杭電ACM 找迴圈節 std::ios::sync_with_stdio(false);

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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).InputInput 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).OutputFor 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)

代碼：

/**Copyright (c)2014,煙台大學電腦與控制工程學院*All rights reserved.*檔案名稱：test.cpp*作    者：冷基棟*完成日期：2015年2月6日*版 本 號：v1.0*問題描述：找迴圈節*程式輸入：一個整數n，和n個整數*程式輸出：相關的各組數*/#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;}

運行結果：

知識點總結：

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的輸入輸出緩衝，可以節省許多時間，使效率與scanf與printf相差無幾。std::ios::sync_with_stdio(false);（偷窺無罪 哈哈）

學習心得：

好好學習 天天向上



杭電ACM 找迴圈節 std::ios::sync_with_stdio(false);