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; 
}

1 10

3 100

1

5

等比数列二分求和：

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstring>

#include<sstream>

#include<algorithm>

#include<queue>

#include<deque>

#include<iomanip>

#include<vector>

#include<cmath>

#include<map>

#include<stack>

#include<set>

#include<fstream>

#include<memory>

#include<list>

#include<string>

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

#define MAXN 18

#define N 33

#define MOD 10000007

#define INF 1000000009

const double eps = 1e-;

const double PI = acos(-1.0);

LL n, m, ans;

/*

等比数列二分求和

*/

LL fpow(LL a, LL b)

{

    LL ret = , tmp = a;

    while (b != )

    {

        if (b & )

            ret = ret*tmp%m;

        tmp = tmp*tmp%m;

        b /= ;

    }

    return ret%m;

}

LL sum(LL a, LL k)

{

    LL t, cur;

    if (k == )

        return a;

    else if (k %  == )

    {

        t = sum(a, k / ),cur = fpow(a, k /  + );

        t = (t + t * cur%m) % m;

        t = (t + cur) % m;

    }

    else

    {

        t = sum(a, k / ), cur = fpow(a, k / );

        t = (t + t * cur % m) % m;

    }

    return t;

}

LL solve(LL num)

{

    if (num == )

        return ;

    else

        return ( + sum(, num)) % m;

}

int main()

{

    while (cin >> n >> m)

    {

        if (n & )

            cout << solve(n/)%m << endl;

        else

            cout << solve((n - )/)*%m << endl;

    }

}