POJ 2369-Permutations（置换群-K次置换后还原有序）

Permutations

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3158		Accepted: 1709

Description

We remind that the permutation of some final set is a one-to-one mapping of the set onto itself. Less formally, that is a way to reorder elements of the set. For example, one can define a permutation of the set {1,2,3,4,5} as follows: 


 

This record defines a permutation P as follows: P(1) = 4, P(2) = 1, P(3) = 5, etc. 

What is the value of the expression P(P(1))? It’s clear, that P(P(1)) = P(4) = 2. And P(P(3)) = P(5) = 3. One can easily see that if P(n) is a permutation then P(P(n)) is a permutation as well. In our example (believe us) 


 

It is natural to denote this permutation by P2(n) = P(P(n)). In a general form the defenition is as follows: P(n) = P1(n), Pk(n) = P(Pk-1(n)). Among the permutations there is a very important one — that moves nothing: 


 

It is clear that for every k the following relation is satisfied: (EN)k = EN. The following less trivial statement is correct (we won't prove it here, you may prove it yourself incidentally): Let P(n) be some permutation of an N elements set. Then there exists
a natural number k, that Pk = EN. The least natural k such that Pk = EN is called an order of the permutation P. 

The problem that your program should solve is formulated now in a very simple manner: "Given a permutation find its order."
Input

In the first line of the standard input an only natural number N (1 <= N <= 1000) is contained, that is a number of elements in the set that is rearranged by this permutation. In the second line there are N natural numbers of the range from 1 up to N, separated
by a space, that define a permutation — the numbers P(1), P(2),…, P(N).
Output

You should write an only natural number to the standard output, that is an order of the permutation. You may consider that an answer shouldn't exceed 109.
Sample Input

5
4 1 5 2 3

Sample Output

6

Source

Ural State University Internal Contest October'2000 Junior Session

题目意思：

计算N个递增有序元素1~N，多少次置换群的运算能够还原成1~N。

解题思路：

据说……置换群中有一个定理：设T为一置换，e为单位置换，T^k=e,那么k的最小正整数解是T的拆分的所有循环长度的最小公倍数。

#include<iostream>
#include<cstdio>
#include<iomanip>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<map>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
#define INF 0xfffffff
#define MAXN 1010
long long a[MAXN];
long long gcd(long long a,long long b)
{
return (a%b!=0?(gcd(b,a%b)):b);
}
long long lcm(long long u,long long v)//最小公倍数
{
long long h;
h=gcd(u,v);
return(u*v/h);
}
int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("G:/cbx/read.txt","r",stdin);
//freopen("G:/cbx/out.txt","w",stdout);
#endif
int n;
scanf("%d",&n);
for(int i=0; i<n; ++i)
{
scanf("%I64d",&a[i]);
--a[i];
}
int res=1;
long long ans=1;
for(int i=0; i<n; ++i)
{
int temp=a[i];
while(1)//求循环节长度
{
if(i==temp)
{
ans=lcm(ans,res);//计算所有循环节的最小公倍数
res=1;
break;
}
++res;
temp=a[temp];
}
}
printf("%I64d\n",ans);
return 0;
}