Poj3233Matrix Power Series)

Poj3233Matrix Power Series)

Matrix Power Series Time Limit:3000 MS Memory Limit:131072KB 64bit IO Format:% I64d & % I64uSubmit Status

Description

GivenN×NMatrixAAnd a positive integerK, Find the sumS=A+A2 +A3 +... +Ak.

Input

The input contains exactly one test case. The first line of input contains three positive integersN(N≤ 30 ),K(K≤ 109) andM(M<104). Then followNLines each containingNNonnegative integers below 32,768, givingA'S elements in row-major order.

Output

Output the elementsSModuloMIn the same wayAIs given.

Sample Input

2 2 40 11 1

Sample Output

1 22 3

First, Set B = A E as the unit matrix.

0 E

Then B ^ (n + 1) = A ^ n + A * (n-1)... A + 1

0 E Note: When the remainder is obtained, it must be judged. If I = j at this time, we should first-1 and then get the remainder and then + 1, because there is an additional unit matrix.

#include 
 
  #include 
  
   #include using namespace std ;#define LL __int64struct node{    LL k[32][32] ;    int n ;};struct node1{    node a , b , c , d ;};node mul(node p,node q,int m){    int i , j , l ;    node s ;    s.n = p.n ;    for(i = 0 ; i < p.n ; i++)        for(j = 0 ; j < p.n ; j++)        {            s.k[i][j] = 0 ;            for(l = 0 ; l < p.n ; l++)            {                if(i == j)                    s.k[i][j] = (s.k[i][j] + p.k[i][l]*q.k[l][j]-1)%m+1 ;                else                    s.k[i][j] = (s.k[i][j] + p.k[i][l]*q.k[l][j])%m ;            }        }    return s ;}node add(node p,node q,int m){    int i , j ;    for(i = 0 ; i < p.n ; i++)        for(j = 0 ; j < p.n ; j++)            {                if(i == j)                    p.k[i][j] = (p.k[i][j] + q.k[i][j]-1 )%m+1;                else                    p.k[i][j] = (p.k[i][j] + q.k[i][j] )%m;            }    return p ;}node1 pow(node1 o,int k,int m){    if( k == 1 )        return o ;    node1 temp = pow(o,k/2,m) , s ;    node p , q ;    p = mul(temp.a,temp.a,m) ; q = mul(temp.b,temp.c,m) ;    s.a = add( p,q,m ) ;    p = mul(temp.a,temp.b,m) ; q = mul(temp.b,temp.d,m) ;    s.b = add( p,q,m ) ;    p = mul(temp.c,temp.a,m) ; q = mul(temp.d,temp.c,m) ;    s.c = add( p,q,m ) ;    p = mul(temp.c,temp.b,m) ; q = mul(temp.d,temp.d,m) ;    s.d = add( p,q,m ) ;    temp = s ;    if( k%2 )    {        p = mul(temp.a,o.a,m) ; q = mul(temp.b,o.c,m) ;        s.a = add( p,q,m ) ;        p = mul(temp.a,o.b,m) ; q = mul(temp.b,o.d,m) ;        s.b = add( p,q,m ) ;        p = mul(temp.c,o.a,m) ; q = mul(temp.d,o.c,m) ;        s.c = add( p,q,m ) ;        p = mul(temp.c,o.b,m) ; q = mul(temp.d,o.d,m) ;        s.d = add( p,q,m ) ;    }    return s ;}int main(){    int n , k , m ;    int i , j ;    node1 p , s ;    while( scanf("%d %d %d", &n, &k, &m) != EOF )    {        p.a.n = p.b.n = p.c.n = p.d.n = n ;        for(i = 0 ; i < n ; i++)        {            for(j = 0 ; j < n ; j++)            {                scanf("%I64d", &p.a.k[i][j]) ;                p.b.k[i][j] = p.c.k[i][j] = p.d.k[i][j] = 0 ;            }            p.b.k[i][i] = p.d.k[i][i] = 1 ;        }        s = pow(p,k+1,m) ;        for(i = 0 ; i < n ; i++)        {            for(j = 0 ; j < n ; j++)            {                if( i == j )                    printf("%I64d", s.b.k[i][j]-1) ;                else                    printf("%I64d", s.b.k[i][j]);                if( j == n-1 )                    printf("\n") ;                else                    printf(" ") ;            }        }    }    return 0;}
  
 

Method 2

When n is an even number, A + A ^ 2 + A ^ 3 ...... A ^ n = (A + A ^ 2 + A ^ 3... A ^ (k/2) + A ^ (k/2) * (A + A ^ 2 + A ^ 3... A ^ (k/2 ))

When n is an odd number, A + A ^ 2 + A ^ 3 ...... A ^ n = (A + A ^ 2 + A ^ 3... A ^ (k/2) + A ^ (k/2 + 1) + A ^ (k/2 + 1) * (A + A ^ 2 + A ^ 3... A ^ (k/2 ))

#include 
 
  #include 
  
   #include using namespace std ;#define LL __int64struct node{    LL a[32][32] ;    int n ;};node mul(node p,node q,int m){    node s ;    s.n = p.n ;    int i , j , k ;    for(i = 0 ; i < p.n ; i++)        for(j = 0 ; j < p.n ; j++)        {            s.a[i][j] = 0 ;            for(k = 0 ; k < p.n ; k++)                s.a[i][j] = ( s.a[i][j] + p.a[i][k]*q.a[k][j] ) % m ;        }    return s ;}node add(node p,node q,int m){    int i , j ;    node s ;    s.n = p.n ;    for(i = 0 ; i < p.n ; i++)        for(j = 0 ; j < p.n ; j++)            s.a[i][j] = ( p.a[i][j] + q.a[i][j] ) % m ;    return s ;}node pow(node p,int k,int m){    if( k == 1 )        return p ;    node s = pow(p,k/2,m) ;    s = mul(s,s,m) ;    if( k%2 )        s = mul(s,p,m) ;    return s ;}node f(node p,int k,int m){    if( k == 1 )        return p ;    node s = f(p,k/2,m) , q , temp ;    int i , j ;    for(i = 0 , temp.n = p.n; i < p.n ; i++)        for(j = 0 ; j < p.n ; j++)        {            if( i == j )                temp.a[i][j]  = 1 ;            else                temp.a[i][j] = 0 ;        }    if( k%2 )    {        q = pow(p,k/2+1,m) ;        s = add( q, mul( s,add(q,temp,m),m ) ,m ) ;    }    else    {        q = pow(p,k/2,m) ;        s = mul(s, add(q,temp,m) ,m);    }    return s ;}int main(){    int n , k , m ;    int i , j ;    node p , s ;    while( scanf("%d %d %d", &n, &k, &m) != EOF )    {        p.n = n ;        for(i = 0 ; i < n ; i++)            for(j = 0 ; j < n ; j++)                scanf("%I64d", &p.a[i][j]) ;        s = f(p,k,m) ;        for(i = 0 ; i < n ; i++)        {            for(j = 0 ; j < n ; j++)            {                if( j == n-1 )                    printf("%I64d\n", s.a[i][j]) ;                else                    printf("%I64d ", s.a[i][j]) ;            }        }    }    return 0;}