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2 years ago
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
const int N = 1010;
const int INF = 0x3f3f3f3f;
int n, k;
int a[N], b[N];
const double eps = 1e-8;
double d[N];
bool check(double x) {
for (int i = 0; i < n; i++) d[i] = a[i] - x * b[i];
sort(d, d + n, greater<double>()); //由大到小排序
double sum = 0;
for (int i = 0; i < n - k; i++) sum += d[i];
return sum >= 0; //从大到小选n-k个看ans是否为可行的解
}
int main() {
while (cin >> n >> k && (n || k)) {
for (int i = 0; i < n; i++) cin >> a[i];
for (int i = 0; i < n; i++) cin >> b[i];
//浮点数二分
double l = 0, r = INF;
while (r - l > eps) {
double mid = (l + r) / 2; //注意浮点数不能用右移操作
if (check(mid))
l = mid; //向右逼近,使结果更大一些
else
r = mid; //向左逼近,使结果更小一些
}
printf("%.0lf\n", l * 100); // 四舍五入
}
return 0;
}