POJ2976(最大化平均值)

Dropping tests

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9329		Accepted: 3271

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
将条件写为(n&&k)。无限WA。

#include <stdio.h>
#include <algorithm>
#include <math.h>
using namespace std;
const int MAXN=;
const double EPS=1.0e-6;
int n,k;
int a[MAXN],b[MAXN];
bool test(double x)
{
    double y[MAXN];
    for(int i=;i<n;i++)
    {
        y[i]=a[i]-x*b[i];
    }
    sort(y,y+n);
    double sum=;
    for(int i=k;i<n;i++)
    {
        sum+=y[i];
    }
    return sum>=0.0;
}
int main()
{
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        if(n==&&k==)    break;
        for(int i=;i<n;i++)    scanf("%d",&a[i]);
        for(int i=;i<n;i++)    scanf("%d",&b[i]);
        double l=;
        double r=0x3f3f3f3f;
        while(fabs(r-l)>EPS)
        {
            double mid=(l+r)/;
            if(test(mid))    l=mid;
            else r=mid;
        }
        int res=(int)(l*+0.5);
        printf("%d\n",res);
    }
    return ;
}