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