Dropping tests
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 npositive 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).
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int N = 1e3 + 7;
const double eps = 1e-8;
double a[N], b[N], c[N];
int n, k;
int check(double mid) {
for (int i = 1; i <= n; ++i) {
c[i] = a[i] - mid * b[i];
}
sort(c+1, c+n+1);
double f = 0;
for (int i = k+1; i <= n; ++i) {
f += c[i];
}
return f >= 0 ? 1 : 0;
}
int main() {
while (~scanf ("%d %d", &n, &k) && (n + k)) {
for (int i = 1; i <= n; ++i) {
scanf ("%lf", &a[i]);
}
for (int i = 1; i <= n; ++i) {
scanf ("%lf", &b[i]);
}
double l = 0, r = 1, mid;
while (r - l > eps) {
mid = (l + r) /2.0;
if (check(mid)) l = mid;
else r = mid;
}
printf ("%.0f\n", l*100);
}
return 0;
}