hdu1867A + B for you again

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

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>

using namespace std;

char s1[],s2[],s3[],s4[];
int nextt[];
int match[];

int main()
{
    int i,j,k;
    while(scanf("%s%s",s1+,s2+)!=EOF)
    {
        int l1 = strlen(s1+),l2 = strlen(s2+);
        nextt[] = ;
        for(i = ;i<=l2;i++)
        {
            int t = nextt[i-];
            while(t&&s2[i]!=s2[t+]) t = nextt[t];
            if(s2[i] == s2[t+]) t++;
            nextt[i] = t;
        }
        match[] = ;
        for(i = ;i<=l1;i++)
        {
            int t = match[i-];
            while(t&&s1[i]!=s2[t+]) t = nextt[t];
            if(s1[i] == s2[t+]) t++;
            match[i] = t;
        }
        int t1 = match[l1];
        for(i = ;i<=l1;i++)
            s3[i] = s1[i];
        for(i = l1+;i<=l1+l2-t1+;i++)
            s3[i] = s2[i-l1+t1];
        nextt[] = ;
        for(i = ;i<=l1;i++)
        {
            int t = nextt[i-];
            while(t&&s1[i]!=s1[t+]) t = nextt[t];
            if(s1[i] == s1[t+]) t++;
            nextt[i] = t;
        }
        match[] = ;
        for(i = ;i<=l2;i++)
        {
            int t = match[i-];
            while(t&&s2[i]!=s1[t+]) t = nextt[t];
            if(s2[i] == s1[t+]) t++;
            match[i] = t;
        }
        int t2 = match[l2];
        for(i = ;i<=l2;i++)
            s4[i] = s2[i];
        for(i = l2+;i<=l2+l1-t2+;i++)
            s4[i] = s1[i-l2+t2];
        if(t1>t2) printf("%s",s3+);
        else if(t2>t1) printf("%s",s4+);
        else
        {
            int l = l1+l2-t2;
            bool bb = ;
            for(i = ;i<=l;i++)
            {
                if(s3[i]<s4[i])
                {
                    printf("%s",s3+);
                    bb = ;
                    break;
                }
                else if(s3[i]>s4[i])
                {
                    printf("%s",s4+);
                    bb = ;
                    break;
                }
            }
            if(!bb) printf("%s",s3+);
        }
        puts("");
    }
    return ;
}