Power Strings

最新推荐文章于 2023-06-30 19:30:28 发布

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最新推荐文章于 2023-06-30 19:30:28 发布

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原文链接：http://www.cnblogs.com/ffhuguang/archive/2013/03/04/2942905.html

本文介绍了一种用于解决字符串幂次分解问题的高效算法。该算法通过预处理得到字符串的next数组，进而判断一个字符串是否可以由另一个较短字符串重复组成，并给出了具体的实现代码。

Description

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.

#include<iostream>
#include<cstdlib>
#include<stdio.h>
#include<string.h>
using namespace std;
const int maxm = 1000010;     // 模式串的最大长度
char p[maxm];
int m, next[maxm];
void getNext()
{
        int i = 0, j = next[0] = -1;
        while(i < m)
        {
            if (j == -1 || p[i] == p[j])
            {

                ++i; ++j;
                next[i] =j;

            }
            else
            j = next[j];
        }

}

int main()
{
       while(scanf("%s",p)!=EOF)
       {
           if(p[0]=='.') break;
           m=strlen(p);
           getNext();
           int cc=1;
           if(m%(m-next[m])==0)
           cc=m/(m-next[m]);
           cout<<cc<<endl;
       }
}

转载于:https://www.cnblogs.com/ffhuguang/archive/2013/03/04/2942905.html