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 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 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[];
//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=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
inline void P(int x)
{
Num=;if(!x){putchar('');puts("");return;}
while(x>)CH[++Num]=x%,x/=;
while(Num)putchar(CH[Num--]+);
puts("");
}
//************************************************************************************** string s;
string t;
int main()
{ cin>>s>>t;
int flag=;
int n=s.size();
if(s==t)
{
cout<<s<<endl;
return ;
}
for(int i=;i<n;i++)
{
if(s[i]!=t[i])
flag++;
}
if(flag%==)
{
printf("impossible\n");
}
else
{
flag/=;
for(int i=;i<n;i++)
{
if(s[i]!=t[i])
{
if(flag>)
{
cout<<s[i];
flag--;
}
else
cout<<t[i];
}
else
cout<<s[i];
}
}
}