矩阵快速幂（hdu4990）

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

思路：可以得出下面的地推公式

 代码：
 

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#define eps 1e1000000000000
 
using namespace std;

 
typedef long long LL;
 
struct Matrix
{
    LL m[5][5];
}I,B,A,T;//I单位矩阵，B原始矩阵，A取N次幂，T存储结果
 
LL a,b,n,mod;
int ssize=3;
 
Matrix Mul(Matrix a,Matrix b)
{
    int i,j,k;
    Matrix c;
    for(i=1;i<=ssize;i++)
    {
        for(j=1;j<=ssize;j++)
        {
            c.m[i][j]=0;
            for(k=1;k<=ssize;k++)
            {
                c.m[i][j]+=(a.m[i][k]*b.m[k][j]);
                c.m[i][j]%=mod;
            }
        }
    }
    return c;
}
 
Matrix quickpow(Matrix A,int n)
{
    Matrix m=A,b=I;
    while(n)
    {
        if(n&1)
            b=Mul(b,m);
        m=Mul(m,m);
        n>>=1;
    }
    return b;
}
 
int main()
{
 
    LL x,y;
    while(cin >> x >> y)
    {
        memset(I.m,0,sizeof(I.m));
        memset(A.m,0,sizeof(A.m));
        memset(B.m,0,sizeof(B.m));
        for(int i=0;i<=ssize;i++)
        {
            I.m[i][i]=1;
        }
        if(x==1)
        {
            cout << 1%mod << endl;
            continue;
        }
        else if(x==2)
        {
            cout << 2%mod << endl;
            continue;
        }
        mod=y;
        n=x;
        B.m[1][1]=1;
        B.m[1][2]=2;
        B.m[1][3]=A.m[2][1]=A.m[2][2]=A.m[3][2]=A.m[3][3]=1;
        A.m[1][2]=2;
        Matrix tmp;
        tmp=quickpow(A,n-2);
        B=Mul(B,tmp);
        cout << B.m[1][2]%mod << endl;
    }
    return 0;
}