POJ 2976（01分数划分+二分）

                                                                                              Dropping tests

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12221		Accepted: 4273

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

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).

Source

Stanford Local 2005

题意：给你n对 a[i] b[i] 去掉k对  使得 sigms(a)/sigma(b) 最大

后来才知道这是一道板子题

(∑ai*xi)/(∑bi*xi)求极值问题

我们可以化简 (∑ai*xi)-L*(∑bi*xi)=0；

F（L)=(∑ai*xi)-L*(∑bi*xi)  由此可以得到这个一个单调递减的函数 随着L的增大而减小、

假设我们要求的最大值L为 l;

只有当F(L)无限接近0时  L=l;

对于我们如何来确定这个解

比如我们选k对  我们只有判断 这前K大的(ai-L*b[i])和是不是<0;因为这是单调递减的，F(L)<0--> L>l  F(L)>0--> L

而我们这个题是删除k对  所以只有n-k就可以了

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<cstring>
#include<map>
#include<stack>
#include<set>
#include<vector>
#include<algorithm>
#include<string.h>
typedef long long ll;
typedef unsigned long long LL;
using namespace std;
const int INF=0x3f3f3f3f;
const double eps=0.0000000001;
const int N=100000+10;
int n,k;
int a[N],b[N];
double ans[N];
int judge(double x){
    double sum=0;
    for(int i=0;i<n;i++){
        ans[i]=a[i]-x*b[i];
    }
    sort(ans,ans+n);
    for(int i=n-1;i>=n-k;i--)sum=sum+ans[i];
    if(sum>0)return 1;
    else
        return 0;
}
int main(){
    while(scanf("%d%d",&n,&k)!=EOF){
        if(n==0&&k==0)break;
        double maxx=0.0;
        for(int i=0;i<n;i++)scanf("%d",&a[i]);
        for(int i=0;i<n;i++)scanf("%d",&b[i]);
        k=n-k;
        double low=0;
        double high=1.0;
        double mid;
        while(low+eps<high){
            mid=(low+high)/2;
            if(judge(mid)){
                low=mid;
            }
            else{
               high=mid;
            }
        }
        printf("%.0f\n",mid*100);
    }
}

 

 

所以我们二分 L（0,1）