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 INF 0x3f3f3f3f3f3f3f3f
#define LL long long
#define FOR(i,j,k) for(LL (i) = j;(i) <= (k);i++)
using namespace std;
const LL MaN=1e5+10;
LL N,X,M,Ans;
LL Loc[MaN],Pre[MaN];
signed main() {
    cin>>N>>X>>M;
    Ans+=X;
    Pre[1]=X; 
    Loc[X]=1;
    LL Sta=INF;
    FOR(i,2,N) {
        X=X*X%M;
        Pre[i]=Pre[i-1]+X;
        Ans+=X;
        if(Loc[X]) { 
			LL Len=i-Loc[X]; 
			Ans+=(N-i)/Len*(Pre[i]-Pre[Loc[X]]); 
			Sta = N - ( N - i ) % Len+1; 
			break; 
		}
        else Loc[X]=i;
    }
    FOR(i,Sta,N) {
        X = X * X % M;
        Ans += X;
    }
    cout << Ans << endl;
    return 0;
}