[gcd相减]

题意:

https://codeforces.ml/contest/1459/problem/C

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,a2+bj…,an+bj

就是对于每个b[j],求gcd(a1+bj,a2+bj,......an+bj);

思路:

gcd(a1+bj,a2+bj,......an+bj) = gcd(a1+bj,a2-a1,a3-a1,......,an-a1) => gcd(a1+bj,G);

G可以直接求出,然后答案就出来了.

gcd相减挺常见的.

标签：gcd,......,bj,a1,a2,数学,有趣,相减
来源： https://www.cnblogs.com/LaiYiC/p/14725208.html