【CUGBACM15级BC第7场 A】hdu 4985 Little Pony and Permutation

Little Pony and Permutation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 846    Accepted Submission(s): 444

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

52 5 4 3 131 2 3

 

Sample Output

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

 

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！

#include <iostream>#include <set>#include <map>#include <stack>#include <cmath>#include <queue>#include <cstdio>#include <bitset>#include <string>#include <vector>#include <iomanip>#include <cstring>#include <algorithm>#include <functional>#define PI acos(-1)#define eps 1e-8#define inf 0x3f3f3f3f#define debug(x) cout<<"---"<<x<<"---"<<endltypedef long long ll;using namespace std;const int N = 100005;int n, a[100010], vis[100010];int main(){    while (scanf("%d", &n) != EOF)    {        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] == 1)            {                continue;            }            int gg = i;            printf("(%d", gg);            vis[gg] = 1;            gg = a[gg];            while (vis[gg] == 0)            {                vis[gg] = 1;                printf(" %d", gg);                gg = a[gg];            }            printf(")");        }        printf("\n");    }    return 0;}