CF #371 div2 B. Filya and Homework

B. Filya and Homework
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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.

题目大意：
给你一些数a0到an，问是否存在一个数x，使得数列中每个数加一次或减一次x或者不变后整个数组所有数字全部相等
思路：
一开始想的是平均数，觉得把所有数变为整个数组的平均数就可以了，然后就wa哭了，被测试6卡住了
后来看别人博客发现只要数字三种以上就不可能了，一种或者两种一定可以，而三种的话只要满足a[0]+a[2]=2*a[1]就可以了
当时思路完全错了，GG
代码

#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>using namespace std;int main(){    int n;    scanf("%d",&n);    int a[5]={-1,-1,-1,-1,-1};    int item;    for(int i=0,k=0;i<n;i++)    {        scanf("%d",&item);        int flag=0;        for(int j=0;j<k;j++)        {            if(item==a[j])            {                flag=1;                break;            }        }        if(flag==0)        {            a[k]=item;            k++;        }        if(a[3]!=-1)            break;    }    if(a[1]==-1||a[2]==-1)        printf("YES\n");    else if(a[3]!=-1)        printf("NO\n");    else if(a[3]==-1)    {        sort(a,a+3);        if(a[0]+a[2]==2*a[1])            printf("YES\n");        else            printf("NO\n");    }    return 0;}