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.

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.

Print the ultimate string by the book.

  
  

   
   asdf sdfg
asdf ghjk
  
  

  
  

   
   asdfg
asdfghjk
  
  

Wang Ye

2008杭电集训队选拔赛——热身赛

lcy   |   We have carefully selected several similar problems for you:   1866  1277  1358  1865  2057 

题意：

将两个字符串合并将匹配的字符串删去按字符串的字典数小的输出；

#include<stdio.h>
#include<string.h>
int next[100005];
void NEW(char b[])//子串的next
{
    int i,q=0;
    next[0]=0;
    int m1;
    m1=strlen(b);
    for(i=1;i<m1;i++)
    {
        while(q>0&&b[q]!=b[i])
            q=next[q-1];
        if(b[q]==b[i])
            q++;
        next[i]=q;
    }
}
int xiao(char a[],char b[])
{
    int i;
    NEW(b);
    int q=0;
    int m=strlen(a);
    for(i=0;i<m;i++)
    {
        while(q>0&&a[i]!=b[q])
            q=next[q-1];
        if(a[i]==b[q])
            q++;
    }
    return q;
}//母函数模板。
int main()
{
    char a[100005],b[100005];
    while(~scanf("%s%s",a,b))
    {
        int m,n;
        m=xiao(a,b);
        n=xiao(b,a);//比较出那个匹配的最少。
        if(m==n)//
        {
            if(strcmp(a,b)>0)//相等的会看哪个在前面使字典数较小。
           {
              printf("%s",b);
              printf("%sn",a+m);


           }
            else
            {
               printf("%s",a);
                printf("%sn",b+m);
           }
        }
       else if(m>n)//比较出那个匹配使字符串较短。
        {
             printf("%s",a);
            printf("%sn",b+m);
        }
        else
        {
             printf("%s",b);
            printf("%sn",a+n);
        }
    }
    return 0;
}