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 + 10;
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) == 5)
    {
        int sum = 0, cnt = 0;
        for (int i=1; i<=k; ++i)
        {
            scanf ("%d", &a[i]);    sum += a[i];
            if (a[i] < y)    cnt++;
        }

        if (cnt <= n / 2)
        {
            int num1 = min (n / 2 - cnt, n - k);
            int numy = n - k - num1;

            sum += num1 + numy * y;
            if (sum > x)    puts ("-1");
            else
            {
                for (int i=1; i<=num1; ++i)    printf ("%d%c", 1, (numy==0 && i==num1) ? '\n' : ' ');
                for (int i=1; i<=numy; ++i)    printf ("%d%c", y, (i==numy) ? '\n' : ' ');    
            }    
        }
        else    puts ("-1");
    }

    return 0;
}