The goal is to find $$$\sum^{X-1}_{i=0}A^{i}$$$ $$$ mod $$$ $$$m$$$ , and a simple thought is that we can calculate $$$\frac{A^{X}-1}{A-1}$$$. But when we divide $$$A^{X}-1$$$ by $$$A-1$$$ then $$$mod$$$ $$$m$$$, problem occurs: when we calculate $$$A^{X}-1$$$ , it's already $$$mod$$$ $$$m$$$ ,and there's maybe no reverse to it.

To solve this problem , I use a simple method: store $$$A^i$$$ with two numbers : $$$k$$$ and $$$r$$$ ,satisfying

When we calculate $$$A^{x}$$$,we refresh $$$k$$$ and $$$r$$$ by multiplication ,then refresh $$$k$$$ by modulo $$$m$$$ , and refresh $$$r$$$ when $$$r \ge A-1 $$$ : we do $$$k$$$ $$$+=$$$ $$$r/(A-1)$$$ and $$$r$$$ $$$=$$$ $$$r$$$ % $$$(A-1)$$$ ,like a carry.

We can prove that $$$r=0$$$ when $$$A=2$$$ and $$$r=1$$$ when $$$A=1$$$.if $$$A=1$$$ , we can NOT use $$$k$$$ and $$$r$$$ to store this num for $$$A-1=0$$$ . But this case is easy to solve . The result is obviously $$$X \bmod m$$$ .

Since we get

we can get

So the answer is $$$k_{X}$$$ . Noted that this method is free to change the denominator(in this problem it's $$$A-1$$$).

Code here: ~~~~~

include<stdio.h>

define yes printf("Yes\n")

define no printf("No\n")

//WA but score 500 :)

struct num{ // store number in form k*deno+r ( 0 <= r < deno) // to solve (number^b / deno) mod Mod = ? long long k; long long r; };

struct num qtimes(struct num n,long long b,long long deno,long long mod){ //Warning :deno != 0 if(b>=2){ long long tmpk = (((n.k*n.k%mod)*deno)%mod + (2*n.r*n.k) % mod + n.r*n.r/deno)%mod; long long tmpr = n.r*n.r % deno; if (b%2==0) return qtimes({tmpk,tmpr},b/2,deno,mod); else{ struct num tmpn = qtimes({tmpk,tmpr},b/2,deno,mod); long long tk2 = (((n.k*tmpn.k)%mod*deno)%mod + (n.r*tmpn.k+n.k*tmpn.r) % mod + n.r*tmpn.r/deno)%mod; long long tr2 = n.r*tmpn.r % deno; return {tk2,tr2};
} } else if(b==1) return {(n.k%mod+n.r/deno)%mod,n.r%deno}; else if(b==0) return {0,1}; }

int main(){ long long a,x,m; scanf("%lld%lld%lld",&a,&x,&m); if (a==1) { printf("%lld",x%m); return 0; } struct num result = qtimes({1,1},x,a-1,m); printf("%lld",result.k+(result.r-1)/(a-1)); // result = (A^x) ,--> {ans , 1} } ~~~~~