HUST 1010 The Minimum Length

1010 - The Minimum Length

Time Limit: 1s Memory Limit: 128MB

Submissions: 1382 Solved: 519

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 abcdefgabcdefgabcdefgabcdefg.... 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

这个题目是给定一个由一段字符串A重复组成的字符串B的一部分，且至少包含一个字符串A，求出那个串A的长度

这个题目再次彰显KMP的强大。。。

因为是至少包含一个完整的A，所以开头第一个字母与后面组成的一个串长为要求的串A的长度的序列一定与A是同构的，那么由KMP算法求得的第m+1位的那个字符的next值与给出的字符串总长的差一定是这个要求的串A的长度。

代码如下：

/*************************************************************************
	> File Name: The_Minimum_Length.cpp
	> Author: Zhanghaoran
	> Mail: chilumanxi@xiyoulinux.org
	> Created Time: Wed 25 Nov 2015 12:17:00 AM CST
 ************************************************************************/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>

using namespace std;
int Max = 0;
void preKmp(char x[], int m, int kmp_Next[]){
    int i, j;
    j = -1;
    kmp_Next[0] = -1;
    i = 0;
    while(i < m){
        if(j == -1 || x[i] == x[j]){
            kmp_Next[++i] = ++ j;
        }
        else
            j = kmp_Next[j];
        if(Max < j)
            Max = j;
    }
    kmp_Next[0] = 0;
}
int nexti[1000010];
char a[1000010];
int main(void){
    while(~scanf("%s", a)){
        preKmp(a, strlen(a), nexti);
        cout << strlen(a) - nexti[strlen(a)] << endl;
        Max = 0;
    }
}