hdu 2955 01 背包（浮点数）

http://acm.hdu.edu.cn/showproblem.php?pid=2955

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

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

又见01背包，按照以往的思路，背包的容量为不被抓到的最大概率，价值为能抢走的最多的钱数，但是这里有一个问题概率是浮点数无法用来做数组的下标，曾想着将其化为整数，但是小数位数不一定是2位。所以，可以换种思路。将银行看做物品，将被抓的概率当做价值，将银行的钱数总钱数看作是背包容量。最后，输出能是概率在规定以上的所需背包的容量，就是答案。

#include 
#include 
#include 
using namespace std;
int a[10005];
double b[10005],dp[10005];
int main()
{
    int n,T;
    double m;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lf%d",&m,&n);
        int sum=0;
       for(int i=0;i=a[i];j--)
          {
              dp[j]=max(dp[j],dp[j-a[i]]*(1-b[i]));
          }
      /*for(int i=0;i<=sum;i++)
          cout << dp[i]<< " ";
      printf("\n");*/
      for(int j=sum;j>=0;j--)
      {
          if(dp[j]>=1-m)
          {
              printf("%d\n",j);
              break;
          }
      }
    }
     return 0;
}