POJ 2406 - Power String

最新推荐文章于 2023-06-25 10:43:41 发布

Sureina

最新推荐文章于 2023-06-25 10:43:41 发布

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分类专栏： ACM POJ 文章标签： ACM POJ

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本文介绍了一种通过分析输入字符串来确定其是否为周期性重复的算法，并计算出构成该字符串的基本子串的重复次数。利用模式匹配的思想，通过预处理获取最长公共前后缀的长度，进而判断并计算出循环次数。

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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

题意：给出一个周期字符串，求出其中子字符串循环的次数。

把第一个子字符串作为模式，求出长度，再进行计算即可。

#include <cstdio>
#include <cstring>

char str[1000005];
int tar[1000005];

int main()
{
    while (scanf("%s", str) != EOF)
    {
        if (strcmp(str, ".") == 0)
            break;
        int len = strlen(str);
        tar[0] = 0;

        int cur = -1;
        for (int i = 1; i < len; ++i)
        {
            while (cur > 0 && str[cur+1] != str[i])
                cur = -1;
            if (str[cur+1] == str[i])
                cur++;
            tar[i] = cur;
        }
        if (len % (len - tar[len - 1] -1) == 0)
            printf("%d\n", len / (len - tar[len - 1] - 1));
        else
            printf("1\n");
    }
    return 0;
}