POJ2976：Dropping tests（01分数规划入门）

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

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：有几门成绩，现在要求选N-K门，使得平均成绩最高，求最高平均成绩。

思路：01分数规划。

有三种需要掌握的01分数规划，之前最大流的时候遇到过，现在有些想法，有空再来实现。

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const double eps=1e-6;
int N,K;
struct in
{
    double x,y;
    double res;
    bool friend operator <(in a,in b){
        return a.res>b.res;
    }
}a[1010];
bool check(double Mid)
{
    for(int i=1;i<=N;i++) a[i].res=a[i].x-Mid*a[i].y;
    sort(a+1,a+N+1);
    double tx=0;
    for(int i=1;i<=K;i++) tx+=a[i].res;
    if(tx>=-eps) return true;
    return false;
}
int main()
{
    while(~scanf("%d%d",&N,&K)&&(N||K)){
        K=N-K;
        for(int i=1;i<=N;i++) scanf("%lf",&a[i].x),a[i].x*=100.0;
        for(int i=1;i<=N;i++) scanf("%lf",&a[i].y);
        double L=0,R=100,Mid,ans=0;
        while(R-L>eps){
            Mid=(L+R)/2;
            if(check(Mid)) ans=Mid,L=Mid+eps;
            else R=Mid-eps;
        }
        printf("%.0lf
",ans);
    }
    return 0;
}