AtCoder Beginner Contest 109 C

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C - Skip

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Time limit : 2sec / Memory limit : 1024MB

Score : 300 points

Problem Statement

There are N cities on a number line. The i-th city is located at coordinate xi.

Your objective is to visit all these cities at least once.

In order to do so, you will first set a positive integer D.

Then, you will depart from coordinate X and perform Move 1 and Move 2 below, as many times as you like:

• Move 1: travel from coordinate y to coordinate y+D.
• Move 2: travel from coordinate y to coordinate y−D.

Find the maximum value of D that enables you to visit all the cities.

Here, to visit a city is to travel to the coordinate where that city is located.

题意：给出n个在坐标轴上的点，并给出每个点的坐标xi,你的初始位置是X,你有两种遍历方式。

1.向前走D步。

2.向后走D步。

你需要遍历到每个点至少一次。

现在需要求出最大的D满足条件。

Constraints

• All values in input are integers.
• 1≤N≤105
• 1≤X≤109
• 1≤xi≤109
• xi are all different.
• x1,x2,…,xN≠X

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Input

Input is given from Standard Input in the following format:

N X
x1 x2 … xN

Output

Print the maximum value of D that enables you to visit all the cities.

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Sample Input 1

3 3
1 7 11

Sample Output 1

2

Setting D=2 enables you to visit all the cities as follows, and this is the maximum value of such D.

• Perform Move 2 to travel to coordinate 1.
• Perform Move 1 to travel to coordinate 3.
• Perform Move 1 to travel to coordinate 5.
• Perform Move 1 to travel to coordinate 7.
• Perform Move 1 to travel to coordinate 9.
• Perform Move 1 to travel to coordinate 11.

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Sample Input 2

3 81
33 105 57

Sample Output 2

24

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Sample Input 3

1 1
1000000000

Sample Output 3

999999999

我们不难发现，想要从X走到每个点,我们离每个点的距离必定是D的整数倍。

所以显而易见，我们直接求出初始点到每个点的距离，D的最大值就是n个距离的最大公约数。

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<iostream>
using namespace std;
int maxn=-1,n,m,a[100005],minn=1926081700;
int main()
{
    std::ios::sync_with_stdio(false);
    cin>>n>>m;
    for(register int i=1;i<=n;++i)
    {
        cin>>a[i];
        a[i]=abs(a[i]-m);
        minn=min(minn,a[i]);
    }
    int k=sqrt(minn);
    bool flag;
    for(register int i=minn;i>=k;--i)
    {
        flag=0;
        if(!(minn%i))
        {
            for(register int j=1;j<=n;++j)
                if(a[j]%i)
                {
                    flag=1;
                    break;
                }
            if(!flag)
            {
                cout<<i;
                return 0;
            }
            flag=0;
            int q=minn/i;
            for(register int j=1;j<=n;++j)
                if(a[j]%q)
                {
                    flag=1;
                    break;
                }
            if(!flag)
                maxn=max(maxn,i);
        }
    }
    if(maxn!=-1)
        cout<<maxn;
    else
        cout<<1;
    return 0;
}