POJ 2976-Dropping tests（01分数规划）

分数规划求最高平均分

最新推荐文章于 2019-10-29 14:59:23 发布

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本文介绍了一种利用01分数规划算法解决课程测试中如何通过放弃部分成绩来获得最高累积平均分的问题。输入包括测试次数、允许放弃的测试数量及各次测试的成绩和总分，输出则是经过最优选择后的最高平均分。

Dropping tests

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13941		Accepted: 4874

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 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

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).

Source

Stanford Local 2005

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关于01分数规划，有一篇讲的不错的博客：http://www.cnblogs.com/perseawe/archive/2012/05/03/01fsgh.html

#include<stdio.h>
#include<algorithm>
using namespace std;
int n,k,a[1005],b[1005];
double c[1005];
int check(double x)
{
	double sum=0;
	for(int i=0;i<n;i++)
		c[i]=a[i]-x*b[i];
	sort(c,c+n);
	for(int i=k;i<n;i++)
		sum+=c[i];
	return sum>=0;
}
void work()
{
	double l=0,r=1.0;
	for(int i=0;i<100;i++)
	{
		double mid=(l+r)/2.0;
		if(check(mid))
			l=mid;
		else
			r=mid;
	}
	printf("%.0f\n",l*100);
}
int main(void)
{
	while(scanf("%d%d",&n,&k)!=EOF)
	{
		if(n==k && n==0)
			break;
		for(int i=0;i<n;i++)
			scanf("%d",&a[i]);
		for(int i=0;i<n;i++)
			scanf("%d",&b[i]);
		work();
	}
	return 0;
}