C - Reading comprehension 二分法 求等比数列前N项和

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. 
TechnicalSpecification 
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

等比数列二分求和：

（1）当时，

（2）当时，那么有

    

（3）当时，那么有

    

 

#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-9;
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 = 1, tmp = a;
    while (b != 0)
    {
        if (b & 1)
            ret = ret*tmp%m;
        tmp = tmp*tmp%m;
        b /= 2;
    }
    return ret%m;
}
LL sum(LL a, LL k)
{
    LL t, cur;
    if (k == 1)
        return a;
    else if (k % 2 == 1)
    {
        t = sum(a, k / 2),cur = fpow(a, k / 2 + 1);
        t = (t + t * cur%m) % m;
        t = (t + cur) % m;
    }
    else
    {
        t = sum(a, k / 2), cur = fpow(a, k / 2);
        t = (t + t * cur % m) % m;
    }
    return t;
}
LL solve(LL num)
{
    if (num == 0)
        return 1;
    else
        return (1 + sum(4, num)) % m;
}
int main()
{
    while (cin >> n >> m)
    {
        if (n & 1)
            cout << solve(n/2)%m << endl;
        else
            cout << solve((n - 1)/2)*2%m << endl;
    }
}