贪心 Codeforces Round #301 (Div. 2) B. School Marks

题目传送门

 /*
     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
             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
 */
 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cstring>
 using namespace std;

 const int MAXN = 1e3 + ;
 const int INF = 0x3f3f3f3f;
 int a[MAXN];

 int main(void)        //Codeforces Round #301 (Div. 2) B. School Marks
 {
     //freopen ("B.in", "r", stdin);

     int n, k, p, x, y;
     while (scanf ("%d%d%d%d%d", &n, &k, &p, &x, &y) == )
     {
         int sum = , cnt = ;
         for (int i=; i<=k; ++i)
         {
             scanf ("%d", &a[i]);    sum += a[i];
             if (a[i] < y)    cnt++;
         }

         if (cnt <= n / )
         {
             int num1 = min (n /  - cnt, n - k);
             int numy = n - k - num1;

             sum += num1 + numy * y;
             if (sum > x)    puts ("-1");
             else
             {
                 for (int i=; i<=num1; ++i)    printf ("%d%c", , (numy== && i==num1) ? '\n' : ' ');
                 for (int i=; i<=numy; ++i)    printf ("%d%c", y, (i==numy) ? '\n' : ' ');
             }
         }
         else    puts ("-1");
     }

     return ;
 }