Problem:
Score : 500 points

Problem Statement

Let us denote by f(x, m) the remainder of the Euclidean division of x by m.

Let A be the sequence that is defined by the initial value A_1=X and the recurrence relation A_{n+1} = f(A_n^2, M).
Find \displaystyle{\sum_{i=1}^N A_i}.

Constraints

1 \leq N \leq 10^{10}

0 \leq X < M \leq 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N X M

Output

Print \displaystyle{\sum_{i=1}^N A_i}.

Sample Input 1

6 2 1001

Sample Output 1

1369

The sequence A begins 2,4,16,256,471,620,\ldots Therefore, the answer is 2+4+16+256+471+620=1369.

Sample Input 2

1000 2 16

Sample Output 2

6

The sequence A begins 2,4,0,0,\ldots Therefore, the answer is 6.

Sample Input 3

10000000000 10 99959

Sample Output 3

492443256176507

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); i++)
#define irep(i, n) for (int i = (n); i >= 0; i--)
using namespace std;
using ll = long long;
using P = pair<int, int>;
const int INF = 1 << 25;
const int MOD = 1e9+7;
ll n,m,x;
ll f(ll a) {
  return a*a % m;
}  

int main() {
  cin >> n >> x >> m;
  vector<int> id(m,-1);
  vector<int> seen;
  ll tot = 0;
  int len = 0;
  while (id[x] == -1)
  {
    seen.push_back(x);
    id[x] = len;
    len++;
    tot += x;
    x = f(x);
  }
  
  int c = len - id[x];
  ll s =  0;
  for (int i = id[x]; i < len; i++) s += seen[i];

  ll ans = 0;
  if (n <= len) {
    rep(i,n) ans += seen[i];
  } else {
    ans += tot;
    n -= len; 
    ans += s*(n/c);
    n %= c;
    rep(i,n) ans += seen[id[x]+i];
  }
  cout << ans << endl;
  return 0;
}