zoukankan      html  css  js  c++  java
  • POJ 2752 Seek the Name, Seek the Fame(next数组运用)

    Seek the Name, Seek the Fame
    Time Limit: 2000MS        Memory Limit: 65536K
    Total Submissions: 24000        Accepted: 12525
    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
    题意:给出一个字符串str,求出str中存在多少子串,既是str的前缀,又是str的后缀,从小到大输出长度(开一个数组倒序输出即可)

    #include<iostream>
    #include<string.h>
    using namespace std;
    char s[411111];
    int pp,nex[411111],nexx[411111];
    void getnext()
    {
        int i=0,k=-1;
        nex[0]=-1;
        while(i<pp)
        {
            if(k==-1||s[i]==s[k])
                nex[++i]=++k;//,cout<<nex[i];
            else
                k=nex[k];
        }
    }
    int main()
    {
        int j;
        while(~scanf("%s",s))
        {
            j=0;
            pp=strlen(s);
            getnext();
            for(int i=pp;nex[i]!=-1;i=nex[i])
                nexx[j++]=i;
            for(int i=j-1;i>=0;i--)
                i?printf("%d ",nexx[i]):printf("%d
    ",nexx[i]);
        }
        return 0;
    }
    
  • 相关阅读:
    SQL Select Case
    Fxcop:Globlization warning/全球化警告
    How to get free(available) physical memory/获取可用的内存
    CA2000: Dispose objects before losing scope/在退出作用域前Dispose object
    并行编译Parallel Building
    VS2010 WebInstall/web安装
    欧洲天空访问
    .NET Framework Client Profile/.net framework 客户端配置
    新书推荐:Practical Code Generation in .NET: Covering Visual Studio 2005, 2008, and 2010
    Disable floppy driver(禁用软盘)
  • 原文地址:https://www.cnblogs.com/aeipyuan/p/9502677.html
Copyright © 2011-2022 走看看