SGU 455 Sequence analysis（Cycle detection，floyd判圈算法）

题目链接：http://acm.sgu.ru/problem.php?contest=0&problem=455

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 x_i and all other values.

Let‘s call a sequence element x_p repeatable if it occurs later in the sequence — meaning that there exists such q, q > p, that x_q = x_p. The first repeatable element M of the sequence is such an element x_m that x_m is repeatable, and none of the x_p 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.

题目大意：给出x[0] = 1，还有A、B、C，有x[i+1] = (A * x[i] + x[i] % B) % C。求第二个循环节出现的位置。

思路：http://en.wikipedia.org/wiki/Cycle_detection#Tortoise_and_hare （floyd判圈算法，龟兔算法？）

给一个初始值x[0]，有一个函数x[i + 1] = f(x[i])。若x始终在有限集合中运算，那么这些x值会构成一个环。

在此题中，整数模一个C后，显然得到的值是有限的。

设第一个循环节从x[μ]开始，循环节长度为λ。

那么对任意i = kλ ≥ μ，一定会有x[kλ] = x[kλ + kλ]，即x[i] = x[2i]。

于是我们可以用以下代码，找出满足x[v] = x[2v]的第一个ν，显然在[μ, μ + λ]间存在一个v = kλ，则此步复杂度为O(μ+λ)。

    long long x = f(x0), y = f(f(x0));
    int v = 1;
    while(x != y)
        x = f(x), y = f(f(y)), v++;

由于对任意i ≥ μ，有x[i] = x[i + λ] = x[i + kλ]，那么同意有x[μ] = x[μ + λ] = x[μ + kλ] = x[μ + v]。

在上面的代码中，我们已经求得了x[v]，其中两个变量都等于x[v]。

那么基于x[μ] = x[μ + v]的事实。我们可以用下面的代码，循环μ次，求出x[μ]。

    x = x0;
    int mu = 0;
    while(x != y)
        x = f(x), y = f(y), mu++;

上述代码已经求出μ与x[μ]。最后，只要再循环λ遍，即可求出循环节长度：

    int lam = 1;
    y = f(x);
    while(x != y) y = f(y), lam++;

总时间复杂度为O(μ+λ)，空间复杂度为O(1)。

对于本题，输出μ+λ即可。

至于本题的计算过程可能溢出的事情我没想……我发现大家都没管，我也不管了……

代码（312MS）：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4
 5 const int MAXR = 2e6;
 6
 7 LL A, B, C;
 8
 9 LL next(LL x) {
10     return (A * x + x % B) % C;
11 }
12
13 int main() {
14     scanf("%I64d%I64d%I64d", &A, &B, &C);
15     LL x = next(1), y = next(x);
16     int v = 1;
17     while(v <= MAXR && x != y)
18         x = next(x), y = next(next(y)), v++;
19     if(v > MAXR) {
20         puts("-1");
21         return 0;
22     }
23
24     x = 1;
25     int mu = 0;
26     while(x != y)
27         x = next(x), y = next(y), mu++;
28
29     int lam = 1;
30     y = next(x);
31     while(mu + lam <= MAXR && x != y) y = next(y), lam++;
32
33     if(mu + lam <= MAXR) printf("%d\n", mu + lam);
34     else puts("-1");
35 }