[2406]Power Strings (POJ) KMP

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/jiangyongy/p/3971593.html

本文介绍了一种解决字符串幂次问题的算法实现，通过KMP算法中的next数组来高效计算给定字符串是否能表示为某个基串的幂次。文章提供了一个完整的C语言代码示例，展示了如何通过输入一个字符串并输出该字符串所能表示的最大幂次数。

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Language: Default

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 32710		Accepted: 13635

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 <stdio.h>
#include <string.h>
#include <stdlib.h>
char a[1000010],b[1000010];
int next[1000010];
int n,num=1;
void get_next()//求next数组
{
    int len=strlen(a);
    int i=0,j=-1;
    next[0]=-1;
    while(i<len)
    {
        if(j==-1||a[i]==a[j])
        {
            i++;
            j++;
            next[i]=j;
        }
        else
            j=next[j];
    }
}
int main()
{
    while(gets(a))
    {
        if(a[0]=='.')
            break;
        int len=strlen(a);
        get_next();
        int n=len-next[len];//n为循环节的长度
        if(len%n==0)
            printf("%d\n",len/n);
        else
            printf("1\n");
    }
    return 0;
}

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转载于:https://www.cnblogs.com/jiangyongy/p/3971593.html