Equivalent Strings
Description
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two stringsaandbof equal length are calledequivalentin one of the two cases:
- They are equal.
- If we split stringainto two halves of the same sizea1anda2, and stringbinto two halves of the same sizeb1andb2, then one of the following is correct:
- a1is equivalent tob1, anda2is equivalent tob2
- a1is equivalent tob2, anda2is equivalent tob1
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 from1to200 000and 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
aaba
abaa
YES
aabb
abab
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".
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef unsigned long long LL; 4 const int maxn=200010; 5 const int B=123; 6 char a[maxn],b[maxn]; 7 int cmp(int len,char x[],char y[]) 8 { 9 for(int i=0;i<len;i++) 10 if(x[i]<y[i])return -1; 11 else if(x[i]>y[i])return 1; 12 return 0; 13 } 14 void get(int len,char *s) 15 { 16 if(len&1)return ; 17 len/=2; 18 get(len,s); 19 get(len,s+len); 20 if(cmp(len,s,s+len)>0) 21 { 22 for(int i=0;i<len;i++) 23 swap(s[i],s[i+len]); 24 } 25 } 26 int main() 27 { 28 while(scanf("%s%s",a,b)!=EOF) 29 { 30 get(strlen(a),a); 31 get(strlen(b),b); 32 if(strcmp(a,b)==0)printf("YES\n"); 33 else printf("NO\n"); 34 } 35 return 0; 36 }View Code
一开始我也是在傻傻的暴力，然而并没有什么软用。。
仔细想想就知道会超时。。
其实我们只要按他的string比较规则把字符串 一层层比较然后排序，
在最后return string a==string b 就行了
这样是不是剪短了简答树的很多子叶呢？？
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