HUST 1010 The Minimum Length-优快云博客

本文介绍了一种利用KMP算法解决寻找字符串最短循环节的问题，通过实例讲解了如何从给定的长字符串中找出其由何种最短字符串重复构成。输入为多个测试用例，每个用例包含一个长字符串，输出为该字符串的最短循环节长度。

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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.

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

kmp求最小循环节。

代码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char str[1000001];
int Next[1000001],len,k;
void findNext() {
    int j = -1,i = 0;
    Next[0] = -1;
    while(i < len) {
        if(j == -1 || str[i] == str[j]) {
            Next[++ i] = ++ j;
        }
        else j = Next[j];
    }
}
int main() {
    while(~scanf("%s",str)) {
        len = strlen(str);
        findNext();///先确立Next数组

        printf("%d\n",len - Next[len]);
    }
}