Robberie

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.

输入

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

输出

2
4
6

没有给背包，所以可以将得到的钱当做背包容量，将被抓的风险当做价值，背包大小就得自己枚举（手搓）
f[j]=max(f[j],f[j-v[i]]+(1-w[i]));
最后一点，没事别自己瞎搓函数，会闹笑话

点击查看代码

#include<iostream>
#include<cstring>
using namespace std;
int v[300];
double w[300];
double f[300000];
int n,sum=0;
double m;
double max(double x,double y){
	if(x>=y) return x;
	else return y;
}
int main(){
	int T;
	cin>>T;
	while(T--){
		cin>>m>>n;
		for(int i=1;i<=n;i++){
			cin>>v[i]>>w[i];
			sum+=v[i];
		}
		memset(f,0,sizeof(f));
		f[0]=1.0;//没抢的时候最安全 
		for(int i=1;i<=n;i++){
			for(int j=sum;j>=v[i];j--){
				f[j]=max(f[j],f[j-v[i]]*(1.0-w[i]));//计算在取j的钱时的安全值 
			}
		}
		for(int i=sum;i>=0;i--){
			if(f[i]>1-m){//安全值大于规定的安全值才能取钱 
				cout<<i;break;
			}
		}
		sum=0;
	}
	return 0;
}