hdu1203 01揹包

至少得反面是一個沒有 所以方程爲 max(dp[j], 1-(1-dp[j-v[i]])*(1-c[i]))

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

double dp[10050], c[10005];
int v[10010];

int main()
{
    int n, m;
    while(scanf("%d %d", &n, &m) != EOF)
    {
        if(n+m == 0) break;
        for(int i = 0; i < m; i++) scanf("%d %lf", &v[i], &c[i]);
        memset(dp, 0, sizeof(dp));
        for(int i = 0; i < m; i++)
            for(int j = n; j >= v[i]; j--)
                dp[j] = max(dp[j], 1-(1-dp[j-v[i]])*(1-c[i]));
        dp[n] = dp[n]*100;
        printf("%.1lf%%\n", dp[n]);
    }
}