描述：
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.
輸入：
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 .
輸出：
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.
樣例輸入：
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
樣列輸出：
2
4
題目大意：先輸入一個正整數表示測試樣例，每一個測試樣例包含一個正整數n（表示種數）和一個實數p（表示概率），接下去n行包含兩個正整數表示價值與概率。求概率小於等於p情況下所能得到的最大價值。
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct asd
{
int value;
double p;
}gw[105];
int main()
{
int t,n,allvalue,k;
double a,dp[10005];
scanf("%d",&t);
while(t--)
{
scanf("%lf %d",&a,&n);
a=1-a; //安全的概率
allvalue=0;
for(int i=1;i<=n;i++)
{
scanf("%d %lf",&gw[i].value,&gw[i].p);
gw[i].p=1-gw[i].p;
allvalue+=gw[i].value;
}
memset(dp,0,sizeof(dp));
dp[0]=1;
for(int i=1;i<=n;i++)
{
for(int j=allvalue;j>=gw[i].value;j--)
{
dp[j]=max(dp[j],dp[j-gw[i].value]*gw[i].p); //最大情況出現的兩種可能性
}
}
for(int i=allvalue;i>=0;i--)
{
if(dp[i]>=a)
{
k=i;
break;
}
}
printf("%d\n",k);
}
return 0;
} 
