题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4985

置换群：http://baike.baidu.com/view/1879054.htm?fr=aladdin

Little Pony and Permutation

Problem Description

As a unicorn, the ability of using magic is the distinguishing feature among other kind of pony. Being familiar with composition and decomposition is the fundamental course for a young unicorn. Twilight Sparkle is interested in the decomposition of permutations.
A permutation of a set S = {1, 2, ..., n} is a bijection from S to itself. In the great magician —— Cauchy‘s two-line notation, one lists the elements of set S in the first row, and then for each element, writes its image under the permutation below it in
the second row. For instance, a permutation of set {1, 2, 3, 4, 5} σ can be written as:

Here σ(1) = 2, σ(2) = 5, σ(3) = 4, σ(4) = 3, and σ(5) = 1.

Twilight Sparkle is going to decompose the permutation into some disjoint cycles. For instance, the above permutation can be rewritten as:

Help Twilight Sparkle find the lexicographic smallest solution. (Only considering numbers).

Input

Input contains multiple test cases (less than 10). For each test case, the first line contains one number n (1<=n<=10^5). The second line contains n numbers which the i-th of them(start from 1) is σ(i).

Output

For each case, output the corresponding result.

Sample Input

5
2 5 4 3 1
3
1 2 3

Sample Output

(1 2 5)(3 4)
(1)(2)(3)

Source

BestCoder Round #7

题意：

题意较难理解，其实就是一个置换群！

看代码就会理解题意！

代码如下:

/*#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;

const int N = 100005;

int n, a[N], vis[N];

int main()
{
    while (~scanf("%d", &n))
    {
        memset(vis, 0, sizeof(vis));
        for (int i = 1; i <= n; i++)
            scanf("%d", &a[i]);
        for (int i = 1; i <= n; i++)
        {
            if (vis[i])
                continue;
            int t = i;
            printf("(%d", t);
            vis[t] = 1;
            t = a[t];
            while (vis[t] == 0)
            {
                vis[t] = 1;
                printf(" %d", t);
                t = a[t];
            }
            printf(")");
        }
        printf("\n");
    }
    return 0;
}*/

#include <cstdio>
#include <cstring>
int main()
{
    int n;
    int a[100017], f[100017];
    while(~scanf("%d",&n))
    {
        memset(f,0,sizeof(f));
        for(int i = 1; i <= n; i++)
        {
            scanf("%d",&a[i]);
        }
        for(int i = 1; i <= n; i++)
        {
            if(f[i])
                continue;
            int t = i;
            f[t] = 1;
            printf("(%d",t);
            while(f[a[t]] == 0)
            {
                printf(" %d",a[t]);
                f[a[t]] = 1;
                t = a[t];
            }
            printf(")");
        }
        printf("\n");
    }
    return 0;
}