快速幂

#include<iostream>
#include<cstdio>
#define int unsigned long long
using namespace std;
int a,b,q;
int qsm(int a,int b,int q)//快速幂
{
    int ans=1;
    while(b)
    {
        if(b&1)ans=(ans*a)%q;
        a=(a*a)%q;
        b>>=1;  //二进制运算，相当于b/=2;
    }
    return ans%q;
}
signed main()
{
    cin>>a>>b>>q;
    cout<<a<<"^"<<b<<" mod "<<q<<"="<<qsm(a,b,q);
    return 0;
}

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cctype>
#include<cstring>
#include<cmath>
using namespace std;
inline long long read(){
	long long x=0;bool w=0;char c=getchar();
	while(!isdigit(c))w|=c=='-',c=getchar();
	while(isdigit(c))x=x*10+(c^48),c=getchar();
	return w?-x:x;
}
namespace star
{
	const int maxn=105,mod=1e9+7;
	struct mat{
		int n,a[maxn][maxn];
		mat(){}
		mat(const int &x){n=x;memset(a,0,sizeof a);}
		inline void set(){for(int i=1;i<=n;i++) a[i][i]=1;}
		inline int* operator [] (const int &x) {return a[x];}
		inline const int* operator [] (const int &x) const {return a[x];}
		inline mat operator * (const mat &b) const{
			mat ans(n);
			for(int i=1;i<=n;i++)
			for(int j=1;j<=n;j++)
			for(int k=1;k<=n;k++)
			ans[i][j]=(ans[i][j]+1ll*a[i][k]*b[k][j]%mod)%mod;
			return ans;
		}
	}now;
	inline mat fpow(mat a,long long b){
		mat ans(a.n);
		ans.set();
		for(;b;b>>=1,a=a*a)if(b&1)ans=ans*a;
		return ans;
	}
	int n;
	long long k;
	inline void work(){
		n=now.n=read();k=read();
		for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) now[i][j]=read();
		now=fpow(now,k);
		for(int i=1;i<=n;i++) {
			for(int j=1;j<=n;j++) printf("%d ",now[i][j]);
			puts("");
		}
	}
}
signed main(){
	star::work();
	return 0;
}