Codeforces Round #371 (Div. 2)B. Filya and Homework（水题 分类讨论）

传送门：B. Filya and Homework

描述：

B. Filya and Homework time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

Today, hedgehog Filya went to school for the very first time! Teacher gave him a homework which Filya was unable to complete without your help.

Filya is given an array of non-negative integers a1, a2, ..., an. First, he pick an integer x and then he adds x to some elements of the array (no more than once), subtract x from some other elements (also, no more than once) and do no change other elements. He wants all elements of the array to be equal.

Now he wonders if it's possible to pick such integer x and change some elements of the array using this x in order to make all elements equal.

Input

The first line of the input contains an integer n (1 ≤ n ≤ 100 000) — the number of integers in the Filya's array. The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109) — elements of the array.

Output

If it's impossible to make all elements of the array equal using the process given in the problem statement, then print "NO" (without quotes) in the only line of the output. Otherwise print "YES" (without quotes).

Examples input 5 1 3 3 2 1 output YES input 5 1 2 3 4 5 output NO Note

In the first sample Filya should select x = 1, then add it to the first and the last elements of the array and subtract from the second and the third elements.

题意：

给定一个序列，对于每一个元素，只能 + 或者 - 一个数val。这个数一旦选定，就不能改。

问能否变成全部数字都一样。

思路：

先用set维护有几个不同数字，然后分类讨论，特别要注意有三个不同数字的时候一定要满足a[0]+a[2]==2*a[1]  

代码：

#include <bits/stdc++.h> using namespace std; set<int>s; int a[4]; int main(){ std::ios::sync_with_stdio(false); std::cin.tie(0); int n; cin>>n; int x; for(int i=0; i<n ;i++){ cin>>x; s.insert(x); } if(s.size()>3)puts("NO"); else if(s.size()==3){ int now=0; for(set<int>::iterator it=s.begin(); it!=s.end(); it++){ a[now++]=*it; } sort(a, a+3); if(a[0]+a[2]==2*a[1])puts("YES"); else puts("NO"); } else puts("YES"); return 0; }