HDOJ 题目1867A + B for you again（KMP） A + B for you again

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概述

A + B for you again

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4407 Accepted Submission(s): 1132

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

Author

Wang Ye

Source

2008杭电集训队选拔赛——热身赛

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有重复的就覆盖掉然后输出。不过这个得比较字符串str1 str2还得再比较str2 str1

ac代码

#include<stdio.h>
#include<string.h>
int next[100010];
void getnext(char *s)
{
	int len,j=0,k=-1;
	len=strlen(s);
	next[0]=-1;
	while(j<len)
	{
		if(k==-1||s[k]==s[j])
		{
			j++;
			k++;
			next[j]=k;
			/*if(s[j]!=s[k])
				next[j]=k;
			else
				next[j]=next[k];*/
		}
		else
			k=next[k];
	}
}
int kmp(char *s1,char *s2)
{
	int i=0,j=0,k;
	int len1=strlen(s1),len2=strlen(s2);
	getnext(s2);
	while(i<len1&&j<len2)
	{
		if(j==-1||s1[i]==s2[j])
		{
			i++;
			j++;
		}
		else
			j=next[j];
	}
	if(i==len1)
		return j;
	return 0;
}
int main()
{
	char s1[100010],s2[100010];
	while(scanf("%s%s",s1,s2)!=EOF)
	{
		int x=kmp(s1,s2);
		int y=kmp(s2,s1);
		if(x==y)
		{
			if(strcmp(s1,s2)>0)
			{
				printf("%s",s2);
				printf("%sn",s1+y);
			}
			else
			{
				printf("%s",s1);
				printf("%sn",s2+x);
			}
		}
		else
			if(x>y)
			{
				printf("%s",s1);
				printf("%sn",s2+x);
			}
			else
			{
				printf("%s",s2);
				printf("%sn",s1+y);
			}
	}
}