【luogu P3390 矩阵快速幂】 模板

题目链接：https://www.luogu.org/problemnew/show/P3390

首先要明白矩阵乘法是什么

对于矩阵A m*p  与  B p*n 的矩阵 得到C m*n 的矩阵

矩阵乘法满足结合律，但不满足交换律（所以可以套快速幂的板子）

进行矩阵乘法时要么重载*号，或者是写一个矩阵相乘的函数

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
struct Matrix{
    long long m[110][110];
}A,E;
long long n,k,mod = 1000000007;
Matrix mul(Matrix A,Matrix B)
{
    Matrix C;
    for(long long i = 0; i < n; i++)
        for(long long j = 0; j < n; j++)
        {
            C.m[i][j] = 0;
            for(long long k = 0; k < n; k++)
                C.m[i][j] = (C.m[i][j]+(A.m[i][k]*B.m[k][j])%mod)%mod;    
        }
    return C;
}
Matrix fast(Matrix A, long long k)
{
    Matrix S = E;
    while(k)
    {
        if(k&1) S = mul(S,A);
        A = mul(A,A);
        k = k>>1;
    }
    return S;
}
int main(){
    
    scanf("%lld%lld",&n,&k);
    for(long long i = 0; i < n; i++)
        for(long long j = 0; j < n; j++)
        scanf("%lld",&A.m[i][j]);
    
    for(long long i = 0; i < n; i++) E.m[i][i] = 1;
    Matrix ans = fast(A,k);
    for(long long i = 0; i < n; i++)
    {
        for(long long j = 0; j < n-1; j++)
        printf("%lld ",(ans.m[i][j])%mod);
        printf("%lld\n",(ans.m[i][n-1])%mod);
    }
    return 0;
}