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>
	long long a[200000]={0};
	long long b[200000]={0};
	long long c[200000]={0};
    long long sum[200000]={0};
int main(){
	long long n;
	int x,m;
	scanf("%lld %d %d",&n,&x,&m);
	long long ans=0;

	a[1]=x;
	sum[1]=x;
	b[x]=1;
	c[x]=1;
	int i=2;
	while (i<=n){
		a[i]=(a[i-1]*a[i-1])%m;
		sum[i]=sum[i-1]+a[i];
		if (b[a[i]]==0){
			b[a[i]]=1;
			c[a[i]]=i;
			i++;
			continue;
		}else{
		  	int len=i-c[a[i]];
		  	ans=sum[c[a[i]]-1];
		  	long long everycirclesum=sum[i-1]-sum[c[a[i]]-1];
		  	long long times=(n-c[a[i]]+1)/len;
		  	ans+=everycirclesum*times;
		  	long long rest=(n-c[a[i]]+1)%len;
		  	ans=ans+sum[c[a[i]]-1+rest]-sum[c[a[i]]-1];
		  	printf("%lld",ans);
		  	return 0;
		}		
	}
   ans=sum[n];
   printf("%lld",ans);
}