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

题目传送门

 /*
     题意：找到一个字符串p，使得它和s，t的不同的总个数相同
     贪心：假设p与s相同，奇偶变换赋值，当是偶数，则有答案
 */
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <cmath>
 #include <iostream>
 using namespace std;

 const int MAXN = 1e5 + ;
 const int INF = 0x3f3f3f3f;
 int p[MAXN];
 char s[MAXN], t[MAXN];

 int main(void)        //Codeforces Round #303 (Div. 2) B. Equidistant String
 {
     //freopen ("B.in", "r", stdin);

     while (scanf ("%s%s", s+, t+) == )
     {
         memset (p, , sizeof (p));
         int len = strlen (s+);        bool flag = false;
         for (int i=; i<=len; ++i)
         {
             if (s[i] == t[i])    p[i] = ;
             else
             {
                 if (flag)    {p[i] = t[i] - '';    flag = !flag;}
                 else    {p[i] = s[i] - '';    flag = !flag;}
             }
         }
         if (flag)    puts ("impossible");
         else
         {
             for (int i=; i<=len; ++i)
                 printf ("%d", p[i]);
             puts ("");
         }
     }

     return ;
 }