hust 1010 最小循环节

最新推荐文章于 2024-06-16 16:55:13 发布

黑码

最新推荐文章于 2024-06-16 16:55:13 发布

阅读量360

点赞数

CC 4.0 BY-SA版权

分类专栏： kmp

本文链接：https://blog.youkuaiyun.com/Littlewhite520/article/details/77506630

kmp 专栏收录该内容

13 篇文章

订阅专栏

本文介绍了一种通过给定的切割后的字符串B来求原始字符串A的最短可能长度的方法。利用KMP算法预处理得到的next数组，可以高效地找到最小循环节的长度。

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

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
就是求最小循环节

#include <bits/stdc++.h>
using namespace std;

char s[1000100];
int f[1000100];

void getfill(char *s,int len)
{
    memset(f,0,sizeof(f));
    f[0]=-1;
    for(int i=1;i<=len;i++)
    {
        int j=f[i];
        while(j&&s[i]!=s[j])
            j=f[j];
        f[i+1]=(s[i]==s[j])?j+1:0;
    }
}

int main()
{
    while(scanf("%s",s+1)!=EOF)
    {
        int len=strlen(s+1);
        getfill(s+1,len);
        int le=len-f[len];
        printf("%d\n",le );
    }
}