HDU-1157 Who's in the Middle

最新推荐文章于 2020-04-03 22:43:09 发布

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本文介绍了一种算法，用于解决在给定的奇数个奶牛中找到中位数牛奶产量的问题。通过排序所有奶牛的产量，可以确定至少一半的奶牛产奶量高于或等于中位数，另一半低于或等于中位数。

See the article on https://dyingdown.github.io/2019/12/15/HDU-1157%20Who’s-in-the-Middle/

Who’s in the Middle

Problem Description

FJ is surveying his herd to find the most average cow. He wants to know how much milk this ‘median’ cow gives: half of the cows give as much or more than the median; half give as much or less.

Given an odd number of cows N (1 <= N < 10,000) and their milk output (1…1,000,000), find the median amount of milk given such that at least half the cows give the same amount of milk or more and at least half give the same or less.

Input

* Line 1: A single integer N

* Lines 2…N+1: Each line contains a single integer that is the milk output of one cow.

Output

* Line 1: A single integer that is the median milk output.

Sample Input

Sample Output

Analysis

Just sort all the numbers, since the input numbers are odds, so the middle number is n / 2 + 1.

Code

#include<bits/stdc++.h>

using namespace std;

int a[2000000];
int main() {
	int n;
	while(cin >> n){
		for(int i = 0; i < n; i ++) {
			cin >> a[i];
		}
		sort(a, a + n);
		cout << a[n / 2] << endl;
	}
}