A + B for you again

Posted 2020-10-13 给杰瑞一块奶酪~

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.

InputFor 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.OutputPrint the ultimate string by the book.Sample Input

asdf sdfg
asdf ghjk

Sample Output

asdfg
asdfghjk

合并字符串
很奇怪，一开始提交总是不对，以为写错了，换c++提交就对了
代码：

#include <stdio.h>
#include <string.h>
char a[100005],b[100005];
int next[100005];
void getnext(int n,char *s)
{
    int i = 0,k = -1;
    next[i] = -1;
    while(i < n)
    {
        if(k == -1 || s[k] == s[i])
        {
            next[++ i] = ++k;
        }
        else k = next[k];
    }
}
int kmp(char *p,char *q)
{
    int n = strlen(p),m = strlen(q);
    getnext(m,q);
    int i = -1,k = -1;
    while(i < n)
    {
        if(k == -1 || p[i] == q[k])
        {
            i ++;
            k ++;
        }
        else k = next[k];
    }
    return k;
}
int main()
{
    while(~scanf("%s%s",a,b))
    {
        int d = kmp(a,b),e = kmp(b,a);//保存a后缀和b前缀的最大重叠 以及 反过来 比较哪个大 就重叠那个 按相应顺序输出，如果相等就按字典序
        if(d > e)printf("%s%s",a,b+d);
        else if(d < e)printf("%s%s",b,a+e);
        else
        {
            if(strcmp(a,b) > 0)printf("%s%s",b,a+e);
            else printf("%s%s",a,b+d);
        }
        putchar(‘\n‘);
    }
}