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)
代码：
/*
*Copyright (c)2014,烟台大学计算机与控制工程学院
*All rights reserved.
*文件名称：test.cpp
*作 者：冷基栋
*完成日期：2015年2月6日
*版 本 号：v1.0
*问题描述：找循环节
*程序输入：一个整数n，和n个整数
*程序输出：相关的各组数
*/
#include <iostream>
#include <cstdio>
#include <cstring>
const int NUM=100001;
using namespace std;
int main()
{
std::ios::sync_with_stdio(false);
int i,j,n,m;
int a[NUM];
while(cin>>n)
{
for(i=1; i<=n; i++)
cin>>a[i];
for(i=1; i<=n; i++)
{
while(a[i])
{
cout<<"("<<i;
j=a[i];
a[i]=0;
while(a[j])
{
cout<<" "<<j;
m=a[j];
a[j]=0;
j=m;
}
cout<<")";
}
}
cout<<endl;
}
return 0;
}
运行结果：
知识点总结：
cin，cout之所以效率低，是因为先把要输出的东西存入缓冲区，再输出，导致效率降低，而这段语句可以来打消iostream的输入输出缓存，可以节省许多时间，使效率与scanf与printf相差无几。std::ios::sync_with_stdio(false);（偷窥无罪
哈哈）
学习心得：
好好学习 天天向上