POJ2976

Dropping tests Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 12121 Accepted: 4231

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 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).

有一点不同的是本题是要求去掉 c 个，所以那个循环是 i < n - c 。

#include<cstdio> #include<iostream> #include<algorithm> using namespace std; const int maxn = 1010; int n, c; int a[maxn], b[maxn]; double dis[maxn];// a - mid * b bool C(double x) { for(int i= 0; i< n; i++) dis[i] = a[i] - x * b[i]; sort(dis, dis+n); double sum = 0; for(int i= c; i< n; i++) sum += dis[i]; //如果最大的 n-c 项和为正， 说明 x 合法 return sum >= 0; } int main () { while(1){ scanf("%d %d", &n, &c); if(n == 0 && c == 0) break; for(int i= 0; i< n; i++) scanf("%d", a+i); for(int i= 0; i< n; i++) scanf("%d", b+i); double lb = 0.0, ub = 1.0; for(int k= 0; k< 100; k++){ double mid = (lb + ub) / 2; if(C(mid)) lb = mid; else ub = mid; } //这里的处理方式要学习 printf("%.0f\n", 100 * lb); } return 0; }