time limit per test 1 second
memory limit per test 256 megabytes
input standard input
output standard output
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 test(s)
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.

【题意】

给定两个01串s和t，构造一个01串p，使得p与s和t不相同的部分相同，统计s和t不相同的位置个数，如果为奇数，直接输出impossible返回，如果为偶数，，则不相同的地方一半输出s，一般输出t，剩下的部分相同，输出s就行，，可以加个特判，，不过没有什么影响，都是62ms。

#include <bits/stdc++.h>
using namespace std;
int main()
{
     string s,t;
     cin>>s>>t;
     int n=s.size();
     int flag=0;
     if(s==t)
     {
         cout<<s<<endl;
         return 0;
     }
     for(int i=0;i<n;i++)
     {
         if(s[i]!=t[i])
            flag++;
     }
     if(flag%2==1)
     {
         puts("impossible");
     }
     else
     {
         flag/=2;
         for(int i=0;i<n;i++)
         {   
             if(s[i]!=t[i])
             {
             if(flag>0)
             {
                cout<<s[i];
                flag--;
             }
             else
                cout<<t[i];
             }
             else
                cout<<s[i];
         }
     }
     return 0;
}