CodeForces 545B

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 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.
Sample Input
Input

0001
1011

Output

0011

Input

000
111

Output

impossible

Hint
In the first sample different answers are possible, namely — 0010, 0011, 0110, 0111, 1000, 1001, 1100, 1101.
사고방식: 같은 길이의 두 문자열 s1, s2를 입력하고 한 문자열을 찾아 s1, s2의 싱크로율과 같다. 예를 들어 s1과 같은 두 개의 문자열이 있으면 s2와도 같은 두 개가 필요하다.s1, s2의 다른 문자는 짝수이어야만 그들과 싱크로율이 같은 것을 찾을 수 있다.

#include
#include
#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
");
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("
");
 }


  return 0;

}