道思路了就很简单
#include <iostream>using namespace std;void initarr(double **&arr, int n){arr = new double*[n];for (int i = 0; i<n; i++)arr[i] = new double[n];for (int i = 0; i < n; i++)memset(arr[i], 0, n * sizeof(double));}int test2(){int n;cin >> n;int *p = new int[n];for (int i = 0; i<n; i++)cin >> p[i];int low = (3 * n + 4) / 5;//至少要做对的题目//前i题对j个的概率 ★有这句话就能写出动态规划double **res;initarr(res, n + 1);res[0][0] = 1;for (int i = 1; i <= n; i++){res[i][0] = res[i - 1][0] * (100 - p[i - 1]) / 100;for (int j = 1; j <= i; j++)res[i][j] = res[i - 1][j - 1] * p[i - 1] / 100 + res[i - 1][j] * (100 - p[i - 1]) / 100;}double sum = 0;for (int i = low; i <= n; i++)sum += res[n][i];printf("%.5f", sum);return 0;}/*450 50 50 50*/
