博客里面还有以前写的解题报告，表示没看懂。这里直接计算其没有被录取的概率，每次去小的值即可

代码如下：

#include 
     
      #include 
      
       #include 
       
        #define MAXN 1005using namespace std;int N, M, s[MAXN];double p[MAXN], dp[10005];inline double min(double x, double y){    return x < y ? x : y;}void zobag(int x){    for (int i = N; i >= s[x]; --i) {        dp[i] = min(dp[i], dp[i-s[x]]*p[x]);    } }double DP(){    for (int i = 1; i <= M; ++i) {        zobag(i);    }    return 100-100*dp[N];}int main(){    while (scanf("%d %d", &N, &M), N|M) {        for (int i = 0; i <= N; ++i) {            dp[i] = 1;        }        for (int i = 1; i <= M; ++i) {             scanf("%d %lf", &s[i], &p[i]);            p[i] = 1.0-p[i];        }         printf("%.1lf%%\n", DP());    }    return 0;}