POJ
来源:互联网 发布:农业技术网络书屋 编辑:程序博客网 时间:2024/05/22 07:16
Seek the Name, Seek the Fame
Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 20666 Accepted: 10782
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:)
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.
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
ababcababababcababaaaaa
Sample Output
2 4 9 181 2 3 4 5
保留 前缀与后缀 相同的 子串的 长度 ,从小到大输出。。
KMP的next运用,貌似hash也可以。
#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>using namespace std;const int len=4e5+5;char a[len];int kmp_next[len],ans[len];void KMP_next(){ int i=0,j=-1; kmp_next[0]=-1; while(a[i]) if(j==-1||a[i]==a[j]) kmp_next[++i]=++j; else j=kmp_next[j];}int main(){ while(~scanf("%s",a)) { KMP_next(); int n=strlen(a),flag=0; ans[flag++]=n; while(n) ans[flag++]=n=kmp_next[n]; for(int i=flag-2; i>=0; i--) if(i) printf("%d ",ans[i]); else printf("%d",ans[i]); cout<<endl; } return 0;}
阅读全文
0 0
- POJ
- poj
- POJ
- POJ
- poj
- poj
- POJ
- POJ
- poj
- POJ
- POJ
- POJ
- POJ
- POJ
- POJ
- POJ
- POJ
- POJ
- 树莓派上的软件安装和卸载命令汇总
- Unity塔防类游戏用协程实现产生怪的波数和个数
- Accelerated 11 Vec (Defining abstract data types)
- Core Java Volume I 读书笔记
- bzoj 1040: [ZJOI2008]骑士 树形dp
- POJ
- springboot -dubbo-ssm的最新开源商城
- java中queue的使用
- 518. Coin Change 2
- 插值与拟合
- 剑指offer 面试题5: 替换空格
- 趣味博弈论——斐波那契博弈
- netty
- Codeforces 742 B. Arpa’s obvious problem and Mehrdad’s terrible solution