[kuangbin带你飞]专题十六 KMP & 扩展KMP & Manacher :G - Power Strings POJ - 2406（kmp简单循环节）...

红太狼 2023-08-17 15:41 76阅读 0赞

### \[kuangbin带你飞\]专题十六 KMP & 扩展KMP & Manacher ###

## G - Power Strings ##

POJ - 2406

题目：

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：求字符串的循环节个数**

**思路：要注意如果该字符串如果不满足都是循环组成的则输出-1，若是都是循环节组成的，那么n-nextt\[n\]就是循环节长度，用总长度除以这个长度就是循环节长度**

// 
    // Created by HJYL on 2019/8/15.
    //
    #include <iostream>
    #include <vector>
    #include <map>
    #include <string>
    #include <queue>
    #include <stack>
    #include <set>
    #include <algorithm>
    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <cstdlib>
    using namespace std;
    const int maxn=1e7+10;
    char str[maxn];
    int nextt[maxn];
    void getnextt()
    {
        int i=0,j=-1;
        nextt[0]=-1;
        int n=strlen(str);
        while(i<n)
        {
            if(j==-1||str[i]==str[j])
            {
                i++,j++;
                if(str[i]!=str[j])
                    nextt[i]=j;
                else
                    nextt[i]=nextt[j];
            } else
                j=nextt[j];
        }
    }
    int main()
    {
        //freopen("C:\\Users\\asus567767\\CLionProjects\\untitled\\text","r",stdin);
        while(~scanf("%s",str))
        {
            if(str[0]=='.')
                break;
            int len=strlen(str);
            getnextt();
            if(len%(len-nextt[len])==0)
             printf("%d\n",len/(len-nextt[len]));
            else
                printf("1\n");
        }
        return 0;
    }

转载于:https://www.cnblogs.com/Vampire6/p/11360900.html