Equidistant String

Little Susie loves strings. Today she calculates distances between them. As Susie is a small girl after all, her strings contain only digits zero and one. She uses the definition of Hamming distance:

We will define the distance between two strings s and t of the same length consisting of digits zero and one as the number of positions i, such that si isn’t equal to ti.

As besides everything else Susie loves symmetry, she wants to find for two strings s and t of length n such string p of length n, that the distance from p to s was equal to the distance from p to t.

It’s time for Susie to go to bed, help her find such string p or state that it is impossible.

Input

The first line contains string s of length n.

The second line contains string t of length n.

The length of string n is within range from 1 to 105. It is guaranteed that both strings contain only digits zero and one.

Output

Print a string of length n, consisting of digits zero and one, that meets the problem statement. If no such string exist, print on a single line “impossible” (without the quotes).

If there are multiple possible answers, print any of them.

Examples

Input
0001
1011
Output
0011
Input
000
111
Output
impossible

Note

In the first sample different answers are possible, namely — 0010, 0011, 0110, 0111, 1000, 1001, 1100, 1101.

ac代码

#include<bits/stdc++.h>
using namespace std;
 
const int maxn=1e5+10;
 
string a,b;
int vis[maxn];
int cnt=0;
 
int main(){
	 
	cin>>a;
	cin>>b; 
	for(int i=0;i<a.size();i++){
		if(a[i]!=b[i]){
			cnt++;
			vis[i]=1;
		}
	}	
	
	if(cnt%2==0){
		int len=cnt/2;
		int cntt=0;
		int i=0;
		for(;i<a.size();i++){
			if(vis[i]){
				if(a[i]=='0')
					cout<<1;
				else
					cout<<0;	
				cntt++;			
			}else{
				cout<<a[i];
			} 
			if(cntt==len){
				i++;
				break;
			}
		}
		for(;i<a.length();i++){
			cout<<a[i];
		} 
	}else
		cout<<"impossible";
	return 0;
}

题目大意

就是给你两个用0和1组成的两个长度相同的字符串，要求一个字符串，使得这个字符串跟两个字符串 相同位置不同字符的个数 相同。

题目解析

首先统计两个字符串相同位置不同字符的个数有多少个，如果这个数是偶数的话，就可以求出来这个串，否则直接输出impossib。相同的字母不用管，因为第三个串直接照搬就行，就看这个不同的地方，只要让一半不同的地方改掉就可以

一个技巧

输出的时候，在不同的地方找len个字符变成相反的，剩下的照搬。怎么做到呢，就是先定义一个cnt记录改变的个数，每次判断一下，如果这个cnt等于了len，就break，否则继续下面的。

来源：CSDN

作者：Smi_l_e_

链接：https://blog.csdn.net/Smi_l_e_/article/details/104017478

标签

string

impossible