Codeforces 545B Equidistant String【思维】

B. Equidistant String

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.

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.

题目大意：

汉明距离：两个字符串位子相等且字符相等的个数。

现在给你一个字符串S，给你一个字符串T，问你能否找到一个字符串P,使得p和S的汉明距离==p和T的汉明距离。

思路：

1、对应字符串S和T，两个字符串相等的位子如果字符相等，那么我们对应在字符串P等同位子上，也设定为相同的字符。

2、如果s【i】！=t【i】，那么我们肯定考虑将这个位子设定为s【i】或者t【i】的那个，那么如果设定为s【i】，那么很肯定，p到S的汉明距离此时就要小于P到T的汉明距离。

那么我们统计s【i】!=t【i】的个数，如果个数为奇数，那么就不存在字符串P，否则就存在，并且按照统计s【i】！=t【i】的个数的奇偶来设定字符串P即可（cont%2==1的时候设定为s【i】，那么cont%2==0的时候就要设定为t【i】）。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
char a[100060];
char b[100060];
char ans[100060];
int main()
{
while(~scanf("%s%s",a,b))
{
int cont=0;
int n=strlen(a);
for(int i=0;i<n;i++)
{
if(a[i]==b[i])continue;
else cont++;
}
if(cont%2==0)
{
for(int i=0;i<n;i++)
{
if(a[i]==b[i])printf("%c",a[i]);
else
{
if(cont%2==0)printf("%c",a[i]);
else printf("%c",b[i]);
cont--;
}
}
printf("\n");
}
else printf("impossible\n");
}
}