poj Dropping tests

Dropping tests

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9338 Accepted: 3272

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 15 0 25 1 64 21 2 7 95 6 7 90 0

Sample Output

83100

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).

首先要注意这题不能用单位价值的排序来求答案是错的，然后二分求得时候要注意精度，终止条件是俩个均不为0

#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
int n, m;
const int N = 1100;
int  a[N], b[N];
double  v[N];
const int inf = 1000000010;
int judge(double k);


int main()
{
    while(scanf("%d %d", &n, &m),n!=0||m!=0)
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        for(int i=0;i<n;i++)
        {
            scanf("%d",&b[i]);
        }
        m=n-m;
        double l=0, r=inf, mid;
        for(int i=0;i<100;i++)
        {
            mid=(l+r)/2;
            if(judge(mid))
            {
                l=mid;
            }
            else
            {
                r=mid;
            }
        }
        r=r*100;
        printf("%.0f\n",r);
    }
    return 0;
}


int judge(double k)
{
    for(int i=0;i<n;i++)
    {
        v[i]=1.000*a[i]-1.000*b[i]*k;
    }
    sort(v,v+n);
    double sum=0.000;
    for(int i=0;i<m;i++)
    {
        sum+=v[n-i-1];
    }
    if(sum>=0.0001) return 1;
    else       return 0;
}