zoukankan      html  css  js  c++  java
  • poj2752 Seek the Name, Seek the Fame

    地址:http://poj.org/problem?id=2752

    题目:

    Seek the Name, Seek the Fame
    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 18711   Accepted: 9607

    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

    思路:next数组的应用
     1 #include <cstdio>
     2 #include <cstring>
     3 #include <iostream>
     4 
     5 using namespace std;
     6 
     7 #define MP make_pair
     8 #define PB push_back
     9 typedef long long LL;
    10 const double eps=1e-8;
    11 const int K=1e6+7;
    12 const int mod=1e9+7;
    13 
    14 int nt[K],ans[K];
    15 char sa[K],sb[K];
    16 void kmp_next(char *T,int *next)
    17 {
    18     next[0]=0;
    19     for(int i=1,j=0,len=strlen(T);i<len;i++)
    20     {
    21         while(j&&T[i]!=T[j]) j=next[j-1];
    22         if(T[i]==T[j])  j++;
    23         next[i]=j;
    24     }
    25 }
    26 int kmp(char *S,char *T,int *next)
    27 {
    28     int ans=0;
    29     int ls=strlen(S),lt=strlen(T);
    30     kmp_next(T,next);
    31     for(int i=0,j=0;i<ls;i++)
    32     {
    33         while(j&&S[i]!=T[j]) j=next[j-1];
    34         if(S[i]==T[j])  j++;
    35         if(j==lt)   ans++;
    36     }
    37     return ans;
    38 }
    39 int main(void)
    40 {
    41     while(1==scanf("%s",sa))
    42     {
    43         kmp_next(sa,nt);
    44         int t=strlen(sa)-1,cnt=1;
    45         ans[cnt++]=t+1;
    46         while(nt[t])
    47         {
    48             ans[cnt++]=nt[t];
    49             t=nt[t-1];
    50         }
    51         for(int i=cnt-1;i;i--)
    52             printf("%d ",ans[i]);
    53         printf("
    ");
    54     }
    55     return 0;
    56 }
  • 相关阅读:
    sql 2012中获取表的信息,包含字段的描述
    C#如何创建泛型类T的实例
    C# 之 DataReader 和 DataSet 的区别
    C#进阶系列——WebApi 接口参数不再困惑:传参详解
    IIS事件查看器_WebServer事件查看器_帮助查看IIS-Web服务器事件执行日志
    SQL分页查询的几种方式
    freeRTOS中文实用教程3--中断管理之延迟中断处理
    freeRTOS中文实用教程2--队列
    freeRTOS中文实用教程1--任务
    UML和模式应用5:细化阶段(5)---系统顺序图
  • 原文地址:https://www.cnblogs.com/weeping/p/6669757.html
Copyright © 2011-2022 走看看