p455.ans1
======================
4

=================
p455.ans2
======================
5

=================
p455.ans3
======================
4

=================
p455.cpp
======================
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int64_t A, B, C;

int64_t f(int64_t x) {
    int64_t res = A;
    res *= x;
    res += x % B;
    res %= C;
    return res;
}

void read() { cin >> A >> B >> C; }

int hare_tortoise() {
    int t = 0;
    int64_t hare = f(f(1));
    int64_t tortoise = f(1);
    do {
        tortoise = f(tortoise);
        hare = f(f(hare));
        t++;
        if(t > 2000042) {
            return -1;
        }
    } while(tortoise != hare);

    int mu = 0;
    tortoise = 1;
    while(tortoise != hare) {
        tortoise = f(tortoise);
        hare = f(hare);
        mu++;
    }

    int lambda = 1;
    hare = f(tortoise);
    while(tortoise != hare) {
        hare = f(hare);
        lambda++;
    }

    return mu + lambda;
}

void solve() {
    // The sequence x_{i+1} = (A * x_i + x_i % B) % C is a functional graph
    // iteration f, so it eventually enters a cycle (rho shape). The first
    // repeatable element is the first one that lies on the cycle, i.e. the
    // entry point of the tail of length mu into the cycle of length lambda. Its
    // second occurrence is exactly mu + lambda steps from the start.
    //
    // We run Floyd's tortoise and hare: advance until they meet inside the
    // cycle, then find mu by walking both one step at a time from x_0 and the
    // meeting point until they coincide, and find lambda by walking one pointer
    // around the cycle. If no meeting happens within the search bound the index
    // exceeds 2 * 10^6, so we report -1.

    int ans = hare_tortoise();
    if(ans > 2000000) {
        ans = -1;
    }
    cout << ans << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}

=================
p455.in1
======================
2 2 9

=================
p455.in2
======================
2305843009213693951 1 9223372036854775807

=================
p455.in3
======================
-2 1 5

=================
statement.txt
======================
455. Sequence analysis
Time limit per test: 1 second(s)
Memory limit: 4096 kilobytes
input: standard
output: standard

Due to the slow 'mod' and 'div' operations with int64 type, all Delphi solutions for the problem 455 (Sequence analysis) run much slower than the same code written in C++ or Java. We do not guarantee that Delphi solution exists.


You are given a sequence of signed 64-bit integers defined as follows:
x0 = 1,
,
where mod is a remainder operator. All arithmetic operations are evaluated without overflow checking. Use standard "remainder" operator for programming languages (it differs from the mathematical version; for example  in programming, while  in mathematics). Use "long long" type in C++, "long" in Java and "int64" in Delphi to store xi and all other values.

Let's call a sequence element xp repeatable if it occurs later in the sequence — meaning that there exists such q, q > p, that xq = xp. The first repeatable element M of the sequence is such an element xm that xm is repeatable, and none of the xp where p < m are repeatable.

Given A, B and C, your task is to find the index of the second occurence of the first repeatable element M in the sequence if the index is less or equal to 2 · 106. Per definition, the first element of the sequence has index 0.

Input
The only line of input contains three signed 64-bit integers: A, B and C (B > 0, C > 0).

Output
Print a single integer  — the index of the second occurence of the first repeatable member if it is less or equal to 2 · 106. Print -1 if the index is more than 2 · 106.

Example(s)
sample input
sample output
2 2 9
4

sample input
sample output
2305843009213693951 1 9223372036854775807
5

sample input
sample output
-2 1 5
4



Note
In the first sample test the sequence starts with the following numbers: 1, 3, 7, 6, 3, 7. The first repeatable element is 3. The second occurence of 3 has index 4.

In the second sample test the sequence starts with the following numbers: 1, 2305843009213693951, -4611686018427387903, 6917529027641081855, 0, 0, 0. The first repeatable element is 0. The second occurence of 0 has index 5.

In the third sample test the sequence starts with the following numbers: 1, -2, 4, -3, 1, -2, 4. The first repeatable element is 1. The second occurence of 1 has index 4.

=================
