POJ 2369 Permutations【置换群】

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

题意：给定n个元素的序列a[]，一次变换后a[i] = a[a[i]]，问经过多少次变换使得序列变回原来的样子。‘

基础置换群，求出所有群的周期，求其最小公倍数’

#include <iostream>  #include<cstdio>#include<iomanip>#include<cmath>#define INF 0x3f3f3f3f  #define MAXN (1000+10)   #define mem(a, b) memset(a, (b), sizeof(a))  #define MOD 1000000007  #define LL long long  using namespace std;int gcd(int a, int b) {    return b == 0 ? a : gcd(b, a%b);}int lcm(int a, int b) {    return a / gcd(a, b) * b;}int a[MAXN]; bool vis[MAXN];int main(){    int n;    while (scanf("%d",&n) != EOF)    {        for (int i = 1; i <= n; i++)            scanf("%d",&a[i]);        int ans = 1; mem(vis, false);        for (int i = 1; i <= n; i++)        {            if (vis[i]) continue;            vis[i] = true;            int j = a[i]; int cnt = 1;            while (i != j)            {                cnt++;                j = a[j];            }            ans = lcm(ans, cnt);        }        printf("%d\n",ans);    }    return 0;}