POJ 2976 -- Dropping tests

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

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).

Source

Stanford Local 2005

 

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
#define Max 1008
#define eps 1e-4
bool cmp(double a,double b)
{
    return a>b;
}
int main()
{
    int n,k,i;
    double left,mid,right,ans;
    double score[Max],a[Max],b[Max];
    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]);
        for(left=0,right=100;right-left>eps;)
        {
            mid=(left+right)/2.0;
            for(i=0;i<n;i++)
                score[i]=a[i]*100.0-b[i]*mid;
            sort(score,score+n,cmp);
            ans=0;
            for(i=0;i<n-k;i++)
                ans+=score[i];
            if(ans>0)
                left=mid;
            else
                right=mid;
        }
        cout<<(int)(left+.5)<<endl;
    }return 0;
}