poj 3233 Matrix Power Series

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Description

Given a n × n matrix A and a positive integer k, find the sum S = A + A² + A³ + … + A^k.

Input

The input contains exactly one test case. The first line of input contains three positive integers n (n ≤ 30), k (k ≤ 10⁹) and m (m < 10⁴). Then follow n lines each containing n nonnegative integers below 32,768, giving A’s elements in row-major order.

Output

Output the elements of S modulo m in the same way as A is given.

Sample Input

2 2 40 11 1

Sample Output

1 22 3

代码：

#include <iostream>#include <cstdio>#include <cstring>using namespace std;struct mat{    int t[65][65];    void set()    {        memset(t,0,sizeof(t));    }} a,b;mat multiple(mat a,mat b,int n,int p){    int i,j,k;    mat temp;    temp.set();    for(i=0; i<n; i++)        for(j=0; j<n; j++)        {            if(a.t[i][j]>0)                for(k=0; k<n; k++)                    temp.t[i][k]=(temp.t[i][k]+a.t[i][j]*b.t[j][k])%p;        }    return temp;}mat quick_mod(mat b,int n,int m,int p){    mat t;    t.set();    for(int i=0;i<n;i++) t.t[i][i]=1;    while(m)    {        if(m&1)        {            t=multiple(t,b,n,p);        }        m>>=1;        b=multiple(b,b,n,p);    }    return t;}void init(int n,int m,int p){    a.set();    b.set();    for(int i=0;i<n;i++)    for(int j=0;j<n;j++)    {        scanf("%d",&b.t[i][j]);        b.t[i][j+n]=b.t[i][j];        a.t[i][j]=b.t[i][j];    }    for(int i=n;i<2*n;i++)    for(int j=n;j<2*n;j++)    if(i==j) b.t[i][j]=1;    for(int i=n;i<2*n;i++)    for(int j=0;j<n;j++)    if(j+n==i) a.t[i][j]=1;    b=quick_mod(b,2*n,m-1,p);    mat temp;    temp.set();    for(int i=0; i<n; i++)        for(int j=0; j<n; j++)        {                for(int k=0; k<2*n; k++)                    temp.t[i][j]=(temp.t[i][j]+b.t[i][k]*a.t[k][j])%p;   //错误一点，无限WA        }   for(int i=0;i<n;i++)   {       for(int j=0;j<n;j++)       printf("%d ",temp.t[i][j]);       puts("");   }}int main(){    int n,m,p;    while(cin>>n>>m>>p)    {        init(n,m,p);    }    return 0;}

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