Dropping tests(01分数规划)

Dropping tests

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8176		Accepted: 2862

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
题解：给你n个数，让求删除k个数后

的最大值；01分数规划；

代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
const int MAXN=10010;
struct Node{
    int a,b;
};
Node dt[MAXN];
double d[MAXN];
int n,k;
bool fsgh(double R){
    double sum=0;
    for(int i=0;i<n;i++)d[i]=dt[i].a-R*dt[i].b;
    sort(d,d+n);
    for(int i=n-1;i>=n-k;i--)sum+=d[i];
    return sum>0?true:false;
}
double erfen(double l,double r){
    double mid;
    while(r-l>1e-6){
        mid=(l+r)/2;
        if(fsgh(mid))l=mid;
        else r=mid;
    }
    return mid;
}
int main(){
    while(scanf("%d%d",&n,&k),n|k){
        double mx=0;
        k=n-k;
        for(int i=0;i<n;i++)scanf("%d",&dt[i].a);
        for(int i=0;i<n;i++)scanf("%d",&dt[i].b),mx=max(1.0*dt[i].a/dt[i].b,mx);
        printf("%.0f\n",erfen(0,mx)*100);
    }
    return 0;
}