POJ 2976 Dropping tests （0/1分数规划）

Dropping tests

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4654		Accepted: 1587

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

 

 

给定ai和bi，求100*sigma(ai*xi)/sigma(bi*xi)的最大值，xi=0或1，且只能有k个为0。

即可以去掉k个(ai,bi)对。

 

y=100*sigma(ai)/sigma(bi)

t(y)=100*sigma(ai)-y*sigma(bi)

求t(y0)=0；

当t(y)<0时，y太大，y0<y；

当t(y)>0时，y太小，y<y0；

 

所以可以二分y的值。

 

去哪k个更优了？

要使y更大就应该是t(y)>0这种情况更可能的出现，所以对100*ai-bi排序，去掉小的k个。使t(y)>0城里更有可能。

 

#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<cmath>



using namespace std;



const int N=1010;

const double eps=1e-10;



int n,k;

double a[N],b[N],score[N];



int main(){



    //freopen("input.txt","r",stdin);



    while(~scanf("%d%d",&n,&k)){

        if(n==0 && k==0)

            break;

        for(int i=0;i<n;i++)

            scanf("%lf",&a[i]);

        for(int i=0;i<n;i++)

            scanf("%lf",&b[i]);

        double l=0,r=100,mid,sum;

        while(r-l>eps){

            mid=(l+r)/2;

            for(int i=0;i<n;i++)

                score[i]=100*a[i]-mid*b[i];

            sort(score,score+n);

            sum=0;

            for(int i=k;i<n;i++)

                sum+=score[i];

            if(sum>0)

                l=mid;

            else

                r=mid;

        }

        printf("%.0f\n",l);

    }

    return 0;

}