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A + B for you again Time Limit: 5000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3811????Accepted Submission(s): 970 Problem DescriptionGenerally 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 Inputasdf sdfg asdf ghjk Sample Outputasdfg asdfghjk

#include<stdio.h>#include<string.h>const int n=100001;int next[n];int len1,len2;void getnext(char s[]){int i=0,j=-1;next[0]=-1;while(i<=len2){if(j==-1||s[i]==s[j]){i++;j++;next[i]=j;}elsej=next[j];}}int kmp(char s1[],char s2[]){int i=0,j=0;len1=strlen(s1);len2=strlen(s2);getnext(s2);while(i<len1&&j<len2){if(j==-1||s1[i]==s2[j]){i++;j++;}elsej=next[j];}if(i>=len1)return j;elsereturn 0;}int main(){char str1[n],str2[n];while(~scanf("%s%s",str1,str2)) ? ? ? ? ? ? ? ? ? ?/*這一題不用str1+1會更加方便，上一題剪布條的其實也可以直接用s而不用s+1，只要把getnext（）和kmp（）里面的一些值改一下就可以 ?*/{int a=kmp(str1,str2);int b=kmp(str2,str1);if(a==b){if(strcmp(str1,str2)>0){printf("%s",str2);printf("%s\n",str1+a);}else{printf("%s",str1);printf("%s\n",str2+a);}}else if(a>b){printf("%s",str1);printf("%s\n",str2+a);}else{printf("%s",str2);printf("%s\n",str1+b);}}}

轉載于:https://www.cnblogs.com/tanjianwen/p/5245430.html

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