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

解题思路：

找循环节，用while循环就可以了，注意输出格式。

代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int maxn=100010;
int num[maxn];
bool vis[maxn];
int n;

void getCircle()
{
    memset(vis,0,sizeof(vis));
    int t;
    bool ok=0;
    for(int i=1;i<=n;i++)
    {
        int temp=i;
        if(!vis[temp])
            printf("(");
        while(!vis[temp])
        {
            t=temp;
            vis[temp]=true;
            temp=num[temp];
            if(!vis[temp])
                printf("%d ",t);
            else
            {
                printf("%d",t);
                ok=1;
            }
        }
        if(ok)
        {
             printf(")");
             ok=0;
        }
    }
    printf("\n");
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&num[i]);
        getCircle();
    }
    return 0;
}