int v[MAXN];//每个数的最小质因子
int prime[MAXN];
int cnt;
void get_list(){
    for(int i=2;i<=maxn;i++){
        if(!not_prime[i]){
            primes[++cnt] = i;
        }
        for(int j = 1;i<=cnt&&i*prime[j]<MAX;j++){
            not_prime[i*prime[j]] = true;
            if(i%primes[j]==0){//即比一个合数大的质数和该合数的乘积可用一个更大的合数和比其小的质数相乘得到
                break;
            }
        }
    }
}

int factors[log2N];
int pows[log2N];
void divide(int x){
    int cnt = 0;
    for(int i = 2;i*i<=x;i++){
        if(x%1==0){
            factor[++cnt] = i;
            pows[cnt]=0;
            while(x%i==0){
                x/=i;
                pows[cnt]++;//去除所有的i
            }
        }
    }
    if(x>1){//若此时x已经被分解来只剩一个质数了
        factor[++cnt]=x;
        pows[cnt]=1;
    }
    for(int i = 1;i<=cnt;i++){
        cout<<prime[i]<<'^'<<pows[i]<<endl;//分解完了，输出就行
    }
}

vector<int> factor[N];
for(int i = 1;i<=n;i++)
    for(int j = 1;j<=n/i;j++)
        facotr[i*j].push_back(i);

int gcd(int a,int b){
    int temp = b;
    while(b!=0){
        temp = b;
        b = a%b;
        a = temp;
    }
    return a;
}

long long x0,y0;
long long d;
void extend_gcd(long long a,long long b){
	if(b==0){
		x0 = 1;
		y0 = 0;
		d = a;
	}else{
		extend_gcd(b,a%b);
		long long t = x0;
		x0 = y0;
		y0 = t-(a/b)*y0;
	}
}

# include<iostream>
# include<string.h>
using namespace std;
long long e[105][105];
long long m[105][105];
long long temp[105][105];
int n;
const int mod = (int)1e9+7;
void mul(long long a[105][105],long long b[105][105]){
	memset(temp,0,sizeof(temp));
	for(int i = 1;i<=n;i++){
		for(int j = 1;j<=n;j++){
			for(int k = 1;k<=n;k++){
				temp[i][j] += (a[i][k]*b[k][j])%mod;
				temp[i][j]%=mod;
			}
		}
	}	
	for(int i = 1;i<=n;i++)
		for(int j = 1;j<=n;j++)
			a[i][j] = temp[i][j];
}
int main(void){
	ios::sync_with_stdio(false);
	long long k;
	cin>>n>>k;
	for(int i = 1;i<=n;i++)
		for(int j = 1;j<=n;j++){
			cin>>m[i][j];
			e[i][i] = 1;
		}
		
	while(k){
		if(k&1){
			mul(e,m);
		}
		mul(m,m);
		k>>=1;
	}
	for(int i = 1;i<=n;i++){
		for(int j = 1;j<=n;j++)
			cout<<e[i][j]<<' ';
	cout<<endl;	
	}
	return 0;
}

void pre()  
{  
    fac[0] = 1;  
    for(int i = 1; i <= MAX; i++)  
        fac[i] = (fac[i - 1] * i) % MOD; //处理f[n]的逆元 
    inv_fac[MAX] = qpow(fac[MAX], MOD - 2);  
    for(int i = MAX - 1; i >= 0; i--)  
        inv_fac[i] = (inv_fac[i + 1] * (i + 1)) % MOD;   
} 
--------------------- 
作者：超超越 
来源：CSDN 
原文：https://blog.csdn.net/kiwi_berrys/article/details/54834977 
版权声明：本文为博主原创文章，转载请附上博文链接！