HDU

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1203

题意：

该题要求得到一份offer的最大概率，在例子中的0.44 = 1-(1-0.2)*(1-0.3)这样求得。

状态方程则为：max[j]=max{max[j],1-(1-max[j-mon[i]])*(1-pop[i])};

Sample Input
10 3
4 0.1
4 0.2
5 0.3
0 0
 

Sample Output
44.0%

Hint

You should use printf("%%") to print a '%'.

************************************************

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<queue>
#include<algorithm>
#include<cmath>
#include<iostream>

using namespace std;
typedef long long LL;

#define INF 0x3f3f3f3f
#define N 22000
#define MAXN 100000000
#define mod 1000000007

double dp[N],g[N];
int p[N];

int main()
{
    int n,m,i,j;

    while(scanf("%d %d", &n,&m),n+m)
    {
        memset(dp,0,sizeof(dp));
        memset(p,0,sizeof(p));
        memset(g,0,sizeof(g));

        for(i=1;i<=m;i++)
            scanf("%d %lf", &p[i], &g[i]);

        for(i=1;i<=m;i++)
            for(j=n;j>=p[i];j--)
            dp[j]=max(dp[j],1-(1-dp[j-p[i]])*(1-g[i]));

        printf("%.1f%%
", dp[n]*100);
    }
    return 0;
}