POJ

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).

二分基础题

#include <iostream>
using namespace std;
#include<string.h>
#include<set>
#include<stdio.h>
#include<math.h>
#include<queue>
#include<map>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
#include <cstdio>
#include <cstdlib>
#include<stack>
#include<vector>
double a[1100];
double b[1100];
double c[1100];
const double MIN=1e-7;
int n,m;
int main()
{
    while(cin>>n>>m)
    {
        if(n==0&&m==0)
            break;
        for(int i=1;i<=n;i++)
                cin>>a[i];
        for(int i=1;i<=n;i++)
                cin>>b[i];
        double kaishi=0.0,jieshu=1.0;
        double mid;
        while(jieshu-kaishi>MIN)
        {
            mid=(kaishi+jieshu)/2.0;
            //cout<<mid<<"_"<<endl;
            for(int i=1;i<=n;i++)
                c[i]=a[i]-mid*b[i];
            sort(c+1,c+1+n);
            double sum=0;
            for(int i=m+1;i<=n;i++)
                sum+=c[i];
            if(sum>=0)
                kaishi=mid;
            else
                jieshu=mid;
        }
        int qqq=mid*1000;
        if(qqq%10>=5)
            qqq+=10;
        qqq/=10;
        cout<<qqq<<endl;
    }
    return 0;
}