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 test(s)
Input
automaton tomat
Output
automaton
Input
array arary
Output
array
Input
both hot
Output
both
Input
need tree
Output
need tree
题目大意：
给定两字符串，如果串2可以通过串1删除若干个字符得到的话，输出automaton，如果串1可以改变顺序则输array，两个都用到为both，否则为need tree
有点技巧的主要是判断除第一种之外的情况。
#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
#include<cmath>
#include<algorithm>
#define LL int
#define inf 0x3f3f3f3f
using namespace std;
char a[101],b[101];
bool bj[1000];
int main()
{
LL n,m,i,j,k,l1,l2;
while(scanf("%s",a)!=EOF)
{
bool vis=false;
scanf("%s",b);
l1=strlen(a);l2=strlen(b);
k=0;
for(i=0;i<l1;i++)
{
if(a[i]==b[k])
k++;
if(b[k]=='\0')
{
vis=true;
break;
}
}
if(vis)
{
printf("automaton\n");
continue;
}
memset(bj,false,sizeof(bj));
k=0;
for(i=0;i<l2;i++)
{
for(j=0;j<l1;j++)//之所以将第一个串放在内层，主要是对串1操作来得到串2
{
if(!bj[j])//比较完后且有相同匹配的字符则标记下来。
{
if(a[j]==b[i])
{
k++;
bj[j]=true;
break;
}
}
}
}
if(k==l2)//如果标记串1中的数量和串2相同的话，是以下2种情况
{
if(l1==l2)
{
printf("array\n");
}
else
printf("both\n");
}
else<span id="transmark"></span>
printf("need tree\n");
}
return 0;
}
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