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 .

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.

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.

To avoid ambiguities due to rounding errors, the judge tests have been constructed so that all answers are at least 0.001 away from a decision boundary (i.e., you can assume that the average is never 83.4997).

给定ai和bi，求100*sigma(ai*xi)/sigma(bi*xi)的最大值，xi=0或1，且只能有k个为0。

即可以去掉k个(ai,bi)对。

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

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

求t(y0)=0；

当t(y)<0时，y太大，y0<y；

当t(y)>0时，y太小，y<y0；

所以可以二分y的值。

去哪k个更优了？

要使y更大就应该是t(y)>0这种情况更可能的出现，所以对100*ai-bi排序，去掉小的k个。使t(y)>0城里更有可能。

#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<cmath>

using namespace std;

const int N=;

const double eps=1e-;

int n,k;

double a[N],b[N],score[N];

int main(){

    //freopen("input.txt","r",stdin);

    while(~scanf("%d%d",&n,&k)){

        if(n== && k==)

            break;

        for(int i=;i<n;i++)

            scanf("%lf",&a[i]);

        for(int i=;i<n;i++)

            scanf("%lf",&b[i]);

        double l=,r=,mid,sum;

        while(r-l>eps){

            mid=(l+r)/;

            for(int i=;i<n;i++)

                score[i]=*a[i]-mid*b[i];

            sort(score,score+n);

            sum=;

            for(int i=k;i<n;i++)

                sum+=score[i];

            if(sum>)

                l=mid;

            else

                r=mid;

        }

        printf("%.0f\n",l);

    }

    return ;

}