hust 1010 The Minimum Length （KMP 最短循环节）

最新推荐文章于 2022-07-02 16:58:00 发布

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最新推荐文章于 2022-07-02 16:58:00 发布

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分类专栏：字符串—KMP

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本文介绍了一道KMP算法的经典问题：通过给定的切割后的字符串B，反推原始字符串A的最短长度。文章提供了完整的解析思路及C语言实现代码。

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1、http://acm.hust.edu.cn/problem.php?id=1010

2、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

3、KMP模板题，调用求子模板串的模式值f[n]的函数，void getFail(char *p, int *f)

4、题目：

The Minimum Length

Time Limit: 1 Sec Memory Limit: 128 MB
Submissions: 385 Solved: 124

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

HINT

Source

[ Submit][ Status][ Discuss] 4、代码：

#include<stdio.h>
#include<string.h>
#define N 1000005
char str[N];
int f[N];
void getFail(char *p, int *f)
{
    int m=strlen(p);
    f[0]=f[1]=0;
    for(int i=1; i<m; ++i)
    {
        int j=f[i];
        while(j && p[i]!=p[j])j=f[j];
        f[i+1] = p[i]==p[j]?1+j:0;
    }
}
int main()
{
    while(scanf("%s",str)!=EOF)
    {
        int len=strlen(str);
        getFail(str,f);
        printf("%d\n",len-f[len]);

    }
    return 0;
}
/*
bcabcab
efgabcdefgabcde
*/