lightoj 1079 - Just another Robbery（01背包）

As Harry Potter series is over, Harry has no job. Since he wants to make quick money, (he wants everything quick!) so he decided to rob banks. He wants to make a calculated risk, and grab as much money as possible. But his friends - Hermione and Ron have decided upon a tolerable probability P of getting caught. They feel that he is safe enough if the banks he robs together give a probability less than P.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case contains a real number P, the probability Harry needs to be below, and an integer N (0 < N ≤ 100), the number of banks he has plans for. Then follow N lines, where line j gives an integer M_j(0 < M_j ≤ 100) and a real number P_j . Bank j contains M_j millions, and the probability of getting caught from robbing it is P_j. A bank goes bankrupt if it is robbed, and you may assume that all probabilities are independent as the police have very low funds.

Output

For each case, print the case number and the maximum number of millions he can expect to get while the probability of getting caught is less than P.

Sample Input

Output for 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

Case 1: 2

Case 2: 4

Case 3: 6

给你一个实数，表示在被抓住概率最大为这么大的时候最多能抢多少钱。

求被抓住的概率不好求，所以反过来求不被抓住的概率。

dp[i]表示在抢的钱为i的时候的不被抓住的概率。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;int main(void){    int T,n,i,j;    double P;    double dp[11000];    double p[110];    int a[110];    scanf("%d",&T);    int cas = 1;    while(T--)    {        scanf("%lf%d",&P,&n);        int sum = 0;        for(i=1;i<=n;i++)        {            scanf("%d%lf",&a[i],&p[i]);            p[i] = 1 - p[i];            sum += a[i];        }        for(i=0;i<=sum;i++)            dp[i] = 0;        dp[0] = 1;        for(i=1;i<=n;i++)            for(j=sum;j>=a[i];j--)                dp[j] = max(dp[j],dp[j-a[i]]*p[i]);        for(i=sum;i>=0;i--)            if(P >= 1 - dp[i])            {                printf("Case %d: %d\n",cas++,i);                break;            }    }    return 0;}