M - Power Strings

刺骨的言语ヽ痛彻心扉 2022-03-22 15:26 178阅读 0赞

### 题目描述： ###

Given two strings a and b we define a*b to be their concatenation. For example, if a = “abc” and b = “def” then a*b = “abcdef”. If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = “” (the empty string) and a^(n+1) = a\*(a^n).  
Input  
Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.  
Output  
For each s you should print the largest n such that s = a^n for some string a.  
Sample Input  
abcd  
aaaa  
ababab  
.  
Sample Output  
1  
4  
3  
Hint  
This problem has huge input, use scanf instead of cin to avoid time limit exceed.  
题意：给一个字符串S长度不超过10^6，求最大的n使得S由n个相同的字符串a连接而成，如："ababab"则由n=3个"ab"连接而成，"aaaa"由n=4个"a"连接而成，"abcd"则由n=1个"abcd"连接而成。

##### 分析： #####

本题的第一反应就是循环节的使用。我们可以发现本题无非就是求出最小循环节，且要保证len% ans==0。如果不等于0，就输出1。  
参考博客：[http://www.cppblog.com/Davidlrzh/articles/115751.html][http_www.cppblog.com_Davidlrzh_articles_115751.html]  
代码如下：

#include"stdio.h"
    #include"string.h"
         char text[1000001];
         int next[1000001];
    void Get_next(char text[],int next[])
    {
        int i,j,len;
        next[0]=-1;
        i=0;j=-1;
        len=strlen(text);
        while(i<len)
        {
            if(j==-1||text[i]==text[j])
            {
                i++;j++;
                if(text[i]==text[j])
                    next[i]=next[j];
                else
                    next[i]=j;
            }
            else
                j=next[j];
        }
    
    }
    int main()
    {
    
         int len,i,ans;
         while(~scanf("%s",text))
         {
             if(text[0]=='.')
                break;
             len=strlen(text);
             Get_next(text,next);
             ans=len-next[len];//得到最小循环节
             if(ans!=len&&len%ans==0)//保证此字符串正好由多于一个循环节构成
                printf("%d\n",len/ans);//求出有几个循环节
             else
                printf("%d\n",1);
    
         }
    
    }

[http_www.cppblog.com_Davidlrzh_articles_115751.html]: http://www.cppblog.com/Davidlrzh/articles/115751.html