#691 (Div. 2) C. Row GCD（最大公约数的性质）

题目描述

You are given two positive integer sequences a1,…,an and b1,…,bm. For each j=1,…,m find the greatest common divisor of a1+bj,…,an+bj.

Input

The first line contains two integers n and m (1≤n,m≤2⋅105).
The second line contains n integers a1,…,an (1≤ai≤1018).
The third line contains m integers b1,…,bm (1≤bj≤1018).

Output

Print m integers. The j-th of them should be equal to GCD(a1+bj,…,an+bj).

Example

input
4 4
1 25 121 169
1 2 7 23
output
2 3 8 24

题目大意

给你n个数a[]，和m个数b[]。问 $gcd(b[i]+a[1],b[i]+a[2],\dots \dots ,b[i]+a[n])$ 的值（1<=i<=m）。

题目分析

我们知道 $gcd(a,b)=gcd(a,b-a)$。
因此：$gcd(a+x,b+x,c+x,d+x)$
=$gcd(a+x,b+x-a-x,c+x-b-x,d+x-c-x)$
=$gcd(a+x,b-a,c-b,d-c)$

这样答案就出来了。
$gcd(b[i]+a[1],b[i]+a[2],\dots \dots ,b[i]+a[n])$
=$gcd(b[i]+a[1],a[2]-a[1],\dots \dots ,a[n]-a[n-1])$

代码如下

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <map>
#include <unordered_map>
#include <queue>
#include <vector>
#include <set>
#include <bitset>
#include <algorithm>
#define LL long long
#define PII pair<int,int>
#define x first
#define y second
using namespace std;
const int N=2e5+5,mod=1e9+7;
LL a[N],b[N]; 
int main()
{
    
    
	int n,m;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
	for(int i=1;i<=m;i++) scanf("%lld",&b[i]);
	LL res=0;
	for(int i=2;i<=n;i++) res=__gcd(res,a[i]-a[i-1]);
	
	for(int i=1;i<=m;i++)
	{
    
    
		LL ans=__gcd(res,a[1]+b[i]);
		printf("%lld ",abs(ans));
	}
	return 0; 
}