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.
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 .
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.
代码：
首先得转换思路，因为概率值是浮点数，没有办法作为背包的容量。
这里选择抢劫的钱数作为背包容量，dp[i]代表抢劫的钱数为i时，不被抓住的最大的概率（这里比较绕-_-）。只要有一次被抓住，就算被抓住，所以选择计算不被抓住比较方便，只需将1-p[i]累乘即可。当不被抓住的概率最大时，被抓住的概率就最小，使得i值尽可能的大，得到最优解。
初始化时，只有dp[0]=1，其余均为0，因为dp代表确实抢到的数目，而不是最大值，也就是这里要求恰好装满。
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<stdlib.h>
#include <math.h>
using namespace std;
#define N 105
#define M 10000 int main(){
int T,n,money[N],sum;
double p,dp[M],pos[N];//pos为抢劫一个银行被抓的概率，dp为抢劫该数目的钱一次都不被抓的概率
scanf("%d",&T);
while( T-- ){
scanf("%lf%d",&p,&n);
sum=;
for( int i=;i<=n;i++ ){
scanf("%d%lf",&money[i],&pos[i]);
sum+=money[i];
}
memset(dp,0.0,sizeof(dp));
dp[]=;//只有没有抢到钱，此时有解的概率为1。其余均为0，代表无解
for( int i=;i<=n;i++ ){
for( int j=sum;j>=money[i];j-- ){
dp[j]=max(dp[j],dp[j-money[i]]*(-pos[i]));//j值代表抢钱的数目（并不是最大的数目，而是确实抢到的数目），j值一定，希望dp值越大越好
}
}
for( int i=sum;i>=;i-- ){
if( dp[i]>=-p ){//不被抓的概率大于预期
printf("%d\n",i);
break;
}
}
}
system("pause");
return ;
}