zoukankan      html  css  js  c++  java
  • hdoj_1867A + B for you again

    A + B for you again

    Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 2989    Accepted Submission(s): 730


    Problem Description
    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<iostream>
    #include<cstdio>
    #include<cstring>
    using namespace std;
    #pragma warning(disable : 4996)
    int Next[100005];
    void get_next(char *B, int m)
    {
    	char pat[100005] = {0};
    	strcpy(pat + 1, B);
    	Next[1] = 0;
    	int i, j = 0;
    	for(i = 2; i <= m; i++)
    	{
    		while(j > 0 && pat[j+1] != pat[i])
    		{
    			j = Next[j];
    		}
    		if(pat[j+1] == pat[i])
    		{
    			j += 1;
    		}
    		Next[i] = j;
    	}
    }
    int kmp(char *A, char *B)
    {
    	char text[100005] = {0};
    	char pat[100005] = {0};
    	strcpy(text + 1, A);
    	strcpy(pat + 1, B);
    	int n = strlen(text + 1);
    	int m = strlen(pat + 1);
    	get_next(B, m);
    	int i, j = 0;
    	for(i = 1; i <= n; i++)
    	{
    		while(j > 0 && pat[j+1] != text[i])
    		{
    			j = Next[j];
    		}
    		if(pat[j+1] == text[i])
    		{
    			j += 1;
    		}
    		if(i == n)
    		{
    			return j;
    		}
    	}
    	return 0;
    }
    int main()
    {
    	freopen("in.txt", "r", stdin);
    	char text[100005], pat[100005];
    	int x, y;
    	while(scanf("%s%s", text + 1, pat + 1) != EOF)
    	{
    		x = kmp(text + 1, pat + 1);
    		y = kmp(pat + 1, text + 1);
    	//	cout << x << " " << y << endl;
    		if(x == y)
    		{
    			if(strcmp(text + 1, pat + 1) > 0)
    			{
    				printf("%s", pat + 1);
    				printf("%s\n", text + x + 1);
    			}
    			else if(strcmp(text + 1, pat + 1) < 0)
    			{
    				printf("%s", text + 1);
    				printf("%s\n", pat + x + 1);
    			}
    			else
    			{
    				printf("%s\n", text + 1);
    			}
    		}
    		else if(x > y)
    		{
    			printf("%s", text + 1);
    			printf("%s\n", pat + x + 1);
    		}
    		else
    		{
    			printf("%s", pat + 1);
    			printf("%s\n", text + y + 1);
    		}
    	}
    }
    这个模版好搓=。=
    #include <iostream>
     #include <cstring>
     #include <cstdio>
     const int N=100001;
     using namespace std;
     int nextt[N];
     void next(char s[])
     {
         int i=1,j=0;
         int len=strlen(s);
         nextt[0]=-1;
         while(i<len)
         {
             if(j==-1||s[i]==s[j])
             {
                 ++i;
                 ++j;
                 if(s[i]!=s[j])
                     nextt[i]=j;
                 else
                     nextt[i]=nextt[j];
             }
             else
                 j=nextt[j];
         }
     }
     int kmp(char ss[],char s[])
     {
         int len1=strlen(ss);
         int len2=strlen(s);
         next(s);
         int i=0,j=0;
         while(i<len1&&j<len2)
         {
             if(j==-1||ss[i]==s[j])
             {
                 ++i;
                 ++j;
             }
             else
                 j=nextt[j];
         }
         if(i==len1)  
             return j; 
         return 0;
     }
     int main()
     {
         char str1[N],str2[N];
         while(~scanf("%s%s",str1,str2))
         {
             int x=kmp(str1,str2);
             int y=kmp(str2,str1);
             if(x==y)
             {
                 if(strcmp(str1,str2)>0)
                 {
                     printf("%s",str2);
                     printf("%s\n",str1+x);
                 }
                 else
                 {
                     printf("%s",str1);
                     printf("%s\n",str2+x);
                 }
             }
             else if(x>y)
             {
                 printf("%s",str1);
                 printf("%s\n",str2+x);
             }
             else
             {
                 printf("%s",str2);
                 printf("%s\n",str1+y);
             }
         }
         return 0;
    }



  • 相关阅读:
    PHPStrom 设置终端字体大小
    PostgreSQL 9 夸库查询
    弹性布局
    sql中的 where 、group by 和 having 用法解析
    数据库面试中常问的几个问题
    SQL中 UNION 和 UNION ALL 操作符小结
    SQL里 inner JOIN、 left (OUTER) JOIN 、right (OUTER) JOIN、 full (OUTER) JOIN 之间的区别
    MYSQL中LIMIT用法
    Java集合框架小结
    jQuery$.each循环遍历详解,各种取值对比,$.each遍历数组、对象、Dom元素、二维数组、双层循坏、类json数据等等
  • 原文地址:https://www.cnblogs.com/lgh1992314/p/5835018.html
Copyright © 2011-2022 走看看