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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

int main() {
    ll N, X, M;
    cin >> N >> X >> M;
    ll ans = X;
    map<ll, int> exist;
    exist[X] = 0;
    vector<ll> cumul;
    cumul.push_back(ans);
    for(int i = 1; i < N; i++){
        X = ((X % M) * (X % M)) % M;
        if(exist.count(X)){
            int temp = exist[X];
            ans = cumul[temp - 1] + ((cumul[i - 1] - cumul[temp - 1]) * ((N - temp) / (i - temp)));
            ans += cumul[(N - temp) % (i - temp) + temp - 1] - cumul[temp - 1];
            break;
        }
        ans += X;
        cumul.push_back(ans);
        exist[X] = i;
    }
    cout << ans << endl;
    return 0;
}