poj 2976 Dropping tests

Dropping tests

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4559		Accepted: 1542

Description

In a certain course, you take n tests. If you get a_i out of b_i questions correct on test i, your cumulative average is defined to be

.

Given your test scores and a positive integer k, determine how high you can make your cumulative average if you are allowed to drop any k of your test scores.

Suppose you take 3 tests with scores of 5/5, 0/1, and 2/6. Without dropping any tests, your cumulative average is . However, if you drop the third test, your cumulative average becomes .

Input

The input test file will contain multiple test cases, each containing exactly three lines. The first line contains two integers, 1 ≤ n ≤ 1000 and 0 ≤ k < n. The second line contains n integers indicating a_i for all i. The third line contains n positive integers indicating b_i for all i. It is guaranteed that 0 ≤ a_i ≤ b_i ≤ 1, 000, 000, 000. The end-of-file is marked by a test case with n = k = 0 and should not be processed.

Output

For each test case, write a single line with the highest cumulative average possible after dropping k of the given test scores. The average should be rounded to the nearest integer.

//0-1 分数规划
//详见： http://www.cnblogs.com/perseawe/archive/2012/05/03/01fsgh.html


#include <iostream>
#include <math.h>
#include <vector>
#include <queue>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int N=1010;
const double eps=1e-6;
double a[N],b[N];
struct node
{
    double z;
    int i;
    bool operator<(const node &b) const
    {
        return z<b.z;
    }
}oz[N];

int main()
{
    int n,k;
    int i;
    while(scanf("%d %d",&n,&k)==2&&n|k)
    {
        for(i=0;i<n;i++)
         scanf("%lf",&a[i]);
        for(i=0;i<n;i++)
         scanf("%lf",&b[i]);
       double ans,temp=0;
       double p,q;
       while(1)
       {
           ans=temp;
           for(i=0;i<n;i++)
           {
                oz[i].i=i;
                oz[i].z=a[i]-temp*b[i];
           }
           sort(oz,oz+n);
          p=q=0;
          for(i=k;i<n;i++)
          {
              p+=a[oz[i].i];
              q+=b[oz[i].i];
          }
          temp=p/q;
          if(fabs(temp-ans)<eps) break;
       }
       printf("%.0lf\n",ans*100);
    }
    return 0;
}