Codeforces Round #303 (Div. 2) B. Equidistant String 水题

B. Equidistant String

Time Limit: 20 Sec  Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/545/problem/B

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

0001
1011

Sample Output

0011

HINT

 

题意

给你两串只含01的字符串，然后让你构造出一组字符串，让他们的和构造出来的字符串差异都相同

题解：

首先看这两组有多少个位置不一样，如果是奇数，直接输出不可能，如果是偶数，那么就前一半为s，后一半为t就好了

代码：

 

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200001
#define mod 10007
#define eps 1e-9
int Num;
char CH[20];
//const int inf=0x7fffffff;   //нчоч╢С
const int inf=0x3f3f3f3f;
/*

inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
*/
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
//**************************************************************************************

string s;
string t;
int main()
{

    cin>>s>>t;
    int flag=0;
    int n=s.size();
    if(s==t)
    {
        cout<<s<<endl;
        return 0;
    }
    for(int i=0;i<n;i++)
    {
        if(s[i]!=t[i])
            flag++;
    }
    if(flag%2==1)
    {
        printf("impossible\n");
    }
    else
    {
        flag/=2;
        for(int i=0;i<n;i++)
        {
            if(s[i]!=t[i])
            {
                if(flag>0)
                {
                    cout<<s[i];
                    flag--;
                }
                else
                    cout<<t[i];
            }
            else
                cout<<s[i];
        }
    }
}

 

转载于:https://www.cnblogs.com/qscqesze/p/4516130.html