【POJ2976】Dropping tests - 01分数规划

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?

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. The end-of-file is marked by a test case with?n?=?k?= 0 and should not be processed.

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("%dn",ans);
	}
	return 0;
}