HUST 1010 The Minimum Length

题目链接：

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

题目大意：

有一个字符串A，一次次的重写A，会得到一个新的字符串AAAAAAAA…..,现在将这个字符串从中切去一部分得到一个字符串B，例如有一个字符串A=”abcdefg”.，复制几次之后得到abcdefgabcdefgabcdefgabcdefg….，现在切去中间红色的部分，得到字符串B，现在只给出字符串B，求出字符串A的长度

思路：

kmp求最小循环节 = len - nxt[len-1] 或 　＝len- nxt[len] 根据具体next数组含义而定。

　　　　　　　　　　　F - The Minimum Length

Time Limit:1000MS Memory Limit:131072KB 64bit IO Format:%lld &
%llu Submit Status Practice HUST 1010

Description

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.

Input

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.

Output

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

Sample Input

bcabcab
efgabcdefgabcde

Sample Output

3
7

/*************************************************************************    > File Name: hust_1010.cpp    > Author: dulun    > Mail: [e-mail]    > Created Time: 2016年03月15日 星期二 21时02分30秒 ************************************************************************/#include<iostream>#include<stdio.h>#include<cstring>#include<cstdlib>#include<algorithm>#define LL long longusing namespace std;const int N = 50086;int nxt[1000006];char a[1000006];void getnxt(int m){    int k = 0;    for(int i = 1; i < m; i++)    {        while(k && a[i] != a[k]) k = nxt[k-1];        if(a[i] == a[k]) k++;        nxt[i] = k;    }}int main(){    while(~scanf("%s", a))    {        memset(nxt, 0, sizeof(nxt));        int m = strlen(a);        getnxt(m);        cout<<m-nxt[strlen(a)-1]<<endl;        memset(a, 0, sizeof(a));    }    return 0;}