01分数规划 POJ2976 Dropping tests

Dropping tests

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12451		Accepted: 4363

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   01分数规划的模板题，讲解的话加一个写得整齐的链接吧。  

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const double acc=1e-7;//accuracy-精度 
int n,k;
double l,r,mid,sum;
double a[1010],b[1010],d[1010];//定义为double 
int main(){
    while(scanf("%d%d",&n,&k)){
        if(!n&&!k) return 0;
        for(int i=1;i<=n;i++) scanf("%lf",&a[i]);
        for(int i=1;i<=n;i++) scanf("%lf",&b[i]);
        l=0.0;
        r=1.0;
        while(r-l>acc){
            mid=(l+r)*1.0/2;
            for(int i=1;i<=n;i++) d[i]=a[i]-mid*b[i];
            sort(d+1,d+n+1);
            sum=0.0;
            for(int i=k+1;i<=n;i++) sum+=d[i];
            if(sum>0) l=mid;
            else r=mid;
        }
        printf("%.0f\n",mid*100);
     }
}