【POJ2976】Dropping tests - 01分数规划

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. 

题目大意

给出$n$个$a$和$b$，让选出$n-k$个使得$\frac{\sum a_i}{\sum b_i}$最大

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 
#include 
#include 
#include 
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;
}