zoukankan      html  css  js  c++  java
  • POJ2752 Seek the Name, Seek the Fame E-kmp

     
    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
    #include <cstdio>
    #include <iostream>
    #include <cstring>
    #include <string>
    #include <algorithm>
    #include <cmath>
    using namespace std;
    char s[400005];
    int nxt[400005];
    int m;
    
    void pre_ekmp(){
        nxt[0] = m;
        int j = 0;
        while(j+1 < m && s[j] == s[j+1]) j++;
        nxt[1] = j;
        int k = 1;
        for (int i = 2;i < m;++i){
            int p = nxt[k] + k - 1;
            int len = nxt[i-k];
            if (i + len < p+1) nxt[i] = len;
            else {
                j = max(0,p-i+1);
                while(i+j < m && s[i+j] == s[j]) j++;
                k = i;
                nxt[i] = j;
            }
        }
    }
    
    int main(){
        while(~scanf("%s",s)){
            m = strlen(s);
            pre_ekmp();
            for (int i = m-1;i > 0;--i){
                if (nxt[i] == m-i)printf("%d ",nxt[i]);
            }
            printf("%d
    ",m);
        }
        return 0;
    }
  • 相关阅读:
    float、定位、inline-block、兼容性需注意的特性总结
    meta 标签 详细说明
    兼容探讨一
    javascript性能优化总结二(转载)
    javascript性能优化总结一(转载人家)
    特效合集(原生JS代码)适合初学者
    svg实现简单沙漏旋转
    SVG制作简单的图形
    SVG的简单介绍
    jQuery之效果
  • 原文地址:https://www.cnblogs.com/mizersy/p/9544219.html
Copyright © 2011-2022 走看看