A + B for you again
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8929 Accepted Submission(s): 2163
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
思路:分别查看一个字符串的后缀是不是另一个字符串的前缀,然后取最大长度的
代码:
#include <map> #include <set> #include <cmath> #include <queue> #include <stack> #include <cstdio> #include <vector> #include <iomanip> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define ll long long #define mod 1000000007 #define mem(a) memset(a,0,sizeof(a)) using namespace std; const int maxn = 100000 + 5 , inf = 0x3f3f3f3f; int Next[maxn]; char str1[maxn],str2[maxn]; void cal_next(char *a){ memset(Next,-1,sizeof(Next)); int len = strlen(a); Next[0] = -1; int k = -1 ; for(int p = 1 ; p < len ; p ++ ){ while(k>-1&&a[k+1]!=a[p]) k = Next[k]; if(a[k+1]==a[p]) Next[p]=++k; } } int kmp(char *s,char *a){ cal_next(a); int slen = strlen(s); int alen = strlen(a); int k = -1; int i = 0; for(; i < slen ; i++ ){ while(k>-1&&s[i]!=a[k+1]) k = Next[k]; if(a[k+1]==s[i]) k++; } if(i==slen){ // cout<<i<<" "<<k<<endl; return k; } return -1; } void solve(){ int x = kmp(str1,str2); int y = kmp(str2,str1); if(x>y){ printf("%s",str1); printf("%s\n",str2+x+1); } else if(x<y){ printf("%s",str2); printf("%s\n",str1+y+1); } else{ if(strcmp(str1,str2)<0){ printf("%s",str1); printf("%s\n",str2+x+1); } else{ printf("%s",str2); printf("%s\n",str1+y+1); } } } int main(){ // freopen("in.txt","r",stdin); // freopen("out.txt","w",stdout); while(scanf("%s %s",str1,str2)!=EOF) solve(); return 0 ; }