zoukankan      html  css  js  c++  java
  • HDU1867

    Generally speaking, there are a lot of problems about strings processing. Now you encounter another such problem. If you get two strings, such as “asdf” and “sdfg”, the result of the addition between them is “asdfg”, for “sdf” is the tail substring of “asdf” and the head substring of the “sdfg” . However, the result comes as “asdfghjk”, when you have to add “asdf” and “ghjk” and guarantee the shortest string first, then the minimum lexicographic second, the same rules for other additions.

    Input

    For each case, there are two strings (the chars selected just form ‘a’ to ‘z’) for you, and each length of theirs won’t exceed 10^5 and won’t be empty.

    Output

    Print the ultimate string by the book.

    Sample Input

    asdf sdfg
    asdf ghjk

    Sample Output

    asdfg
    asdfghjk
    
    #include <stdio.h>
    #include <string.h>
    
    int next[100005];
    
    void getnext(char str[])
    {
        int i = 1,j = 0;
        int len = strlen(str);
        next [0] = -1;
        while(i < len)
        {
            if(j == -1 || str[i] == str[j])
            {
                i++;
                j++;
                next[i] = j;
            }
            else
                j = next[j];
        }
    }
    
    int kmp(char str1[],char str2[])
    {
        int i= 0,j = 0;
        int len1 = strlen(str1),len2 = strlen(str2);
        getnext(str2);
        while(i<len1 && j<len2)
        {
            if(j == -1 || str1[i] == str2[j])
            {
                i++;
                j++;
            }
            else
                j = next[j];
        }
        if(i == len1)
            return j;
        return 0;
    }
    
    int main()
    {
        int x,y;
        char str1[100005],str2[100005];
        while(scanf("%s%s",str1,str2)!=EOF)
        {
            x = kmp(str1,str2);
            y = kmp(str2,str1);
            if(x == y)
            {
                if(strcmp(str1,str2)>0)
                {
                    printf("%s",str2);
                    printf("%s
    ",str1+x);
                }
                else
                {
                    printf("%s",str1);
                    printf("%s
    ",str2+x);
                }
            }
            else if(x>y)
            {
                printf("%s",str1);
                printf("%s
    ",str2+x);
            }
            else
            {
                printf("%s",str2);
                printf("%s
    ",str1+y);
            }
        }
    
        return 0;
    }
  • 相关阅读:
    perf-stat
    perf原理
    ubuntu中Docker的安装与使用
    NVM相关手册及新特性理解
    #2018BIT软件工程基础#结对项目:四则运算题目生成
    #2018BIT软件工程基础#个人项目:数独
    第一篇博文:自我介绍&新学期展望
    越早明白这些道理,越能少走一些弯路
    把知识连接起来就是创意
    【翻译】24款界面精美的免费UI工具包
  • 原文地址:https://www.cnblogs.com/aerer/p/9931013.html
Copyright © 2011-2022 走看看