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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 Mj (0 < Mj ≤ 100) and a real number Pj . Bank j contains Mj millions, and the probability of getting caught from robbing it is Pj. 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
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
Case 1: 2
Case 2: 4
Case 3: 6
Hint
For the first case, if he wants to rob bank 1 and 2, then the probability of getting caught is 0.02 + (1 - 0.02) * .03 = 0.0494 which is greater than the given probability (0.04). That’s why he has only option, just to rob rank 2.

这题可以转化成一个01背包问题
就是倒着用要取的值做背包重量，概率当做背包装的价值

这题需要把被抓的概率转化成不被抓的概率

然后再次明确一下什么时候背包是可用的

两个属性彼此正相关…马上要联想到二分，背包这些算法

#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>using namespace std;double dp[100001];int tu[101];double tt[101];int main(){    int T;    cin >> T;    int n,u=0;    double p;    while (T--)    {        memset(dp, 0, sizeof(dp));        cin >>p>> n;        p = 1 - p;        int he = 0;        for (int a = 1; a <= n; a++)        {            cin >> tu[a] >> tt[a];            tt[a] = 1 - tt[a];            he += tu[a];        }        dp[0] = 1;        for (int a = 1; a <= n; a++)        {            for (int b = he; b >= 0; b--)            {                if (b - tu[a] < 0)break;                dp[b] = max(dp[b], dp[b - tu[a]]*tt[a]);            }        }        int jg = 0;        for (int a = he; a >= 0; a--)        {            if (dp[a] >= p)            {                jg = a;                break;            }        }        printf("Case %d: %d\n", ++u, jg);    }}