学步园

The aspiring Roy the Robber has seen a lot of American movies, and knows that the bad guys usually gets caught in the end, often because they become too greedy. He has decided to work in the lucrative business of bank robbery only
for a short while, before retiring to a comfortable job at a university.

For a few months now, Roy has been assessing the security of various banks and the amount of cash they hold. He wants to make a calculated risk, and grab as much money as possible.

His mother, Ola, has decided upon a tolerable probability of getting caught. She feels that he is safe enough if the banks he robs together give a probability less than this.

The first line of input gives T, the number of cases. For each scenario, the first line of input gives a floating point number P, the probability Roy needs to be below, and an integer N, the number of banks he has plans for. Then
follow N lines, where line j gives an integer Mj and a floating point number Pj .

Bank j contains Mj millions, and the probability of getting caught from robbing it is Pj .

For each test case, output a line with the maximum number of millions he can expect to get while the probability of getting caught is less than the limit set.

Notes and Constraints
0 < T <= 100
0.0 <= P <= 1.0
0 < N <= 100
0 < Mj <= 100
0.0 <= Pj <= 1.0
A bank goes bankrupt if it is robbed, and you may assume that all probabilities are independent as the police have very low funds.

3
0.04 3
1 0.02
2 0.03
3 0.05
0.06 3
2 0.03
2 0.03
3 0.05
0.10 3
1 0.03
2 0.02
3 0.05

2
4
6
解题：用反向的思维，所有给出的概率都是被捉概率，那么想要成功的逃出，则逃出的概率越大越好。以钱为背包，价值为概率。每一个包里的钱可能是一家银行的也可是多家的，那么要完成抢多家银行看成一个事件，那要完成抢都要逃过，那么抢每一家看做一步，组合起来，也就完成了这件事，那么每个概率都是要相乘的，所乘结果也就是完成这件事的概率。
#include<stdio.h>
#define N 10100
#define e 0.00000001
struct nnn
{
    int w;
    double p;
};
int sum;
double P,dp[N];
double max(double a,double b){return a>b?a:b;}
void zeroonepack(int use,double p)
{
    for(int s=sum;s>=use;s--)
    dp[s]=max(dp[s],dp[s-use]*p);
}
int main()
{
    nnn a[N];
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf%d",&P,&n);
        P=1-P;sum=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d%lf",&a[i].w,&a[i].p);
            a[i].p=1-a[i].p;
            sum+=a[i].w;
        }
        dp[0]=1;
        for(int i=1;i<=sum;i++)
            dp[i]=0;
        for(int i=1;i<=n;i++)
        zeroonepack(a[i].w,a[i].p);
        for(int i=sum;i>=0;i--)
        if(dp[i]>=P)
        {printf("%d\n",i);break;}
    }
}