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3thweek——E.暴力求解+DFS

时间:2015-08-02 16:44:59      阅读:105      评论:0      收藏:0      [点我收藏+]

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Description

Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are calledequivalent in one of the two cases:

  1. They are equal.
  2. If we split string a into two halves of the same size a1 and a2, and string b into two halves of the same size b1 and b2, then one of the following is correct:
    1. a1 is equivalent to b1, and a2 is equivalent to b2
    2. a1 is equivalent to b2, and a2 is equivalent to b1

As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.

Gerald has already completed this home task. Now it‘s your turn!

Input

The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200 000 and consists of lowercase English letters. The strings have the same length.

Output

Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.

Sample Input

案例输入输出:

Input
aaba abaa
Output
YES
Input
aabb abab
Output
NO

Hint

In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".

In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That‘s why string "aabb" is equivalent only to itself and to string "bbaa".

分析步骤:

1.先判断两串字符串长度为奇数还是偶数,奇数则要遍历全部的两个字符串,字符和顺序都相同才能输出YES,否则输出NO。

2.如果是偶数,两个字符串完全相同就直接输出YES,否则将两个字符串各切成两个子字符串再比较。

代码如下:

 1 #include<iostream>
 2 #include<cstring>
 3 using namespace std;
 4 const int maxn=200005;
 5 char a[maxn],b[maxn];
 6 
 7 int i,flag=0;
 8 int compare(char* a1,char* b1,int l)  //自定义比较函数
 9 {   
10     for(i=0;i<l;i++)
11     {
12         if(a1[i]!=b1[i])
13         {
14             return 0;
15         }
16     }
17     flag=1;
18     return flag;
19 }
20     
21 
22 int fun(char* a1,char* b1,int l)
23 {   
24     int k=0;
25     if(compare(a1,b1,l))
26            k=1; 
27     if(l%2==1)
28     {
29         if(k==1)
30            return 1;
31         else
32             return 0;
33     }    
34     else
35     {
36         if(k==1)
37             return 1;
38     else
39     {
40             l=l/2;
41           if( (fun(a1,b1,l)  && fun(a1+l,b1+l,l)) || (fun(a1,b1+l,l) && fun(a1+l,b1,l)) )   
42             return 1;
43           else
44               return 0;
45     }
46     }
47 }
48 
49 int main()
50 {
51     cin>>a;
52     cin>>b;
53     int len=strlen(a);
54     if(fun(a,b,len))
55        cout<<"YES"<<endl;
56     else
57         cout<<"NO"<<endl;
58 return 0;
59 }

 

 

 

3thweek——E.暴力求解+DFS

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原文地址:http://www.cnblogs.com/x512149882/p/4695928.html

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