HDU 2955 Robberies（dp） - c++编程基础

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HDU 2955 Robberies（dp）

2015-07-20 17:26:01 来源: 作者: 【大中小】浏览:3次

Problem 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

Sample Output

2
4
6

题意：

输入小偷被抓概率的最大值和银行个数，然后输入每个银行的钱和偷这个银行被抓的概率，求小偷在最大被抓的概率内可以偷到的最多的钱？

#include
  
   
#include
   
     #include
    
      #include
     
       using namespace std; #define N 10005 int va[N]; double c[N],dp[N]; int main() { int t,n,i; double p; scanf("%d",&t); while(t--) { int sum=0; scanf("%lf%d",&p,&n); p=1-p; //不被抓的概率 for(i=0;i
      
       =va[i];v--) dp[v]=max(dp[v],dp[v-va[i]]*(1-c[i])); for(i=sum;i>=0;i--) if(dp[i]-p>0.0000001) break; printf("%d\n",i); } return 0; }


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