Description

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.

 

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

 

 

 

题意：一个人去抢银行，输入它被抓的概率(P)和银行的个数(N)，接下来输入每个银行的钱和被抓的概率....

求这个人可以不被抓最多可以抢钱抢到多少....

 

 

解题思路：  

　　　　　　 既然是求他不被抓又抢到的钱最多，那么就是求他不被抓的概率中抢到的钱最多的...

　　　　　　那dp[j]表示他不被抓的概率，j表示抢到的钱

　　　　　　  dp[j]=max(dp[j],dp[j-N[i]]*(1-P[i]))

　　　　　　这里的N[i]表示银行里的钱，P[i]表示被抓概率.

　　　　　　

　　　　　　求出他抢到的钱不被抓的概率后，只要循环找到满足dp[j]>=1-p的j就好了（这里倒过来循环方便寻找）

 

 

 

 

 

 

 

代码如下：

 

 

　　　　

1 #include 
         
           2 #include 
          
            3 double P[105],dp[10005]; 4 int N[105]; 5 double max(double x,double y) 6 { 7  8     return x>y?x:y; 9 }10 11 int main()12 {13     int T;14     scanf("%d",&T);15     while(T--)16     {17         int n,total=0;18         double p;19         scanf("%lf%d",&p,&n);20         for(int i=0; i
           
            =N[i]; j--)32             {33 34                 dp[j]=max(dp[j],dp[j-N[i]]*(1-P[i]));35 36             }37         }38         for(int i=total; i>=0; i--)39         {40 41             if(dp[i]>=1-p)42             {43                 printf("%d\n",i);44                 break;45             }46 47         }48     }49     return 0;50 }