hdu-2955 Robberies

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.

InputThe 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 .OutputFor 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

说实话，这题写的时候没有看懂题意- -。
题意：给定n个银行和被抓的最大概率p。
下面n个银行，给的是能抢到的利润和被抓的概率。
要我们求的是在不被抓的情况下，能获得最大的利润，就是个01背包问题。
因为要求的是不被抓的概率，所以把背包问题的加法运算改成乘法运算就行了。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int M = 1111111;
double cost[M],bg[M];
int wei[M];
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        double p;
        int n,sum=0;
        memset(bg,0,sizeof(bg));
        scanf("%lf %d",&p,&n);
     //   printf("%d n",n);
        for(int i=0;i<n;i++){
            scanf("%d %lf",&wei[i],&cost[i]);
            sum+=wei[i];
        //    printf("%d
",sum);
        }
        bg[0]=1.0;
        for(int i=0;i<n;i++){
            for(int j=sum;j>=wei[i];j--){
                bg[j]=max(bg[j],bg[j-wei[i]]*(1.0-cost[i]));
            }
        }
        for(int j=sum;j>=0;j--){
           // printf("%d
",j);
            if(bg[j]>(1.0-p)){
                printf("%d
",j);
                break;
            }
        }
    }
    return 0;
}