YT14-HDU-找迴圈節 （關於std::ios::sync_with_stdio(false);的作用和疑問）

不被看好的青春叫成長發表於2015-01-29
iOSFalse

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)

原始碼如下：

#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程式碼如下： 

#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;
}

執行結果：



提交自己的程式碼之後經常是Time Limit Exceeded。

看網上找程式碼的時候別人用的通常都是C語言的scanf和printf輸入輸出，一直以為那是C語言的程式碼，但這次偶然看到了同學的程式碼（和我“抄”得居然是同一個。。。老師教導我們要”抄“之有道）但其用了
於是百度了一下：

cin，cout之所以效率低，是因為先把要輸出的東西存入緩衝區，再輸出，導致效率降低，而這段語句可以來打消iostream的輸入輸出快取，可以節省許多時間，使效率與scanf與printf相差無幾。

又學到了一手，最近做題總算出現時間超限，以後就不用擔心了。不過不知道

std::ios::sync_with_stdio(false);

有什麼弊端，


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