power string kmp算法

最新推荐文章于 2020-08-09 12:50:55 发布

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最新推荐文章于 2020-08-09 12:50:55 发布

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文章标签： kmp

本文链接：https://blog.youkuaiyun.com/a716121/article/details/47906373

本文介绍了一个算法问题PowerStrings，该问题涉及字符串的拼接与幂运算。文章详细解释了如何通过构建next数组来寻找字符串的循环节，进而计算出构成输入字符串的最小重复单元的数量。

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Power Strings

Time Limit: 3000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

Submit Status

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
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#define maxn 1000005
using namespace std;
char str[maxn];
int next[maxn],len;
void get_next();
int main()
{
    while(~scanf("%s",str)&&strcmp(str,"."))
    {
        len=strlen(str);
        get_next();
        int ans=1;
        if(len%(len-next[len])==0)
            ans=len/(len-next[len]);
        printf("%d\n",ans);
    }
}
void get_next()
{
    int i=0,j=-1;
    next[0]=-1;
    while(i<len)
    {
        if(j==-1||str[i]==str[j])
        {
            i++;
            j++;
            if(str[i]!=str[j])//仅仅满足j==-1
                next[i]=j;
            else
                next[i]=next[j];//仅仅满足str[i]==str[j]
        }
        else
            j=next[j];
    }
}
由于我们知道next数组中存的是一个位置(假设next[j]的值为k，对应的字符串为M,如果k>0,那么M[0....k-1]和M[j-k.....j-1]是相同的，并且0...k-1这个序列一定是最长的)，比如a b c a b c d(next值：-1 0 0 0 1 2 3 )，由next[6]=3可知，M[0..2]=M[3..6]，这就找到了循环节，于是我们思考从next数组作为切入点，来找到一种方法来求得循环节的个数。