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.
思路
题目要求 $\frac{\sum a_i}{\sum b_i} \geq x$,$x$的最大值 ,也就是$\sum a_i - x \sum b_i \geq 0$ 二分完把$a_i-x b_i$排序取$n-k$个大的即可
/************************************************ *Author : lrj124 *Created Time : 2018.09.28.20:35 *Mail : [email protected] *Problem : poj2976 ************************************************/ #include <algorithm> #include <iostream> #include <cstdio> #include <cmath> using namespace std; const int maxn = 1000 + 10; int n,k,a[maxn],b[maxn]; double tmp[maxn]; inline bool check(double x) { for (int i = 1;i <= n;i++) tmp[i] = a[i]-x*b[i]; sort(tmp+1,tmp+n+1); double ans = 0; for (int i = k+1;i <= n;i++) ans += tmp[i]; return ans >= 0; } int main() { while (cin >> n >> k) { if (!n && !k) break; for (int i = 1;i <= n;i++) cin >> a[i]; for (int i = 1;i <= n;i++) cin >> b[i]; double l = 0,r = 0x3f3f3f3f; while (fabs(r-l) >= 1e-6) { double mid = (l+r)/2; if (check(mid)) l = mid; else r = mid; } int ans = int(l*100+0.5); printf("%d\n",ans); } return 0; }