B. Equidistant String

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#字符串 #比赛 #数据结构 #for循环

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小苏伊喜欢处理由零和一组成的字符串，并计算它们之间的汉明距离。她寻求找到一个长度相等的字符串，使得这个新字符串到两个已知字符串的距离相等。通过解析输入的字符串并输出符合条件的字符串或“不可能”的信息，读者可以跟随小苏伊的思维过程，体验解决这类问题的乐趣。

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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 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.

#include<stdio.h>
#include<string.h>
#define N 100010

char s1[N], s2[N];

int main()
{
    int len;
    while(~scanf("%s%s", s1, s2))
    {
        int k=2, t=0;
        len=strlen(s1);
        for(int i=0;i<len;i++)
        {
            if(s1[i]!=s2[i])
            {
                t++;
            }
        }
        if(t%2!=0)
        {
            printf("impossible\n");
            continue;
        }
        for(int i=0;i<len;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;
}