poj seek the name,seek the fame

最新推荐文章于 2023-11-07 21:18:30 发布

clover_hxy

最新推荐文章于 2023-11-07 21:18:30 发布

阅读量339

点赞数

分类专栏：字符串处理 KMP

本文链接：https://blog.youkuaiyun.com/clover_hxy/article/details/50698650

版权

字符串处理同时被 2 个专栏收录

69 篇文章

订阅专栏

KMP

16 篇文章

订阅专栏

本文深入探讨了信息技术领域的核心内容，包括但不限于前端开发、后端开发、移动开发、游戏开发、大数据开发、开发工具、嵌入式硬件、音视频基础、AI音视频处理等细分技术领域。通过详细解析每个技术领域，旨在为读者提供全面的技术知识框架。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Seek the Name, Seek the Fame

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15526		Accepted: 7854

Description

The little cat is so famous, that many couples tramp over hill and dale to Byteland, and asked the little cat to give names to their newly-born babies. They seek the name, and at the same time seek the fame. In order to escape from such boring job, the innovative little cat works out an easy but fantastic algorithm:

Step1. Connect the father's name and the mother's name, to a new string S.
Step2. Find a proper prefix-suffix string of S (which is not only the prefix, but also the suffix of S).

Example: Father='ala', Mother='la', we have S = 'ala'+'la' = 'alala'. Potential prefix-suffix strings of S are {'a', 'ala', 'alala'}. Given the string S, could you help the little cat to write a program to calculate the length of possible prefix-suffix strings of S? (He might thank you by giving your baby a name:)

Input

The input contains a number of test cases. Each test case occupies a single line that contains the string S described above.

Restrictions: Only lowercase letters may appear in the input. 1 <= Length of S <= 400000.

Output

For each test case, output a single line with integer numbers in increasing order, denoting the possible length of the new baby's name.

Sample Input

ababcababababcabab
aaaaa

Sample Output

2 4 9 18
1 2 3 4 5

Source

POJ Monthly--2006.01.22,Zeyuan Zhu

题目大意：求字符串s中所以前缀等于后缀的字符串的长度

对于字符串S的前i个字符构成的子串，既是它的后缀又是它的前缀的字符串中（它本身除外），最长的长度记作next[i]

kmp算法与失配函数的应用

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char s[4000003];
int n,m,len,t[4000003],ans[4000003];
void calc_t()
{
  t[0]=-1; int j;
  for (int i=0;i<len;i++)
   {
   	 j=t[i];
   	 while (s[i]!=s[j]&&j!=-1)
   	  j=t[j];
   	 t[i+1]=++j;
   }
} 
int main()
{
  while (gets(s))
   {
   	 len=strlen(s); memset(t,0,sizeof(t));
   	 calc_t();
   	 memset(ans,0,sizeof(ans)); int num=0;
   	 ans[++num]=len;
	 int j=len; 
   	 while (j!=0)
   	  {
   	  	 ans[++num]=t[j];
   	  	 j=t[j];
   	  }
   	 for (int i=num;i>=1;i--)
   	 if (ans[i])
   	  printf("%d ",ans[i]);
   	 printf("\n");
   }
}