解题思路：给出一个临界概率，在不超过这个概率的条件下，小偷最多能够偷到多少钱。因为对于每一个银行都只有偷与不偷两种选择，所以是01背包问题。

这里有一个小的转化，即为f[v]代表包内的钱数为v的时候，小偷不被逮捕的概率，这样我们在用

for(i=1;i<=n;i++)

{

for(v=vol;v>=0;v--)

f[v]=max(f[v],f[v-c[i]]*（1-p[i]）);
}

的过程中，在求出最大的不被抓的概率过程中，记录下了在此过程中的包中的钱数与此时对应的概率，这样最后只需用一个循环判断在概率大于临界值的时候跳出循环，就得到了偷到的钱数

包的容量是给出的n个银行一共的钱(即为不管给出的那个临界概率是多少，最多能偷到的钱)，每一个物品的消耗是该银行存有的钱。每一个物品的价值是（1-p[i]）（即在该银行不被抓的概率）

反思：可耻地看了题解，因为老是转化不过去，概率因为是浮点型的不懂怎么转化，然后包的容量是所给出的所有银行所存的钱的和，也没有想到。

Robberies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 13067    Accepted Submission(s): 4834

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

#include<stdio.h>
#include<string.h>
int c[105];
double p[105] ,f[10010];
double max(double a,double b)
{
	if(a>b)
	return a;
	else
	return b;
}
int main()
{
	int ncase,n,i,v,vol;
	double m;
	scanf("%d",&ncase);
	while(ncase--)
	{
		vol=0;
		scanf("%lf %d",&m,&n);
		for(i=1;i<=n;i++)
		{
			scanf("%d %lf",&c[i],&p[i]);
			p[i]=1-p[i];
			vol+=c[i];
		}
		memset(f,0,sizeof(f));
		f[0]=1;

		for(i=1;i<=n;i++)
		{
			for(v=vol;v>=0;v--)
			{
			f[v]=max(f[v],f[v-c[i]]*p[i]);
			printf("f[%d]=%lf\n",v,f[v]);
		    }
		}
		for(i=vol;i>=1;i--)
		{
			if(f[i]>=1-m)
				break;
		}
			printf("%d\n",i);

	}
}