复数快速幂【模板】

实数范围内的乘法都是满足结合律的，

像 $1+sqrt2,1+2i, 1+2sqrt2 i$ 这些没有直接可用的乘法，但是相乘后形式保持不变，所以可以模拟乘法。

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
const ll mod = 1e9 + 7;
struct Complex{     //a+bi
    ll a, b;
    Complex(ll _a, ll _b)
    {
        a = _a; b = _b;
    }
};

Complex mul(Complex x, Complex y, ll mod)
{
    ll a = ((x.a*y.a - x.b*y.b)%mod + mod) % mod;
    //ll a = ((x.a*y.a - 5*x.b*y.b)%mod + mod) % mod;  //a+bsqrt5_i
    ll b = (x.a*y.b + x.b*y.a)%mod;
    return Complex(a, b);
}

Complex cpow(Complex x, ll n, ll mod)
{
    Complex ret(1, 0);
    while(n)
    {
        if(n&1)  ret = mul(ret, x, mod);
        x = mul(x, x, mod);
        n >>= 1;
    }
    return ret;
}


int main()
{
    Complex a(1, 2);
    Complex ans = cpow(a, 3, 100);
    printf("%lld %lld
", ans.a, ans.b);
}