Dropping tests POJ - 2976

 

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

Hint

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).

1.令F=Σ(a[i])/Σ(b[i]);

2.令G=Σ(a[i])-F*Σ(b[i])=Σ(a[i]-F*(b[i]))

3.当G>0时，存在更优的F；

4.可以把a[i]-F*(b[i])看作每件商品对答案的贡献

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstring>
using namespace std;

int n,k;
int a[1005],b[1005];
double d[1005];

bool check(double tem){
    double sum=0;
    for(int i=0;i<n;i++) d[i]=a[i]-tem*(b[i]);
    sort(d,d+n);
    for(int i=n-1;i>=k;i--) sum+=d[i];
    return sum>0 ? true:false;
}

void d_search(double ma){
    double l=0,r=ma;
    for(int i=0;i<100;i++){
        double mid=(l+r)/2.0;
        if(check(mid)) l=mid;
        else r=mid;
    }
    printf("%.0f\n",l*100);
}
int main()
{   while(~scanf("%d%d",&n,&k),n||k){
        32         for(int i=0;i<n;i++) scanf("%d",&a[i]);
        for(int i=0;i<n;i++) scanf("%d",&b[i]);
        d_search(1.0);
    }
    return 0;
}