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最新推荐文章于 2019-04-09 18:29:06 发布

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本文探讨了如何根据两个由0和1组成的等长字符串s1和s2，找出一个同样长度的字符串p，使得p与s1及s2之间的汉明距离相等。文章提供了一个具体的实现算法，展示了如何通过判断s1和s2中不匹配位的数量是否为偶数来决定解决方案的存在性，并给出了示例。

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Description

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 s_i isn't equal to t_i.

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.

Sample Input

Input

0001
1011

Output

Input

000
111

Output

impossible

Hint

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

思路：输入两个相同长度的字符串s1,s2，找一个字符串和s1,s2的相似度一样，例如和s1有两个一样的，则和s2也需要两个一样的。s1,s2的不一样的字符需是偶数个才能找到和他们相似度一样的。

#include<stdio.h>
#include<string.h>
#define N 100010
int main()
{
 int len,i;
 char s1[N],s2[N];
 while(~scanf("%s%s",s1,s2))
 {
    int k=2;
    int n=0;
   len=strlen(s1);
   for(i=0;i<=len-1;i++)
   {
     if(s1[i]!=s2[i])
     {
      n++;
     }

   }
   if(n%2!=0)
   {
     printf("impossible\n");
continue;
   }
     for(i=0;i<=len-1;i++)
     {
     if(s1[i]==s2[i])
     {
       printf("0");
     }
     else
     {
       if(k%2)
         printf("%c",s1[i]);
       else
         printf("%c",s2[i]);
        k++;
     }
 }
printf("\n");
 }


  return 0;

}