[kuangbin带你飞]专题十六 KMP & 扩展KMP & Manacher F - The Minimum Length HUST - 1010 （kmp循环节）

F - The Minimum Length HUST - 1010

题目链接：https://vjudge.net/contest/70325#problem/F

题目：

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 abcd efgabcdefgabcdefgabcdefg.... Then I cut the red part: efgabcdefgabcde as string B. From B, you should find out the shortest A.

InputMultiply 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.
OutputFor each line, output an integer, as described above.Sample Input

bcabcab
efgabcdefgabcde

Sample Output

3
7
题意：求最小循环节的长度
思路：kmp中n-next[n]就是最小循环节的长度，由于现在该题不支持提交，我就贴一下我写的

// 
// 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=1e6+10;
char str[maxn];
int nextt[maxn];
void getnext()
{
    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))
    {
        getnext();
        int len=strlen(str);
        printf("%d\n",len-nextt[len]);
    }
    return 0;
}