Codeforces 282C XOR and OR【思维】

C. XOR and OR

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The Bitlandians are quite weird people. They do everything differently. They have a different alphabet so they have a different definition for a string.

A Bitlandish string is a string made only of characters "0" and "1".

BitHaval (the mayor of Bitland) loves to play with Bitlandish strings. He takes some Bitlandish stringa, and applies several (possibly zero) operations to it. In one operation the mayor may take any two adjacent characters of a string, define one of them asx and the other one as y. Then he calculates two values p andq: p = x xor y,q = x or y. Then he replaces one of the two taken characters byp and the other one by q.

The xor operation means the bitwise excluding OR operation. Theor operation is the bitwise OR operation.

So for example one operation can transform string 11 to string 10 or to string 01. String 1 cannot be transformed into any other string.

You've got two Bitlandish strings a andb. Your task is to check if it is possible for BitHaval to transform stringa to string b in several (possibly zero) described operations.

Input

The first line contains Bitlandish string a, the second line contains Bitlandish stringb. The strings can have different lengths.

It is guaranteed that the given strings only consist of characters "0" and "1". The strings are not empty, their length doesn't exceed10⁶.

Output

Print "YES" if a can be transformed intob, otherwise print "NO". Please do not print the quotes.

Examples

Input

1110

Output

YES

Input

101

Output

NO

Input

000101

Output

NO

题目大意：

问最终a字符串能否变成b字符串。

我们操作可以拿出相邻两个字符使得其进行亦或或者是或操作：

亦或：10->11 01->11 11->01 11->10 00->00

或：10->11 01->11 11->11 00->00

思路：

很显然，00只能变成00而不能变成其他的情况，而01 10 和11之间可以任意的互换。

那么我们分类 讨论：

①如果一开始两个字符串是相等的，那么就是YES.

②否则判断两个字符串的长度是否相等，如果不等就是NO.

③如果两个字符串长度相同，那么判断是否有一方全都是0.如果存在，就是NO.否则就是YES.

Ac代码：

#include<stdio.h>#include<string.h>using namespace std;char a[1500000];char b[1500000];int main(){    while(~scanf("%s%s",a,b))    {        if(strcmp(a,b)==0)printf("YES\n");        else        {            if(strlen(a)!=strlen(b))            {                printf("NO\n");            }            else            {                int n=strlen(a);                int flag=0;                for(int i=0;i<n;i++)                {                    if(a[i]=='1')flag=1;                }                if(flag==0)                {                    printf("NO\n");                }                else                {                    flag=0;                    for(int i=0;i<n;i++)                    {                        if(b[i]=='1')flag=1;                    }                    if(flag==0)printf("NO\n");                    else printf("YES\n");                }            }        }    }}