Robberies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 24462    Accepted Submission(s): 9037

Problem Description

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

题目要求的是在不被抓的情况下所能抢的最多的钱。

有小数的是概率，第一个输入的小数是最大的被抓的概率，以后如果抢的被抓的概率大于他的话就会被抓

因为每多抢一家银行就要把被抓的概率相乘，这里用的是不被抓的概率，即如果抢银行不被抓的概率大于我们输入的不被抓的概率那这时就可以成功抢到这笔钱

#include <stdio.h>
#include <math.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <queue>
#define maxn 105
using namespace std;
double n,b[maxn],dp[10005];
int a[maxn];
int main()
{
    int T;
    cin >> T;
    while(T--){
        int m;
        cin >> n >> m;
        n = 1 - n;
        int sum = 0;
        for(int i=0;i<m;i++){
            cin >> a[i] >> b[i];
            sum += a[i];
            b[i] = 1 - b[i];
        }
        memset(dp,0,sizeof(dp));
        dp[0] = 1;
        for(int i=0;i<m;i++){
            for(int j=sum;j>=a[i];j--){
                dp[j] = max(dp[j],dp[j-a[i]]*b[i]);//dp出每一个数的不被抓的概率
            }
        }
        for(int j=sum;j>=0;j--){
            if(dp[j]-n>1e-8){//若不被抓的概率大于题目给出的不被抓的概率，则此时所能抢的金额最多（从最大开始的循环）
                cout << j << endl;
                break;
            }
        }
    }
    return 0;
}