【codeforce745A】Hongcow Learns the Cyclic Shift_hongcow is learning to spell! one day, his teacher-优快云博客

Hongcow通过学习拼写单词并对其进行循环移位来创造新词汇。此过程涉及将单词的最后一个字符移动到开头，形成不同的组合。挑战在于确定通过这种移位能产生多少种不同的字符串。

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Hongcow Learns the Cyclic Shift

Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Appoint description:

Description

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 acyclic 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.

Sample Input

Input

abcd

Output

Input

bbb

Output

Input

yzyz

Output

每次将最后一个字母放到最前面，输出会组成多少不同的字符串，也就是直到出现重复的为止

i\j a b c d

d a b c

c d a b

b c d a

a b c d

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char c[55][110],s[110];
int main()
{
	
	int faut;
	//while(~scanf("%s",s)){
	scanf("%s",s);
		int l=strlen(s);
		strcpy(c[0],s);
		int num=1;
	   for(int i=1;i<l;i++){
	   	faut=0;
	   	for(int j=0;j<i;j++)
	   	c[i][j]=s[l-i+j];//存储新的字符串 
	   	for(int n=i;n<l;n++)
	   	c[i][n]=s[n-i];
	   	for(int k=0;k<i;k++){
	   		if(strcmp(c[i],c[k])==0){
	   			faut=1;break;
			   }
		   }
		   if(faut==0)
		   num++;
	   }
	   printf("%d\n",num);
	//}
	return 0;
}