poj 3233 Matrix Power Series Matrix summation

http://poj.org/problem?id=3233

Problem Solving Matrix Quick Power + Bisection Equivalence Sequence Summation

AC code

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <string>
#include <queue>
#include <map>
#include <vector>
using namespace std;
const int maxn = 35;
const int inf = 0x3f3f3f3f,mod = 1000000007;
const doubleepx = 1e- 10 ;
typedef long long ll;
struct Matrix
{
    int m[maxn][maxn]; //Does anyone know why the array will get a Runtime Error if the array is too large
};
Matrix A,I;
int n,m,k;
Matrix add(Matrix a,Matrix b)
{
    Matrix c;
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<n;j++)
        {
            c.m[i][j]=a.m[i][j]+b.m[i][j];
            c.m[i][j]%=m;
        }
    }
    return c;
}
Matrix multi(Matrix a,Matrix b)
{
    Matrix c;
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<n;j++)
        {
            c.m[i][j]=0;
            for(int k=0;k<n;k++)
            {
                c.m[i][j]+=a.m[i][k]*b.m[k][j];
            }
            c.m[i][j]%=m;
        }
    }
    return c;
}
Matrix poww(Matrix a,int n)
{
    Matrix ans=I;
    while(n>0)
    {
        if(n&1)
            years = multi(years,a);
        n=n>>1;
        a=multi(a,a);
    }
    return ans;
}
Matrix sum(Matrix a,int n)
{
    if(n==1) return a;
    Matrix t=sum(a,n/2);
    if(n&1)
    {
        Matrix cur=poww(a,n/2+1);
        t=add(t,multi(t,cur));
        t=add(t,cur);
    }
    else
    {
        Matrix cur=poww(a,n/2);
        t=add(t,multi(t,cur));
    }
    return t;
}
intmain ()
{
    cin>>n>>k>>m;
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<n;j++)
        {
            cin>>A.m[i][j];
            A.m[i][j]%=m;
            I.m[i][j]=(i==j);
        }
    }
    Matrix ans=sum(A,k);
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<n;j++)
        {
            cout<<ans.m[i][j]<<" ";
        }
        cout<<endl;
    }
    return 0;
}