CodeForces 714B. Filya and Homework（模拟）

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刺猬Filya收到了一项家庭作业，需要通过增加或减少一个整数X来调整一组数字，使它们最终全部相等。本篇介绍了如何解决这个问题，并提供了一个具体的算法实现。

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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.

题意：给定N个数字，对于每一个数字可以选择加上X或者减去X或者不变。最后能否找到一个X使得处理后的数字全都相等。

代码

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<map>
using namespace std;
const int maxn=100005;
const int INF=0x3f3f3f3f;
int num[maxn];
int main()
{
    int N;
    scanf("%d",&N);
    int min_num=INF;
    int max_num=0;
    map<int,int>mp;
    for(int i=0; i<N; i++)
    {
        scanf("%d",&num[i]);
        min_num=min(min_num,num[i]);
        max_num=max(max_num,num[i]);
        mp[num[i]]++;
    }
    //检测num中不同数字的数量
    if((int)mp.size()<=2)
    {
        printf("YES\n");
        return 0;
    }

    bool flag=false;//还没找到需要更改的数
    int flag_num;//需要更改的绝对值
    for(int i=0; i<N; i++)
    {
        if(num[i]!=(max_num+min_num)/2)
        {
            if(flag==false)
            {
                flag=true;
                flag_num=(int)abs(num[i]-(max_num+min_num)/2);
            }
            else
            {
                if((int)abs(num[i]-(max_num+min_num)/2)!=flag_num)
                {
                    printf("NO\n");
                    return 0;
                }
            }
        }
    }
    printf("YES\n");
    return 0;
}