[BZOJ1441] Min

传送门

http://www.lydsy.com/JudgeOnline/problem.php?id=1441

题解

• 裴蜀定理
• 存在d=(a1,a2,⋯,an)使得∑ni=1=aixi....
• 也就是扩展欧几里得定理里面的ax+by=c,有解时,c的最小值为gcd(a,b)

var n,i,a,b:longint;function gcd(a,b:longint):longint;begin if b=0 then exit(a) else gcd:=gcd(b,a mod b);end;begin readln(n); read(a); for i:=2 to n do  begin   read(b);   a:=gcd(a,abs(b));  end; writeln(a);end.