Equidistant String （周赛2）

最新推荐文章于 2020-03-05 15:25:39 发布

原创最新推荐文章于 2020-03-05 15:25:39 发布 · 396 阅读

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本文介绍了一个编程问题的解决方法，即计算两个由0和1组成的等长字符串之间的汉明距离，并寻找一个中间字符串，使其到两个给定字符串的距离相等。文章提供了具体的输入输出示例及一个C语言实现方案。

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 andt of the same length consisting of digits zero and one as the number of positionsi, such that s_i isn't equal tot_i.

As besides everything else Susie loves symmetry, she wants to find for two stringss andt of lengthn such stringp of lengthn, that the distance fromp to s was equal to the distance fromp tot.

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 lengthn.

The second line contains string t of lengthn.

The length of string n is within range from1 to10⁵. 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

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：输入两个长度相等的只有1和0的字符串，查找相同位置不相同的个数，如果是奇数，直接输出impossible，否则在循环，如果对应位置相等，输出第一个字符串的对应位置，否则如果为奇数位字符输出第二个字符串对应的位置，偶数位字符输出第一个字符对应的位置（方法不唯一，仅为其中一种~ = =）

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

char str1[100000],str2[100000];

int main()
{
    int num;
    int i,j;
    while(~scanf("%s%s",str1,str2))
    {
        num=0,j=0;
        for(i=0;str1[i]!='\0';i++)
        {
            if(str1[i]!=str2[i])
                num++;
        }

        if(num%2==1)
            printf("impossible\n");

        else
        {
            for(i=0;str1[i]!='\0';i++)
            {
                if(str1[i]==str2[i])
                        printf("%c",str1[i]);
                else
                {
                    if(j%2==0)
                        printf("%c",str1[i]);
                    else
                        printf("%c",str2[i]);

                    j++;
                }
            }
            printf("\n");
        }
    }
    return 0;
}