zoukankan      html  css  js  c++  java
  • Evanyou Blog 彩带

      题目传送门

    Seek the Name, Seek the Fame

    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 23798   Accepted: 12431

    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

     


      分析:

      又是一道神奇的$KMP$题,又是需要掌握$next$数组的正确姿势。(咦,我为什么要说又?)

      首先一个字符串它自己肯定算一个答案。求出$next$数组然后从$next[n]$开始计数,只要当前$next[]$不为$0$,那么就记录新答案为$next[]$值,然后继续。至于正确性可以自己证明,只要深入理解了$next$数组,就不难理解了。

      (当然,这题可以用$hash$做,且复杂度也差不多是$O(n)$的)

      Code:

     

    //It is made by HolseLee on 11th Aug 2018
    //POJ2752
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<cmath>
    #include<iostream>
    #include<iomanip>
    #include<algorithm>
    using namespace std;
    
    const int N=2e6+7;
    int n,nxt[N],k,ans[N],cnt;
    char s[N];
    
    int main()
    {
        while(scanf("%s",s)!=EOF){
            n=strlen(s);
            nxt[0]=nxt[1]=0;k=0;
            for(int i=1;i<n;++i){
                while(k&&s[i]!=s[k])
                k=nxt[k];
                nxt[i+1]=(s[i]==s[k]?++k:0);
            }
            k=nxt[n];cnt=1;ans[1]=n;
            while(k){
                ans[++cnt]=k;
                k=nxt[k];
            }
            for(int i=cnt;i>=1;--i)
            printf("%d ",ans[i]);
            printf("
    ");
        }
        return 0;
    }

     

     

  • 相关阅读:
    win查看所有wifi密码
    vsftp配置详解
    python3.7项目打包为一个exe
    ATT&CK实战系列——红队实战(一)
    PHP SECURITY CALENDAR 2017 (Day 9
    python3安装gmpy2
    [CISCN2019 总决赛 Day2 Web1]Easyweb(预期解)
    python2与python3共存及py2IDLE打不开的解决方案
    [BJDCTF 2nd]
    PHP SECURITY CALENDAR 2017 (Day 1
  • 原文地址:https://www.cnblogs.com/cytus/p/9458985.html
Copyright © 2011-2022 走看看