Robberies

题意：一个强盗要去抢劫银行，对于每个银行来说，都有一个不被抓的概率p，和能抢劫到的钱数money,每个银行最多只可以被抢劫一次。问在不被抓的总概率P下，怎样得到最大价值的钱数。

分析：被抓概率可以转换成安全概率，Roy的安全概率大于1-P时都是安全的。抢劫的金额为0时，肯定是安全的，所以d[0]=1;其他金额初始为最危险的所以概率全为0；

注意：不要误以为精度只有两位。

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.

SampleInput

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

SampleOutput

2
4
6

#include<iostream>
#include<cstdio>
#include<cmath>
#define MAXN 101
#define MAXV 10001

using namespace std;

int cost[MAXN];
double weight[MAXV],d[MAXV];

int main()
{
    int test,sumv,n,i,j;
    double P;
    cin>>test;
    while(test--)
    {
        scanf("%lf %d",&P,&n);
        P=1-P;
        sumv=0;
        for(i=0;i<n;i++)
        {
            scanf("%d %lf",&cost[i],&weight[i]);
            weight[i]=1-weight[i];
            sumv+=cost[i];
        }
        for(i=0;i<=sumv;i++)
            d[i]=0;
        d[0]=1;
        for(i=0;i<n;i++)
        {
            for(j=sumv;j>=cost[i];j--)
            {
                d[j]=max(d[j],d[j-cost[i]]*weight[i]);
            }
        }
        bool flag=false;
        for(i=sumv;i>=0;i--)
        {
            if(d[i]-P>0.000000001)
            {
                printf("%d\n",i);
                break;
            }
        }
    }
    return 0;
}