Robberies（01背包）

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.

Input

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 .

Output

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.

Sample Input

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

Sample Output

2
4
6
题目大意：
给定一个浮点数和整数分别代表最低成功率，以及n户人家，接下来是n行人家的价值以及小偷被抓的概率，求不低于最低成功率能偷到的最大价值。
小偷成功的情况是都要成功，所以概率就是（1-p1）*（1-p2)....

#include <iostream>
#include <cstring>
using namespace std;
double a[],dp[],ans;
int v[];
int n;
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        memset(dp,,sizeof dp);
        cin>>ans>>n;
        int maxn=;
        for(int i=;i<=n;i++)
        {
            cin>>v[i]>>a[i];
            maxn+=v[i];///找到最大可能的得到价值
        }
        dp[]=;
        for(int i=;i<=n;i++)
            for(int j=maxn;j>=v[i];j--)
                dp[j]=max(dp[j],dp[j-v[i]]*(-a[i]));
        for(int i=maxn;i>=;i--)
            if(dp[i]>=(-ans))
                {cout<<i<<'\n';break;}
    }
    return ;
}