周赛-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.
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.
判断字符不同的个数

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-9
#define LL long long
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define CRR fclose(stdin)
#define CWW fclose(stdout)
#define RR freopen("input.txt","r",stdin)
#pragma comment(linker,"/STACK:102400000")
#define WW freopen("output.txt","w",stdout)

const int Max = 1e5;

char s[Max*10];
char t[Max*10];
char p[Max*10];
int num;
int main()
{
    int len;
    scanf("%s",s);
    scanf("%s",t);
    len=strlen(s);
    num=0;
    for(int i=0;i<len;i++)
    {
        if(s[i]==t[i])
        {
            p[i]=s[i];
        }
        if(s[i]!=t[i])
        {
            num++;
            if(num&1)
            {
                p[i]=s[i];
            }
            else
            {
                p[i]=t[i];
            }
        }
    }
    if(num==0)
    {
        printf("%s",s);
    }
    else if(num&1)
    {
        printf("impossible\n");
    }
    else
    {
        p[len]='\0';
        printf("%s\n",p);
    }
    return 0;
}

来源：CSDN

作者：绝尘花遗落

链接：https://blog.csdn.net/huayunhualuo/article/details/47358905

标签

string

impossible