hust 1010 The Minimum Length (KMP 最短循环节)
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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 MBSubmissions: 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
bcabcabefgabcdefgabcde
Sample Output
37
HINT
Source
#include<stdio.h>#include<string.h>#define N 1000005char 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;}/*bcabcabefgabcdefgabcde*/
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