Codeforces Round #313 (Div. 二) D. Equivalent Strings(简单搜索)
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are called equivalent in one of the two cases:
- They are equal.
- 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:
- a1 is equivalent to b1, and a2 is equivalent to b2
- 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!
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.
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
aaba abaa
YES
aabb abab
NO
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".
题目链接:codeforces.com/contest/560/problem/D
题目大意:给出两串长度相等的字符串a,b,问其是否为等价字符串。
判断依据为:1.两个字符串相等。
2.把每个字符串拆分成长度相等的两个子串:a=a1+a2,b=b1+b2,如果a1=b2且a2=b1,则a与b等价
深搜递归判断子串是否等价,子串等价,主串就一定等价。注意子串长度相等的条件,不满足不能拆。
代码如下:
#include <cstdio> #include <cstring> #include <iostream> #include <string> using namespace std; string a,b; bool eq(int len,string x,string y)//用来判断x,y是否等价 { if(x==y) return true; if(len%2||len==0) //当主串不能拆时,不等价 return false; len/=2; string a1=x.substr(0,len); string a2=x.substr(len,len); string b1=y.substr(0,len); string b2=y.substr(len,len); if(eq(len,a1,b2)&&eq(len,a2,b1)) return true; if(eq(len,a1,b1)&&eq(len,a2,b2)) return true; return false; } int main(void) { cin>>a; cin>>b; bool p=true; int len=a.length(); if(eq(len,a,b)) printf("YES\n"); else printf("NO\n"); }
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