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
解题思路
对两个字符串分别KMP，，得到相同前后缀长度。再根据长度判断该如何输出
#include <cstdio>#include <cstring>const int maxn = 100010;int next[maxn];char s1[maxn],s2[maxn];int len1,len2;void get_next(char *a){int len = strlen(a);int k = -1;int j = 0;next[j] = k;while(a[j] != ‘\0’) {if(k == -1 || a[j] == a[k]) {k++;j++;next[j] = k;}else k = next[k];}}int kmp(char *a,char *b){ //a主串 b模式串get_next(b);int i = 0,j = 0;while(a[i] != ‘\0’) {if(j == -1 || a[i] == b[j]) {i++;j++;}else j = next[j];}return j;}int main(){while(scanf(“%s%s”,s1,s2) != EOF) {len1 = strlen(s1);len2 = strlen(s2);int str1 = kmp(s1,s2);int str2 = kmp(s2,s1);if(str1 == str2) {if(strcmp(s1,s2) < 0) printf(“%s%s\n”,s1,s2+str1);\n”,s2,s1+str1);}else if(str1 < str2) {printf(“%s%s\n”,s2,s1+str2);}\n”,s1,s2+str1);}return 0;}
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