POJ 2976 Dropping tests

Dropping tests

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13793		Accepted: 4838

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).

题目大意：一个序列，每个数有两个属性，ai和bi，求去掉k个数后的sum{ai}/sum{bi}最大

题解：01分数规划

设答案为t，若存在sigma ai/ sigma bi >t,那么答案还可以继续扩大。

将公式变形,sigma ai>t*sigma bi,sigma (ai-t*bi)>0,令si=ai-t*bi，然后排序，去掉k个小的后的

sigma si是否大于0，若是则答案还可以继续扩大，否则缩小。

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#define maxn 1009
using namespace std;

int n,k;
double l,r,mid,a[maxn],b[maxn],s[maxn];
double eps=1e-7;

bool check(double p){
    for(int i=1;i<=n;i++)s[i]=a[i]-p*b[i];
    sort(s+1,s+n+1);double all=0.;
    for(int i=n;i>k;i--)all+=s[i];
    return all>0;
}
int main(){
    while(1){
        scanf("%d%d",&n,&k);
        if(!n&&!k)break;
        for(int i=1;i<=n;i++)scanf("%lf",&a[i]);
        for(int i=1;i<=n;i++)scanf("%lf",&b[i]);
        l=0;r=1;
        while(r-l>eps){
            mid=(l+r)/2.;double all=0.;
            if(check(mid))l=mid;
            else r=mid;
        }
        printf("%.0lf
",l*100);
    }
    return 0;
}