Codeforces Round #385 (Div. 2) 745A Hongcow Learns the Cyclic Shift 【字符串】

最新推荐文章于 2022-01-20 00:05:42 发布

Emiya丶

最新推荐文章于 2022-01-20 00:05:42 发布

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分类专栏：简单题 CF题解文章标签： ACM codeforces

本文链接：https://blog.youkuaiyun.com/qq_37405320/article/details/75314936

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本文介绍了一个关于字符串循环移位的算法问题，通过实例演示了如何通过编程来确定由一个初始字符串通过循环移位可以产生的不同字符串的数量。该算法采用暴力求解方式，通过模拟字符串的每次变化并进行对比来实现。

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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.

Examples
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：输入一个长为n的字符串。问你在这一系列变化中，有多少个不同的字符串

思路：我爱暴力

#include<bits/stdc++.h>
using namespace std;
#define m(a) memset(a,0,sizeof(a))

char str[55];
char temp[55][60];

int main()
{
    m(str);

    while(~scanf("%s",str))
    {
        m(temp);
        int len=strlen(str);
        strcpy(temp[0],str);
        int sum=1;

        for(int i=1; i<len; i++)
        {
            int flag=1;
            temp[i][len-1]=temp[i-1][0];

            for(int j=0; j<len-1; j++)///模拟字符串变化
            {
                temp[i][j]=temp[i-1][j+1];
            }
//            printf("%s\n",temp[i]);
            for(int k=0; k<i; k++)
            {
                if(strcmp(temp[k],temp[i])==0)///比较
                {
                    flag=0;
                    break;
                }
            }
            if(flag)
            {
                sum++;
            }

        }
        printf("%d\n",sum);


    }
    return 0;
}