【组队赛三】-C cf448B

Suffix Structures
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
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Status

Practice

CodeForces 448B
Description
Bizon the Champion isn't just a bison. He also is a favorite of the "Bizons" team.

At a competition the "Bizons" got the following problem: "You are given two distinct words (strings of English letters), s and t. You need to transform word s into word t". The task looked simple to the guys because they know the suffix data structures well. Bizon Senior loves suffix automaton. By applying it once to a string, he can remove from this string any single character. Bizon Middle knows suffix array well. By applying it once to a string, he can swap any two characters of this string. The guys do not know anything about the suffix tree, but it can help them do much more.

Bizon the Champion wonders whether the "Bizons" can solve the problem. Perhaps, the solution do not require both data structures. Find out whether the guys can solve the problem and if they can, how do they do it? 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".

<span style="color:#3333ff;">/*
——————————————————————————————————————————————————————————————————————
       
       author    :     Grant Yuan
       time      :     2014.7.22
       algorithm :     字符串匹配
       explain   :     对两个字符串中的每个字符的个数进行统计，如果第二个字符串中有字符的个数
                       比第一个字符串中相应字符的个数多，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，如果能够匹配下来，则为“automaton”，否则，如果两个字符串的长度
                       相等，则为“array”，如果前面条件都不满足，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
");
             else  {
                        int t=0;
                        for(int i=0;i<l1;i++)
                                if(s1[i]==s2[t]) t++;
                        if(t==l2)
                            printf("automaton
");
                        else
                                if(l1==l2) printf("array
");
                        else
                                printf("both
");
             }
    }
    return 0;
}
</span>