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1079 - Just another Robbery


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Time Limit: 4 second(s)

Memory Limit: 32 MB


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

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

Note

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.



大意：强盗知道每个银行有多少钱，和去抢劫该银行被捕的概率 。强盗应该怎样在被捕概率低于 p 的情况下，抢劫一些银行，得到最多的钱。


思路：当强盗碰到某个银行时两个选择，抢或不抢。

若抢，则 dp [ j ] = dp [ j - m [ i ] ] * ( 1 - p [ i ]) 

若不抢，则 dp [ j ] = dp [ j ] (注意：这里等式右边的 dp [ j ] 是上一次循环的结果，如果不抢这个银行，当然就不用更新 dp [ j ] 的值了啊，总之这一切都是因为一维背包的内层循环是逆序的)

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n,m[110];
double P,p[110],dp[10010]; // dp[i] = j：表示抢到 i 元钱，不被捕的概率为 j 
int main()
{
  int t,text=0;
  scanf("%d",&t);
  while(t--)
  {
    scanf("%lf %d",&P,&n);
    int sum=0;
    for(int i=1;i<=n;i++)
    {
      scanf("%d %lf",&m[i],&p[i]);
      sum+=m[i];
    }
    memset(dp,0,sizeof(dp));
    dp[0]=1;
    for(int i=1;i<=n;i++)
    {
      for(int j=sum;j>=m[i];j--)
      {
        dp[j]=max(dp[j],dp[j-m[i]]*(1-p[i]));
      }
    }
    int ans=0;
    for(int i=0;i<=sum;i++)
    {
      if(1-dp[i]<P)
      {
        ans=i;
      }
    }
    printf("Case %d: %d\n",++text,ans);
  }
  return 0;
}