Can they solve it either only with use of suffix automaton or only with use of suffix array or they need both structures?
Note that any structure may be used an unlimited number of times, the structures may be used in any order.
Input The first line contains a non-empty word s. The second line contains a non-empty word t. Words s and t are different. Each word consists only of lowercase English letters. Each word contains at most 100 letters. Output In the single line print the answer to the problem. Print "need tree" (without the quotes) if word s cannot be transformed into word t even with use of both suffix array and suffix automaton. Print "automaton" (without the quotes) if you need only the suffix automaton to solve the problem. Print "array" (without the quotes) if you need only the suffix array to solve the problem. Print "both" (without the quotes), if you need both data structures to solve the problem. It's guaranteed that if you can solve the problem only with use of suffix array, then it is impossible to solve it only with use of suffix automaton. This is also true for suffix automaton. Sample Input Input automaton tomat Output automaton Input array arary Output array Input both hot Output both Input need tree Output need tree Hint In the third sample you can act like that: first transform "both" into "oth" by removing the first character using the suffix automaton and then make two swaps of the string using the suffix array and get "hot"./*—————————————————————————————————————————————————————————————————————— author : Grant Yuan time : 2014.7.22 algorithm : 字符串匹配 explain : 对两个字符串中的每一个字符的个数进行统计,假设第二个字符串中有字符的个数 比第一个字符串中对应字符的个数多。则输出“need tree”。 否则,用字符串B在A中进行单个字符的一一匹配。假设可以匹配下来,则为“automaton”,否则,假设两个字符串的长度 相等。则为“array”,假设前面条件都不满足,则为“both”。
—————————————————————————————————————————————————————————————————————— */ #include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> using namespace std; int a[26]; char s1[103],s2[103]; int l1,l2; int main() { while(~scanf("%s%s",&s1,&s2)){ l1=strlen(s1); l2=strlen(s2); memset(a,0,sizeof(a)); int t=0; for(int i=0;i<l1;i++) { a[s1[i]-'a']++; } for(int i=0;i<l2;i++) a[s2[i]-'a']--; int flag1=1; for(int i=0;i<26;i++){ if(a[i]<0) flag1=0; } if(flag1==0) printf("need tree\n"); else { int t=0; for(int i=0;i<l1;i++) if(s1[i]==s2[t]) t++; if(t==l2) printf("automaton\n"); else if(l1==l2) printf("array\n"); else printf("both\n"); } } return 0; } </span>