poj 2976 Dropping tests

最新推荐文章于 2015-01-27 16:57:53 发布

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最新推荐文章于 2015-01-27 16:57:53 发布

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原文链接：http://www.cnblogs.com/372465774y/archive/2013/05/07/3065684.html

本文介绍了一种使用分数规划的方法来解决课程测试中如何通过丢弃部分成绩来提高累计平均分的问题。通过输入学生的测试成绩及允许丢弃的成绩数量，算法能够计算出最高的可能累计平均分，并将结果四舍五入到最接近的整数。

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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;
}