Luogu P3390 [TEMPLATE] rapid matrix power, luogup3390 Matrix

Luogu P3390 [TEMPLATE] rapid matrix power, luogup3390 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 sample #1:

2 11 11 1

Output sample #1:

1 11 1

Description

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

 1 #include <cstdio> 2 #include <iostream> 3 using namespace std; 4 typedef long long ll; 5 ll x[999][999]; 6 ll ans[999][999]; 7 ll dx[999][999]; 8 const int p=1e9+7; 9 inline void anscf(int n)10 {11     for(int i=1;i<=n;i++)12      for(int j=1;j<=n;j++)13       dx[i][j]=ans[i][j],ans[i][j]=0;14 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        ans[i][j]=(ans[i][j]+(x[i][k]*dx[k][j])%p)%p;19 }20 inline void xcf(int n)21 {22     for(int i=1;i<=n;i++)23      for(int j=1;j<=n;j++)24       dx[i][j]=x[i][j],x[i][j]=0;25 26     for(int i=1;i<=n;i++)27      for(int j=1;j<=n;j++)28       for(int k=1;k<=n;k++)29        x[i][j]=(x[i][j]+(dx[i][k]*dx[k][j])%p)%p;30 }31 inline void fastpow(ll n,ll w)32 {33     while(w)34     {35         if(w%2==1) anscf(n);36         w/=2;37         xcf(n);38     }39 }40 int main()41 {42     ll n,k;43     scanf("%lld%lld",&n,&k);44     for(int i=1;i<=n;i++)45      for(int j=1;j<=n;j++)46       scanf("%d",&x[i][j]),ans[i][j]=x[i][j];47     fastpow(n,k-1);48     for(int i=1;i<=n;i++)49     {50       for(int j=1;j<=n;j++)51        printf("%lld ",ans[i][j]);52       puts("");53     }54 }