HUST 1010 The Minimum Length （kmp求最小循环节）

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1010 - The Minimum Length

时间限制：1秒 内存限制：128兆

题目描述

There is a string A. The length of A is less than 1,000,000. I rewrite it again and again. Then I got a new string: AAAAAA...... Now I cut it from two different position and get a new string B. Then, give you the string B, can you tell me the length of the shortest possible string A.For example, A="abcdefg". I got abcdefgabcdefgabcdefgabcdefg.... Then I cut the red part: efgabcdefgabcde as string B. From B, you should find out the shortest A.

输入

Multiply Test Cases.For each line there is a string B which contains only lowercase and uppercase charactors.The length of B is no more than 1,000,000.

输出

For each line, output an integer, as described above.

样例输入

bcabcabefgabcdefgabcde

样例输出

37

题目大意：给你一串字符串，问这个字符串是由长度多少的字符串重复组成，也就是循环节。用kmp求循环节就好了

#include <stdio.h>#include <string.h>#include <iostream>using namespace std;char st[1000100];int next[1001000];void getnext(char st[],int m){    int j = 0 ,k = -1;    next[0] = -1;    while (j < m)    {        if (k == -1 || st[j] == st[k] )        {            ++j;            ++k;            next[j] = k;        }        else         {            k = next[k];        }        //printf("%d\n",next[j]);    }}int main(){    while (cin >> st)    {        int len = strlen(st);        getnext(st,len);        if(len == 1)             printf("1\n");          printf("%d\n",len - next[len]);    }    return 0;}