A + B for you again

Time Limit : 5000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 3   Accepted Submission(s) : 1

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

 

kmp的应用

代码

1 #include 
       
          2 #include 
        
           3 #include 
         
            4 using namespace std;  5 const int Max=100010;  6 int next[Max];  7 char str1[Max],str2[Max];  8 void get_next(char *pat,int len)  9 {
     10     int i=0,j=-1; 11     next[0]=-1; 12     while(i<=len) 13     {
     14         if(j==-1||pat[i]==pat[j]) 15         {
     16             i++,j++; 17             next[i]=j; 18         } 19         else 20         j=next[j]; 21     } 22 } 23 24 int kmp(char *s,char *pat) 25 {
     26     int i = 0,j = 0,len1 = strlen(s),len2 = strlen(pat); 27     get_next(pat,len2); 28     while(i
          
           
            =len1) 38       return j; 39     else 40       return 0; 41 } 42 int main() 43 {
     44    while(scanf("%s%s",str1,str2)!=EOF) 45     {
     46         int idx1,idx2; 47         idx2 = kmp(str1,str2); 48         idx1 = kmp(str2,str1); 49 // printf("idx2=%d  idx1=%d\n",idx2,idx1); 50         if( idx1 == idx2) 51         {
     52             if(strcmp(str1,str2) < 0) 53             {
     54                 printf("%s%s",str1,str2+idx2); 55             } 56             else 57             {
     58                 printf("%s%s",str2,str1+idx1); 59             } 60         } 61         else if( idx2 > idx1 ) 62         {
     63             printf("%s%s",str1,str2+idx2); 64         } 65         else 66         {
     67             printf("%s%s",str2,str1+idx1); 68         } 69         printf("\n"); 70     } 71     return 0; 72 }