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

bcabcabefgabcdefgabcde

Sample Output

37

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: [e-mail]> 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;    }}