POJ 2976 Dropping tests 【二分：最大化平均值】

最新推荐文章于 2019-10-29 14:59:23 发布

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最新推荐文章于 2019-10-29 14:59:23 发布

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分类专栏： POJ 挑战程序设计—中级篇

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Dropping tests

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12352		Accepted: 4318

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

#include<iostream>  
#include<cstdio>  
#include<cstring>  
#include<string>  
#include<cmath>  
#include<queue>  
#include<stack>  
#include<vector>  
#include<map>  
#include<algorithm>  
using namespace std;  
#define ll long long  
#define ms(a,b)  memset(a,b,sizeof(a))  
const int M=1e5+10;  
const int inf=0x3f3f3f3f;  
const int mod=1e9+7; 
int i,j,k,n,m; 
int a[M]; 
int b[M];
double y[M];

bool cmp(double a,double b)
{ 
	return a>b;
}

bool check(double mid)
{
	for(int i=0;i<n;i++)
		y[i]=a[i]-b[i]*mid;
	sort(y,y+n,cmp);
	double sum=0;
	for(int i=0;i<n-k;i++)
       sum+=y[i];
	return sum>=0;
}


int main()
{
	while(~scanf("%d%d",&n,&k)){
		if(n==0&&k==0)break;
		for(int i=0;i<n;i++)scanf("%d",&a[i]);
		for(int i=0;i<n;i++)scanf("%d",&b[i]);
     double lb=0,ub=1000000000;
	 for(int i=0;i<100;i++){
		 double mid=(lb+ub)/2;
         if(check(mid))lb=mid;
		 else ub=mid;
	 }
	 printf("%.0lf\n",ub*100);
	}
	return 0;
}