POJ 2976 Dropping tests

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本文介绍了一种通过二分查找和贪心策略结合的算法来解决如何最大化考试平均分的问题。该算法能够在限定条件下舍弃部分成绩，从而实现整体平均分的最大化。文章详细解释了算法的实现过程，并提供了一个具体的C++代码示例。

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Dropping tests

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10282		Accepted: 3590

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 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).

Source

Stanford Local 2005

题解：

这道题很容易想到贪心，但是用例子容易验证贪心是不正确的，这里有点没想明白。如果用二分的方法，就要确保二分的最终解一边都是合法取值，一边都是非法取值，这样我们构造C(x):从n个数字中取出n-k个数，能够大于等于x。这样我们就确保了这一点。此外要解决这道题还需要一些数学形式的变换，一直在WA的地方就是最后要求解的是最接近的整数，这一点忘记了。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm> 
using namespace std;
const int maxn=1000+10;
const int inf=1<<30;
int a[maxn],b[maxn],n,k;
double c[maxn];
int judge(double m)
{
	double sum=0;
	for(int i=0;i<n;i++)	c[i]=a[i]-m*b[i];
	sort(c+0,c+n);
//	for(int i=0;i<n;i++) cout<<a[i]<<" "<<m*b[i]<<" "<<c[i]<<endl;
//	cout<<endl;
	for(int i=n-1;i>=n-k;i--) sum+=c[i];
	
	if(sum>=0) return 1;
	else return 0;
}
int main()
{
	//freopen("input.txt","r",stdin);
	while(scanf("%d%d",&n,&k)==2)
	{
		if(n==0&&k==0) break;
		k=n-k;
		for(int i=0;i<n;i++)  scanf("%d",&a[i]);
		for(int i=0;i<n;i++)  scanf("%d",&b[i]);
		double l=0,r=inf;
		//cout<<judge(0.84)<<endl;
		for(int i=1;i<=100;i++)
		{
			double m=(l+r)/2;
			if(judge(m)) l=m;
			else r=m;
		}
		 printf("%.0f\n",l*100); 
	}
	return 0;
 }