【模板】矩阵快速幂（洛谷P3390）

Description

　　给定\(n*n\)的矩阵\(A\)，求\(A^k\)

Input

　　第一行，\(n\),\(k\)
　　第\(2\)至\(n+1\)行，每行\(n\)个数，第\(i+1\)行第\(j\)个数表示矩阵第\(i\)行第\(j\)列的元素

Output

　　输出\(A^k\)
　　共\(n\)行，每行\(n\)个数，第\(i\)行第\(j\)个数表示矩阵第\(i\)行第\(j\)列的元素，每个元素模\(10^9+7\)

Solution

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int Mod=1e9+7;
int n;
long long k,ans[110][110],a[110][110],c[110][110];
long long mo(long long a,long long b)
{
	return a+b>=Mod?a+b-Mod:a+b;
}
int main()
{
	scanf("%d%lld",&n,&k);
	for (int i=1;i<=n;i++) ans[i][i]=1;
	for (int i=1;i<=n;i++) 
		for (int j=1;j<=n;j++)
			scanf("%lld",&a[i][j]);
	while (k)
	{
		if (k&1)
		{
			memset(c,0,sizeof(c));
			for (int i=1;i<=n;i++)
				for (int j=1;j<=n;j++)
					for (int k=1;k<=n;k++)
						c[i][j]=mo(c[i][j],ans[i][k]*a[k][j]%Mod);
			for (int i=1;i<=n;i++)
				for (int j=1;j<=n;j++)
					ans[i][j]=c[i][j];
		}
		k>>=1;
		memset(c,0,sizeof(c));
		for (int i=1;i<=n;i++)
			for (int j=1;j<=n;j++)
				for (int k=1;k<=n;k++)
					c[i][j]=mo(c[i][j],a[i][k]*a[k][j]%Mod);
		for (int i=1;i<=n;i++)
			for (int j=1;j<=n;j++)
				a[i][j]=c[i][j];
	}
	for (int i=1;i<=n;i++)
	{
		for (int j=1;j<n;j++)
			printf("%lld ",ans[i][j]);
		printf("%lld\n",ans[i][n]);
	}
	return 0;
}