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
",l);
    }
    return 0;
}