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

 

题目传送门

 1 /*  2  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  3  贪心：假设p与s相同，奇偶变换赋值，当是偶数，则有答案  4 */  5 #include <cstdio>  6 #include <algorithm>  7 #include <cstring>  8 #include <cmath>  9 #include <iostream> 10 using namespace std; 11 12 const int MAXN = 1e5 + 10; 13 const int INF = 0x3f3f3f3f; 14 int p[MAXN]; 15 char s[MAXN], t[MAXN]; 16 17 int main(void) //Codeforces Round #303 (Div. 2) B. Equidistant String 18 { 19 //freopen ("B.in", "r", stdin); 20 21 while (scanf ("%s%s", s+1, t+1) == 2) 22  { 23 memset (p, 0, sizeof (p)); 24 int len = strlen (s+1); bool flag = false; 25 for (int i=1; i<=len; ++i) 26  { 27 if (s[i] == t[i]) p[i] = 0; 28 else 29  { 30 if (flag) {p[i] = t[i] - '0'; flag = !flag;} 31 else {p[i] = s[i] - '0'; flag = !flag;} 32  } 33  } 34 if (flag) puts ("impossible"); 35 else 36  { 37 for (int i=1; i<=len; ++i) 38 printf ("%d", p[i]); 39 puts (""); 40  } 41  } 42 43 44 return 0; 45 }