将指数转化为用2进制表示的数

实数快速幂

LL qmi(int a, int b, int p)
{
    LL res = 1;
    while (b)
    {
        if (b & 1) res = res * a % p; // 如果幂次的二进制表示的这一位上是1
        a = (LL) a * a % p; // 注意可能爆int范围
        b >>= 1;
    }

    return res;
}

矩阵快速幂

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

typedef long long LL;

const int N = 110, mod = 1e9 + 7;

struct Matrix
{
    int n;
    int v[N][N];

    Matrix operator * (const Matrix &b) const // 重载一下乘法运算符
    {
        Matrix c;
        memset(c.v, 0, sizeof c.v); // 记得一定一定一定要清空
        c.n = n;
        for (int i = 1; i <= n; i ++)
            for (int j = 1; j <= n; j ++)
                for (int k = 1; k <= n; k ++)
                    c.v[i][j] = ((LL) v[i][k] * b.v[k][j] % mod + c.v[i][j]) % mod;
        return c;
    }
}a, res;

int n;
LL k;

int main()
{
    scanf ("%d %lld", &n, &k);

    for (int i = 1; i <= n; i ++)
        for (int j = 1; j <= n; j ++)
        {
            scanf ("%d", &a.v[i][j]);
            if (i == j) res.v[i][i] = 1; // 初始化res为单位矩阵
        }
    a.n = res.n = n;

    while (k)
    {
        if (k & 1) res = a * res;
        a = a * a;
        k >>= 1;
    }

    for (int i = 1; i <= n; i ++)
    {
        for (int j = 1; j <= n; j ++)
            printf ("%d ", res.v[i][j]);
        puts("");
    }

    return 0;
}

矩阵快速幂加速斐波那契数列求解

构造一个基本矩阵,然后将求斐波那契数列时的递推加法变成矩阵快速幂加速

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

typedef long long LL;

const int mod = 1e9 + 7;

struct Fib
{
    int v[3][3];

    Fib operator * (const Fib &b) const
    {
        Fib c;
        memset(c.v, 0, sizeof c.v);

        for (int i = 1; i <= 2; i ++)
            for (int j = 1; j <= 2; j ++)
                for (int k = 1; k <= 2; k ++)
                    c.v[i][j] = ((LL) v[i][k] * b.v[k][j] % mod + c.v[i][j]) % mod;
        return c;
    }
}a, res;

LL n;

void qmi(LL k)
{
    while (k)
    {
        if (k & 1) res = res * a;
        a = a * a;
        k >>= 1;
    }
}

int main()
{
    scanf ("%lld", &n);

    if (n <= 2) 
    {
        printf ("1");
        return 0;
    }

    a.v[1][1] = a.v[1][2] = a.v[2][1] = 1;
    res.v[1][1] = res.v[1][2] = 1;

    qmi(n - 2);

    printf ("%lld", res.v[1][1]);
    return 0;
}