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dropping tests
time limit: 1000ms memory limit: 65536k
total submissions: 9338 accepted: 3272

description

in a certain course, you take n tests. if you get ai out of bi 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 ai for all i. the third line contains n positive integers indicating bi for all i. it is guaranteed that 0 ≤ ai ≤ bi ≤ 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).


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
#include
#include
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         {
            scanf("%d",&a[i]);
        }
        for(int i=0;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     {
        v[i]=1.000*a[i]-1.000*b[i]*k;
    }
    sort(v,v n);
    double sum=0.000;
    for(int i=0;i     {
        sum =v[n-i-1];
    }
    if(sum>=0.0001) return 1;
    else       return 0;
}

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