sgu 455. Sequence analysis （floyd 判圈算法，O(1)空间复杂度求循环节）

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 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_mthat 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.

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.

比赛的时候逗了，往看空间限制了....

直接开了个set判重。。。显然MLE 了。。。

然后这道题的正解是 floyd判圈算法（也叫龟兔算法？）

http://www.cnblogs.com/oyking/p/4286916.html  这份题解讲得很详细

/*************************************************************************
    > File Name: code/2015summer/#5/CC.cpp
    > Author: 111qqz
    > Email: rkz2013@126.com 
    > Created Time: 2015年07月30日 星期四 21时02分17秒
 ************************************************************************/
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define y0 abc111qqz
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define tm crazy111qqz
#define lr dying111qqz
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)  
typedef long long LL;
typedef unsigned long long ULL;
const int inf = 0x7fffffff;
const int LIM = 2E6;
LL a,b,c;
LL next (LL x)
{
    return (a*x+x%b)%c;
}
int main()
{
    cin>>a>>b>>c;
    LL x = next(1);
    LL y = next(next(1));
    int v = 1;
    while (v<=LIM &&x!=y)
    {
    x = next(x);  //乌龟走一步
    y = next(next(y)); //兔子走两步
    v++;
    }
    if (v>LIM)   //在规定范围内找不到循环节（兔子没有和乌龟到同一个位置）
    {
    puts("-1");  
    return 0;
    }

    x = 1;
    int mu = 0;  //找到循环节的起点
    while (x!=y) 
    {
    x = next(x);
    y = next(y);
    mu++;
    }
    
    int lam = 1; //循环节的长度
    y = next(x);
    while (mu+lam <= LIM  &&x!=y)
    {
    y = next(y);
    lam++;
    }
    if (mu+lam<=LIM)
    {
    printf("%d
",mu+lam);
    }
    else
    {
    puts("-1");
    }

    return 0;
}