周赛二 CodeForces 545B 思维题

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 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 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.

FA

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>

using namespace std;

int main()
{
    char s3[100002];
    char s1[100002],s2[100002];
    while(cin>>s1>>s2)
    {
        int len=strlen(s1);
        for(int i=0; i<len; i++)
            s3[i]='0';

        int num=0,num1=0,num2=0;
        for(int i=0; i<len; i++)
        {
            if(s1[i]=='1'&&s2[i]!='1')
                num1++;
            else if(s2[i]=='1'&&s1[i]!='1')
                num2++;
            if(s2[i]==s1[i])
                s3[i]=s1[i];
            else
                num++;
        }

        if(num%2)
            cout<<"impossible\n";
        else
        {
            int sum=0;
            if(num1>num2)
            {
                for(int i=0; i<len; i++)
                {
                    if(s1[i]=='1'&&s2[i]=='0')
                    {
                        s3[i]='1';
                        sum++;
                        if(sum+num2==num/2)
                            break;
                    }
                }
            }
            else if(num1<num2)
                for(int i=0; i<len; i++)
                {
                    if(s1[i]=='0'&&s2[i]=='1')
                    {
                        s3[i]='1';
                        sum++;
                        if(sum+num1==num/2)
                            break;
                    }
                }
            //cout<<sum;
            for(int i=0; i<len; i++)
                cout<<s3[i];
            cout<<endl;
        }
    }
}

先判断一下不相同的个数是否是偶数个，若不是，则不会找到一个距离相同的。若是偶数个，那么只要保证这偶数个不同位中一半是第一个字符串的，另一半是第二个字符串的就可以，可以用奇偶分开选的方法或前一半和后一半分开选的方法