快速冪一般都是用來快速遞推某個矩陣的n次冪的，一些題目的解法可以通過找規律遞推，構建遞推矩陣，就可以進行高次計算。

模板自用：

#include <iostream>
#include <vector>
using namespace std;

vector<vector<unsigned long long>> mul(vector<vector<unsigned long long>> a,vector<vector<unsigned long long>> b){
    int n=a.size();
    
 
int s=a[0].size();
    int m=b[0].size();
    vector<vector<unsigned long long>> c(n,vector<unsigned long long>(m,0));
    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
            for(int k=0;k<s;k++){
                c[i][j]+=a[i][k]*b[k][j];
            }
        }
    }
    
 
return c;
}

vector<vector<unsigned long long>> qmi1(vector<vector<unsigned long long>> a,int k){
    int n=a.size();
    vector<vector<unsigned long long>> res(n,vector<unsigned long long>(n,0));
    for(int i=0;i<n;i++){
        res[i][i]=1;
    }
    while
 
(k){
        if(k&1)res=mul(res,a);
        a=mul(a,a);
        k>>=1;
    }
    return res;
}

unsigned long long qmi2(unsigned long long a,unsigned long long k){
    unsigned long long res=1;
    while(k){
        if(k&1)res=res*1ull*a;
        a=a*1ull*a;
        k>>=1;
    }
    return res;
}

void out(vector<vector<unsigned long long>> a){
    for(int i=0;i<a.size();i++){
        for(int j=0;j<a[0].size();j++){
            cout<<a[i][j]<<" ";
        }
        cout<<endl;
    }
}

記錄一次快速冪