矩阵快速幂

矩阵快速幂

给定n*n的矩阵A，求A^k

Code：

#include <cstdio>
#include <cstring>
#define ll long long
const ll N=102;
const ll mod=1e9+7;
ll k;int n;
struct matrix
{
    ll dx[N][N];
    matrix()
    {
        memset(dx,0,sizeof(dx));
    }
    matrix(ll d)
    {
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                dx[i][j]=d;
    }
    matrix friend operator *(matrix n1,matrix n2)
    {
        matrix n3;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                for(int k=1;k<=n;k++)
                    n3.dx[i][j]=(n3.dx[i][j]+n1.dx[i][k]*n2.dx[k][j])%mod;
        return n3;
    }
}B,ANS;
void quick_pow(matrix A,ll k)
{
    while(k)
    {
        if(k&1)
            ANS=A*ANS;
        A=A*A;
        k>>=1;
    }
}
void init()
{
    scanf("%d%lld",&n,&k);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            scanf("%lld",&B.dx[i][j]);
}
void work()
{
    for(int i=1;i<=n;i++)
        ANS.dx[i][i]=1;
    quick_pow(B,k);
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
            printf("%lld ",ANS.dx[i][j]);
        printf("\n");
    }
}
int main()
{
    init();
    work();
    return 0;
}