codeforces 559B Equivalent Strings
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".
递归~
神奇的题目,直接递归就好了,刚开始T得超夸张,后来加上一句判断现在两个串是否相等就过了……而且还非常快……
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int n,m;
char a[200001],b[200001];
bool dfs(int l1,int r1,int l2,int r2)
{
if(l1==r1) return a[l1]==b[l2];
if((r1-l1+1)%2)
{
for(int i=l1;i<=r1;i++)
if(a[i]!=b[i+l2-l1]) return 0;
return 1;
}
int flag=1;
for(int i=l1;i<=r1;i++)
if(a[i]!=b[i+l2-l1])
{
flag=0;break;
}
if(flag) return 1;
int mid1=l1+r1>>1,mid2=l2+r2>>1;
if(dfs(mid1+1,r1,l2,mid2) && dfs(l1,mid1,mid2+1,r2)) return 1;
return (dfs(l1,mid1,l2,mid2) && dfs(mid1+1,r1,mid2+1,r2));
}
int main()
{
scanf("%s%s",a+1,b+1);
n=strlen(a+1);m=strlen(b+1);
if(n!=m)
{
puts("NO");return 0;
}
if(dfs(1,n,1,m)) puts("YES");
else puts("NO");
return 0;
}
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