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 <stdio.h>
#include <inttypes.h>

int64_t N;
int X, M;

int Acnt = 0;
int A[114514];
int64_t Asum[114514];

int deta[114514];

int main(void) {
	int i;
	if (scanf("%" SCNd64 "%d%d", &N, &X, &M) != 3) return 1;
	if (N == 1) {
		printf ("%d\n", X);
		return 0;
	}
	A[0] = X;
	Asum[0] = A[0];
	deta[X] = 1;
	Acnt = 1;
	for (;;) {
		int tugi = (int)(((int64_t)A[Acnt - 1] * A[Acnt - 1]) % M);
		if (deta[tugi]) {
			int end = Acnt;
			int start = deta[tugi] - 1;
			int count = end - start;
			int offset = start;
			int64_t lcount = N - offset;
			int64_t lnum = lcount / count;
			int lleft = (int)(lcount % count);
			int64_t lsum = Asum[end - 1] - (start > 0 ? Asum[start - 1] : 0);
			int64_t answer = start > 0 ? Asum[start - 1] : 0;
			answer += lsum * lnum;
			if (start + lleft - 1 > 0) {
				answer += Asum[start + lleft - 1] - (start > 0 ? Asum[start - 1] : 0);
			}
			printf("%" PRId64 "\n", answer);
			return 0;
		} else {
			A[Acnt] = tugi;
			Asum[Acnt] = Asum[Acnt - 1] + A[Acnt];
			Acnt++;
			deta[tugi] = Acnt;
			if (Acnt == N) {
				printf("%" PRId64 "\n", Asum[Acnt - 1]);
				return 0;
			}
		}
	}
}