A - Hongcow Learns the Cyclic Shift

CodeForces - 745A

Hongcow is learning to spell! One day, his teacher gives him a word that he needs to learn to spell. Being a dutiful student, he immediately learns how to spell the word.

Hongcow has decided to try to make new words from this one. He starts by taking the word he just learned how to spell, and moves the last character of the word to the beginning of the word. He calls this a cyclic shift. He can apply cyclic shift many times. For example, consecutively applying cyclic shift operation to the word "abracadabra" Hongcow will get words "aabracadabr", "raabracadab" and so on.

Hongcow is now wondering how many distinct words he can generate by doing the cyclic shift arbitrarily many times. The initial string is also counted.

Input
The first line of input will be a single string s (1 ≤ |s| ≤ 50), the word Hongcow initially learns how to spell. The string s consists only of lowercase English letters ('a'–'z').

Output
Output a single integer equal to the number of distinct strings that Hongcow can obtain by applying the cyclic shift arbitrarily many times to the given string.

Example
Input
abcd
Output
4
Input
bbb
Output
1
Input
yzyz
Output
2

Note
For the first sample, the strings Hongcow can generate are "abcd", "dabc", "cdab", and "bcda".

For the second sample, no matter how many times Hongcow does the cyclic shift, Hongcow can only generate "bbb".

For the third sample, the two strings Hongcow can generate are "yzyz" and "zyzy".

题意：一个字符串顺序不变，有几种不同变化。

解法：字符串拼接，将第一个字符放到最后，再放第二个，直到倒数第二个字符，每次变化后用set来存储，去掉重复字符串，最后求set的size为变化个数。

代码：

#include<iostream>
#include<cstring>
#include<string>
#include<set>
using namespace std;
int main()
{
    set<string> a;
    string str,str1,str2;
    cin>>str;
    int n=str.length();
    for(int i=0;i<n;i++){
        str1=str.substr(0,i);
        str2=str.substr(i,n-i);
        a.insert(str2+str1);
    }
    cout<<a.size()<<endl;
    return 0;
}