hdu Dropping tests 0/1分数规划（二分求值）

Problem 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.

 

Sample Input

3 1 5 0 2 5 1 6 4 2 1 2 7 9 5 6 7 9 0 0

 

Sample Output

83 100

***************************************************************************************************************************

给定一个公式

y=100*sigma(ai)/sigma(bi)

t(y)=100*sigma(ai)-y*sigma(bi)

问最后去掉那几个k能得到正好的值，

***************************************************************************************************************************

#include<iostream>
#include<string>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;
double a[1001],b[1001],score[1001];
int n,k,i,j;
const double eps=1e-8;
int main()
{
    while(scanf("%d %d",&n,&k)&&(n+k))
    {
        for(i=0;i<n;i++)
            scanf("%lf",&a[i]);
        for(i=0;i<n;i++)
            scanf("%lf",&b[i]);
        double le=0,ri=100,mid,sum;
        while(ri-le>eps)
        {
            mid=(ri+le)/2;
            for(i=0;i<n;i++)
                score[i]=100*a[i]-b[i]*mid;
            sort(score,score+n);
            sum=0;
            for(i=k;i<n;i++)
                sum+=score[i];
            if(sum>0)
                le=mid;
            else
                ri=mid;
        }
        printf("%.0f
",le);
    }
    return 0;
}