1 /* 2     贪心：首先要注意，y是中位数的要求；先把其他的都设置为1，那么最多有(n-1)/2个比y小的，cnt记录比y小的个数 3             num1是输出的1的个数，numy是除此之外的数都为y，此时的numy是最少需要的，这样才可能中位数大于等于y 4 */ 5 #include 
     
       6 #include 
      
        7 #include 
       
         8 #include 
        
          9 using namespace std;10 11 const int MAXN = 1e3 + 10;12 const int INF = 0x3f3f3f3f;13 int a[MAXN];14 15 int main(void)        //Codeforces Round #301 (Div. 2) B. School Marks16 {17     //freopen ("B.in", "r", stdin);18 19     int n, k, p, x, y;20     while (scanf ("%d%d%d%d%d", &n, &k, &p, &x, &y) == 5)21     {22         int sum = 0, cnt = 0;23         for (int i=1; i<=k; ++i)24         {25             scanf ("%d", &a[i]);    sum += a[i];26             if (a[i] < y)    cnt++;27         }28 29         if (cnt <= n / 2)30         {31             int num1 = min (n / 2 - cnt, n - k);32             int numy = n - k - num1;33 34             sum += num1 + numy * y;35             if (sum > x)    puts ("-1");36             else37             {38                 for (int i=1; i<=num1; ++i)    printf ("%d%c", 1, (numy==0 && i==num1) ? '\n' : ' ');39                 for (int i=1; i<=numy; ++i)    printf ("%d%c", y, (i==numy) ? '\n' : ' ');    40             }    41         }42         else    puts ("-1");43     }44 45     return 0;46 }