POJ2976 Dropping tests (01分数规划)

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 npositive 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：问去掉k个数后∑a[i]/∑b[i]最大。

题解：01分数规划，二分答案就ok了。

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cmath>
#define N 1007
#define eps 0.000001
using namespace std;

int n,k;
double a[N],b[N],t[N];

int main()
{
    while(~scanf("%d%d",&n,&k)&&(n+k))
    {
        for (int i=1;i<=n;i++) scanf("%lf",&a[i]);
        for (int i=1;i<=n;i++) scanf("%lf",&b[i]);
        double l=0.0,r=10.0;
        while(r-l>=eps)
        {
            double mid=(l+r)/2;
            for (int i=1;i<=n;i++)
                t[i]=a[i]-mid*b[i];
            sort(t+1,t+n+1);
            double sum=0;
            for (int i=k+1;i<=n;i++)
                sum+=t[i];
            if (sum>=0) l=mid;
            else r=mid;    
        }
        printf("%.0f
",l*100);
    }
}