POJ：2976 Dropping tests(二分+最大化平均值）

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_ifor 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).

题意：有N个考试，每个考试有ai和bi两个值，最后成绩由上面的公式求得。幸运的是，可以放弃K个科目，求最大化最后的成绩。

思路：这是一道简单的最大化平均值模板题，化简出（ai-mid*bi）>0,求n-k前项。

AC代码：

#include<stdio.h>
#define INF 0x3f3f3f3f
#include<algorithm>
using namespace std;
bool cmp(double x,double y)
{
    return x>y;
}
int n,k;
double y[1100];
int a[1100];
int b[1100];
bool C(double  mid)
{
   for(int i=0 ; i<n ; i++)
   {
       y[i]=a[i]-mid*b[i];
   }
   sort(y,y+n,cmp);
   double  sum=0;
   for(int i=0 ; i<n-k ; i++)
   {
       sum+=y[i];
   }
   if(sum>=0)
    return true;
   return false;
}
int main()
{
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        if(n==0&&k==0)
            break;
        for(int i=0 ; i<n ; i++)
            scanf("%d",&a[i]);
        for(int i=0 ; i<n ; i++)
            scanf("%d",&b[i]);
        double   st=0,en=INF;
        for(int i=1 ; i<=100 ; i++)
        {
            double mid=(st+en)/2;
            if(C(mid))
                st=mid;
            else
                en=mid;
        }

        printf("%.0f
",st*100);
    }
}