Reading comprehension HDU - 4990

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; 
}

InputMulti test cases，each line will contain two integers n and m. Process to end of file. 
[Technical Specification] 
1<=n, m <= 1000000000OutputFor each case，output an integer，represents the output of above program.Sample Input

1 10
3 100

Sample Output

1
5

直接利用源程序暴力打出 1，2，5，10，21，42 找出规律 fn = fn-1 + 2*fn-2+1

数据比较大，直接求矩阵快速幂。

推出 转化矩阵为：

1 2 1
1 0 0
0 0 1

　初始矩阵为

1 2 1

　直接上代码：

//Asimple
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <queue>
#include <vector>
#include <string>
#include <cstring>
#include <stack>
#include <set>
#include <map>
#include <cmath>
#define swap(a,b,t) t = a, a = b, b = t
#define CLS(a, v) memset(a, v, sizeof(a))
#define test() cout<<"============"<<endl
#define debug(a)  cout << #a << " = "  << a <<endl
#define dobug(a, b)  cout << #a << " = "  << a << " " << #b << " = " << b << endl
using namespace std;
typedef long long ll;
const int N=;
//const ll MOD=10000007;
const int INF = ( << );
const double PI=atan(1.0)*;
const int maxn = +;
const ll mod = ;
int n, m, len, ans, sum, v, w, T, num;
int MOD;

struct Matrix {
    long long grid[N][N];
    int row,col;
    Matrix():row(N),col(N) {
        memset(grid, , sizeof grid);
    }
    Matrix(int row, int col):row(row),col(col) {
        memset(grid, , sizeof grid);
    }

    //矩阵乘法
    Matrix operator *(const Matrix &b) {
        Matrix res(row, b.col);
        for(int i = ; i<res.row; i++)
            for(int j = ; j<res.col; j++)
                for(int k = ;k<col; k++)
                    res[i][j] = (res[i][j] + grid[i][k] * b.grid[k][j] + MOD) % MOD;
        return res;
    }

    //矩阵快速幂
    Matrix operator ^(long long exp) {
        Matrix res(row, col);
        for(int i = ; i < row; i++)
            res[i][i] = ;
        Matrix temp = *this;
        for(; exp > ; exp >>= , temp = temp * temp)
            if(exp & ) res = temp * res;
        return res;
    }

    long long* operator[](int index) {
        return grid[index];
    }

    void print() {

        for(int i = ; i <row; i++) {
            for(int j = ; j < col-; j++)
                printf("%d ",grid[i][j]);
            printf("%d\n",grid[i][col-]);
        }
    }
};

void input(){
    ios_base::sync_with_stdio(false);
    while( cin >> n >> MOD ) {
        Matrix A;
        A[][] = A[][] = ;
        A[][] = ;
        A[][] = A[][] = ;
        A = A^n;
        Matrix B;
        B[][] = B[][] = ;
        B[][] = ;
        A = A*B;
        if( n% ) cout << A[][] << endl;
        else cout << A[][] << endl;
    }
}

int main(){
    input();
    return ;
}