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)
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
走自己的路，让别人说去吧
