链接

http://poj.org/problem?id=2769

题解

$a\equiv b(\mod m)\Leftrightarrow a-b=km$
因此只要枚举差的约数就好了

代码

#include <cstdio>
#include <cctype>
#include <cstring>
#define abs(x) ((x)>0?(x):-(x))
#define ll long long
#define eps 1e-8
#define maxn 1000010
#define cl(x) memset(x,0,sizeof(x))
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define dinf 1e100
using namespace std;
ll N, a[maxn], d;
bool mark[maxn];
ll read(ll x=0)
{
	ll c, f=1;
	for(c=getchar();!isdigit(c) and c!=-1;c=getchar())if(c=='-')f=-f;
	for(;isdigit(c);c=getchar())x=x*10+c-48;
	return f*x;
}
int main()
{
	ll i, j, k, T=read();
	while(T--)
	{
		N=read();
		for(i=1;i<=N;i++)a[i]=read();
		cl(mark);
		for(i=1;i<=N;i++)for(j=i+1;j<=N;j++)
		{
			d=abs(a[i]-a[j]);
			for(k=1;k*k<=d;k++)
				if(d%k==0)mark[k]=mark[d/k]=1;
		}
		for(i=1;mark[i];i++);
		printf("%lld\n",i);
	}
	return 0;
}