Dropping tests

Dropping tests

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7201 Accepted: 2499

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

分析：二分最大平均值 

code：

<span style="font-size:18px;">#include<stdio.h>#include<iostream>#include<algorithm>using namespace std;const double eps=1e-10;struct node{    double v,w;    double ave;} nodee[1100];int n,k;bool cmp(node a,node b){    return a.ave<b.ave;}int main(){    int i;    while(scanf("%d%d",&n,&k)!=EOF)    {        if(n==0&&k==0)            break;        for(i=0; i<n; i++)            scanf("%lf",&nodee[i].v);        for(i=0; i<n; i++)            scanf("%lf",&nodee[i].w);        double lb=0.0,ub=100.0,sum1;        while(ub-lb>eps)        {            double mid=(lb+ub)/2;            for(int i=0; i<n; i++)                nodee[i].ave=100*nodee[i].v-mid*nodee[i].w;            sort(nodee,nodee+n,cmp);            sum1=0;            for(int i=k; i<n; i++)                sum1+=nodee[i].ave;            if(sum1>0)                lb=mid;            else                ub=mid;        }        printf("%.0f\n",lb);    }    return 0;}</span><span style="font-size:24px;"></span>