Little Pony and Permutation

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).

5
2 5 4 3 1
3
1 2 3

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

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;
}

[BestCoder Round #7] hdu 4985 Little Pony and Permutation (找循环节）