hdu4990矩阵快速幂

Read the program below carefully then answer the question. 
#pragma comment(linker, "/STACK:1024000000,1024000000") 
#include <cstdio> 
#include<iostream> 
#include <cstring> 
#include <cmath> 
#include <algorithm> 
#include<vector> 

const int MAX=100000*2; 
const int INF=1e9; 

int main() 
{ 
  int n,m,ans,i; 
  while(scanf("%d%d",&n,&m)!=EOF) 
  { 
    ans=0; 
    for(i=1;i<=n;i++) 
    { 
      if(i&1)ans=(ans*2+1)%m; 
      else ans=ans*2%m; 
    } 
    printf("%d\n",ans); 
  } 
  return 0; 
}

Input

Multi test cases，each line will contain two integers n and m. Process to end of file. 
[Technical Specification] 
1<=n, m <= 1000000000

Output

For each case，output an integer，represents the output of above program.

Sample Input

1 10
3 100

Sample Output

1
5

 f[n]=2*f[n-1]+1=f[n-1]+f[n-1]+1=f[n-1]+2*f[n-2]+1

f[n]=2*f[n-1]=f[n-1]+f[n-1]=f[n-1]+2*f[n-2]+1;

#include <cstdio>
#include <cstring>
#include <iostream>
#define ll long long
using namespace std;
ll n,m;
struct matrix
{
    ll mat[15][15];
    matrix()
    {
        memset(mat,0,sizeof(mat));
    }
};
matrix mul(matrix B,matrix A)
{
    int i,j,k;
    matrix C;
    for(i=1;i<=3;i++)
        for(j=1;j<=3;j++)
        for(k=1;k<=3;k++)
        C.mat[i][j]=(C.mat[i][j]+B.mat[i][k]*A.mat[k][j])%m;
    return C;

}
matrix pow(matrix A,ll p)
{
    matrix ans;
    for(int i=1;i<=3;i++)
        ans.mat[i][i]=1;
    while(p)
    {
        if(p&1)
            ans=mul(ans,A);
        A=mul(A,A);
        p>>=1;
    }
    return ans;
}
int main()
{
    while(~scanf("%lld%lld",&n,&m))
    {
        matrix a,b;
        a.mat[1][1]=1;
        a.mat[1][2]=2;
        a.mat[1][3]=1;
        a.mat[2][1]=1;
        a.mat[3][3]=1;
        b.mat[1][1]=1;
        b.mat[2][1]=0;
        b.mat[3][1]=1;
        a=pow(a,n-1);
        b=mul(a,b);
        printf("%lld\n",b.mat[1][1]%m);


    }
    return 0;
}