感觉网上的这个题解比答案给的好多了，至少人家想到了用01背包变形->_->把钱的总和想象成包的容量，容量最多也就是sum嘛~

然后先不考虑概率是否够，递推结束后从大数向小数查找 满足概率就输出即是解

本来以为想明白了就没事了，还有一个梗是dp数组最大应该是钱数总和，所以数组不能开小了，就因为这RE了好多次，这要是比赛不还的后悔死啊—>_—>还是欠练

再就是为了保险起见还是max自己写一个吧~~总的来说就是注意细节

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.

<center> </center>
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 
#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
double pj[1050],dp[100500],p,rate;
int t,m[1050],n,sum;
double max(double a,double b)
{
     if(a<b) return b;
     else return a;
}
int main()
{
   // freopen("cin.txt","r",stdin);
    while(~scanf("%d",&t))
    {
         while(t--)
         {
              memset(dp,0,sizeof(dp));
              dp[0]=1;
              sum=0;
              scanf("%lf%d",&p,&n);
              p=1-p;
              for(int i=0;i<n;i++)
              {
                   scanf("%d%lf",&m[i],&pj[i]);
                   pj[i]=1-pj[i];
                   sum+=m[i];
              }
              for(int i=0;i<n;i++)
              {
                   for(int j=sum;j>=m[i];j--)
                    dp[j]=max(dp[j],dp[j-m[i]]*pj[i]);
              }
              for(int i=sum;i>=0;i--)
              {
                   if(dp[i]>p)
                   {
                        printf("%d\n",i);
                        break;
                   }
              }
         }
    }
    return 0;
}