A + 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): 6276 Accepted Submission(s): 1551

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

给两串字符，输出最长合串（除去相同串）
列：asd asasd
输出 asasd

#include <iostream>#include <cstdio>#include <cstring>#define N 100010using namespace std;int next[100005];void getnext(char *t){    int i=1,j=0;next[1]=0;    int len=strlen(t+1);    while(i<len)    {        if(j==0 || t[i]==t[j])        {            ++i;++j;next[i]=j;        }        else            j=next[j];    }}int kmp(char *s,char *t){    int i,j,len1,len2;    getnext(t);    i=1;j=1;    len1=strlen(s+1);len2=strlen(t+1);    while(i<=len1 && j<=len2)    {        if(j==0 || s[i]==t[j])        {            ++i;++j;        }        else        {            j=next[j];        }    }    if(j>len2 && i<=len1)        return 0;    return j-1;//求出连续相同的串的长度}char str1[100005],str2[100005];int main(){    int mark1,mark2,j;    while(scanf("%s%s",str1+1,str2+1)!=EOF)    {        mark1=kmp(str1,str2);        mark2=kmp(str2,str1);        if(mark1 == mark2)        {            if(strcmp(str1+1,str2+1)>=0)            {                printf("%s",str2+1);                for(j=mark1+1;str1[j]!='\0';j++)                    printf("%c",str1[j]);            }            else            {                printf("%s",str1+1);                for(j=mark2+1;str2[j]!='\0';j++)                    printf("%c",str2[j]);            }        }        else if(mark1>mark2)        {            printf("%s",str1+1);            for(j=mark1+1;str2[j]!='\0';j++)                printf("%c",str2[j]);        }        else if(mark2>mark1)        {            printf("%s",str2+1);            for(j=mark2+1;str1[j]!='\0';j++)                printf("%c",str1[j]);        }        printf("\n");    }    return 0;}