Dropping tests POJ - 2976

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

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).

#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
using namespace std;
const int maxn = 2005;
const int INF = 0x3f3f3f3f;
int n, m;
double a[maxn];
double b[maxn];
double t[maxn];
double judge(double k)
{
	double ans = 0;
	for (int i = 0; i < n; i++)
	{
		t[i] = a[i] - b[i] * k;
	}
	sort(t, t + n);
	for (int i = m; i < n; i++)
		ans += t[i];
	return ans;
}
int main()
{
	while (scanf_s("%d%d", &n, &m) != EOF )
	{
		if (n + m == 0)
			break;
		for (int i = 0; i < n; i++)
			scanf_s("%lf", &a[i]);
		for (int i = 0; i < n; i++)
			scanf_s("%lf", &b[i]);
		double l = 0.0;
		double r = 1.0;
		while (r-l >1e-7)
		{   
			double  mid = (l + r) / 2;
			if (judge(mid) > 0)
			
				l = mid;
			else
				r = mid;
		}
		printf("%1.f\n", l * 100);
	}
	return 0;
}