gcd & lcm &exgcd

int lcm(int a, int b) {
    return a / gcd(a, b) * b;
}
int gcd(int a,int b) {
    return b ? gcd(b,a%b):a;
}

int exgcd(int a,int b,int &x,int &y) {
    if(!b) { x=1; y=0; return a; }
    int r=exgcd(b,a%b,x,y);
    int t=y; y=x-(a/b)*y; x=t;
    return r; //gcd
}

原文地址：https://www.cnblogs.com/wronin/p/11422377.html