快速幂(整数 + 矩阵)

一、整数m ^ n  % k 的快速幂:

ll quickpow(ll m, ll n , ll k){
    ll   ans = 1;
    while(n){
        if(n & 1)//如果n是奇数
            ans = (ans * m) % k;
        n = n >> 1;//位运算“右移1类似除2”
        m = (m * m) % k;
    }
    return ans;
}
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二、矩阵 M ^ n % mod 的快速幂:

struct Matrix
{
    ll m[MAXN][MAXN]; //二维数组存放矩阵
    Matrix(ll num[MAXN][MAXN])
    {
        for(int i = 0 ; i < MAXN ; i++)
            for(int j = 0 ; j < MAXN ; j++)
                m[i][j] = num[i][j];
    }  //对数组的初始化
    Matrix() {}
};

Matrix operator * (Matrix &m1, Matrix &m2)
{
    int i, j, k;
    Matrix temp;
    for (i = 0; i < MAXN; i++)
    {
        for (j = 0; j < MAXN; j++)
        {
            temp.m[i][j] = 0;
            for(k = 0 ; k < MAXN ; k++)
                temp.m[i][j] += (m1.m[i][k] * m2.m[k][j]) % mod;
            temp.m[i][j] %= mod; //注意每一步都进行取模
        }
    }
    return temp;
}

Matrix quickpow(Matrix &M , ll n, ll mod)
{
    Matrix tempans(base);  //初始化为单位矩阵
    while(n)
    {
        if(n & 1)
            tempans = tempans * M; //已经重载了*
        n = n >> 1;
        M = M * M;
    } //快速幂(类似整数)
    return tempans;
}
  
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原文地址:https://www.cnblogs.com/sunus/p/4404818.html