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

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;
}