P3390 [TEMPLATE] rapid matrix power, p3390 Matrix

P3390 [TEMPLATE] rapid matrix power, p3390 Matrix
Background

Rapid matrix power

Description

Given n * n matrix A, evaluate A ^ k

Input/Output Format Input Format:

The first line, n, k

The number of n rows from 2nd to n + 1. The number of j rows in I + 1 indicates the elements in column j of row I in the matrix.

Output Format:

Output A ^ k

N rows in total, n numbers in each row, and j numbers in row I represent the elements in column j of row I. Each element is 10 ^ 9 + 7

Input and Output sample Input example #1:

2 11 11 1

Output sample #1:

1 11 1

Description

N <= 100, k <= 10 ^ 12, | matrix element | <= 1000 algorithm: rapid matrix power

 

Bare question !.

Note that the tmp value is accumulated when the matrix is multiplied.

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #define LL long long  6 using namespace std; 7 const int mod = 1e9+7; 8 LL n,k; 9 LL a[101][101];10 LL tmp[101][101];11 LL ans[101][101];12 void mul(LL a[][101],LL b[][101])13 {14     memset(tmp,0,sizeof(tmp));15     for(int i=1;i<=n;i++)16         for(int j=1;j<=n;j++)17             for(int k=1;k<=n;k++)18                 tmp[i][j]+=a[i][k]*b[k][j]%mod;19             20     for(int i=1;i<=n;i++)21         for(int j=1;j<=n;j++)22             a[i][j]=tmp[i][j]%mod; 23 }24 void fastpow(LL a[][101],LL k)25 {26     27     for(int i=1;i<=n;i++)ans[i][i]=1;28     while(k)29     {30         if(k%2)mul(ans,a);31         mul(a,a);32         k/=2;33     }34     for(int i=1;i<=n;i++)35     {36         for(int j=1;j<=n;j++)37             cout<<ans[i][j]%mod<<" ";38         printf("\n");39     }40         41 }42 int main()43 {44     cin>>n>>k;45     for(int i=1;i<=n;i++)46         for(int j=1;j<=n;j++)47             cin>>a[i][j];48     fastpow(a,k);49     return 0;50 }