10057 - A mid-summer night's dream.

#先對vector 進行sort
#偶數，1 2 6 7
#2 和 6 都算是 nums[mid]/2, 所以要對照 2 和 6 在vector裡面出現多少次, total +=1, diff就是 abs(2-6) +1
#奇數， 1 2 6 7 8
#6 就是 nums[mid], total就是看vector出現多少次 6 , diff的話因為只會有1個6， 如果有三個6也一樣，都是 1，因為 abs(6-6) +1 =1


#include <iostream>
#include <set>
#include<vector>
#include<string>
#include<sstream>
#include<algorithm>
#include<map>
using namespace std;
int main()
{
	int t;
	while (cin >> t) {
		int number;
		vector<int> nums;
		for (int i = 0; i < t; i++) {
			cin >> number;
			nums.push_back(number);
		}
		sort(nums.begin(), nums.end());
		int mid = 0;
		int total = 0;
		int diff = 0;
		int k = 0;
		for (int i = 0; i < t; i++) {
			if (nums.size() % 2 == 1) {
				mid = nums.size() / 2;
				diff = 1;
				if (nums[nums.size() / 2] == nums[i]){
					total++;
				}
			}		
			if (nums.size() % 2 == 0) {
				mid = nums.size() / 2 - 1;
				if (nums[i] == nums[mid]) {
					total++;
				}
				else if (nums[mid + 1] == nums[i]) {
					total++;
				}
				diff = abs(nums[mid] - nums[mid+1]) + 1;			
			}
		}
		cout << nums[mid] << " " << total << " " << diff << endl;
	}
}