模板之快速幂和求逆元

ll mypow(ll a,ll b){
    ll sum=1LL;
    while(b){
        if(b&1) sum=(1LL*sum*a)%mod;
        a=(1LL*a*a)%mod;
        b>>=1;
    }
    return sum;
}
void ext_gcd(ll a, ll b, ll &d, ll &x, ll &y)//拓展欧几里得求逆元
{
    if (!b){
        d=a; x=1LL; y=0LL;
    }
    else{
        ext_gcd(b, a%b, d, y, x);
        y-=x*(a/b);
    }
}
ll inv(ll k)//求逆元
{
    ll x, y, d;
    ext_gcd(k, mod, d, x, y);
    return (x+mod)%mod;
}