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.
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;
}
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