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)
原代码如下:
#include <iostream> #include <cstdio> #include <cstring> const int NUM=100000; using namespace std; int main() { 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; }
AC代码如下:
#include <iostream> #include <cstdio> #include <cstring> const int NUM=100000; 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:
提交自己的代码之后经常是Time Limit Exceeded。
看网上找代码的时候别人用的通常都是C语言的scanf和printf输入输出,一直以为那是C语言的代码,但这次偶然看到了同学的代码(和我“抄”得居然是同一个。。。老师教导我们要”抄“之有道)但其用了
于是百度了一下:
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。
又学到了一手,最近做题总算出现时间超限,以后就不用担心了。不过不知道
std::ios::sync_with_stdio(false);
有什么弊端,