矩阵快速幂

debug:重载乘号的时候要把两个传进来的矩阵用起来

// Problem: P3390 【模板】矩阵快速幂
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P3390
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
using namespace std;
#define ll long long
# define int long long
#define ull unsigned long long
#define pii pair<int,int>

#define baoliu(x, y) cout << fixed << setprecision(y) << x
#define endl  "\n"
#define debug1(x) cerr<<x<<" "
#define  debug2(x) cerr<<x<<endl
const int N = 2e5 + 10;
const int M = 1e6 + 10;
const int inf = 0x3f3f3f3f;
const int mod =1e9+7;
const double eps = 1e-8;
int n, m;
int k;
struct matrix{
	int c[101][101];
	matrix(){memset(c,0,sizeof c);}
};
matrix operator*(matrix &a,matrix &b){
	matrix res;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			for(int k=1;k<=n;k++){
				res.c[i][j]+=(a.c[i][k]*b.c[k][j])%mod;
				res.c[i][j]%=mod;
			}
		}
	}
	return res;
}
matrix pow(matrix &a,int k){
	matrix res;
	for(int i=1;i<=n;i++)res.c[i][i]=1;
	while(k){
		if(k&1)res=(res*a);
		a=a*a;
		k>>=1;
	}
	return res;
}
void solve(){
	cin>>n>>k;
	matrix a;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			cin>>a.c[i][j];
			//cerr<<a.c[i][j]<<" ";
		}
		//cerr<<endl;
	}
	matrix ans=pow(a,k);
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			cout<<ans.c[i][j]<<" ";
		}
		cout<<endl;
	}
	
}
signed main() {
    cin.tie(0);
    
    ios::sync_with_stdio(false);

    int t;
   //cin>>t;
     t=1;
    while (t--) {
solve();
    }
    return 0;
}

__EOF__

本文作者：爱飞鱼
本文链接：https://www.cnblogs.com/mathiter/p/18088356.html
关于博主：评论和私信会在第一时间回复。或者直接私信我。
版权声明：本博客所有文章除特别声明外，均采用 BY-NC-SA 许可协议。转载请注明出处！
声援博主：如果您觉得文章对您有帮助，可以点击文章右下角【推荐】一下。您的鼓励是博主的最大动力！