（贪心）School Marks -- codefor -- 540B

http://codeforces.com/problemset/problem/540/B

 

School Marks

 

Little Vova studies programming in an elite school. Vova and his classmates are supposed to write n progress tests, for each test they will get a mark from 1 to p. Vova is very smart and he can write every test for any mark, but he doesn't want to stand out from the crowd too much. If the sum of his marks for all tests exceeds value x, then his classmates notice how smart he is and start distracting him asking to let them copy his homework. And if the median of his marks will be lower than y points (the definition of a median is given in the notes), then his mom will decide that he gets too many bad marks and forbid him to play computer games.

Vova has already wrote k tests and got marks a1, ..., ak. He doesn't want to get into the first or the second situation described above and now he needs to determine which marks he needs to get for the remaining tests. Help him do that.

Input

The first line contains 5 space-separated integers: n, k, p, x and y (1 ≤ n ≤ 999, n is odd, 0 ≤ k < n, 1 ≤ p ≤ 1000, n ≤ x ≤ n·p,1 ≤ y ≤ p). Here n is the number of tests that Vova is planned to write, k is the number of tests he has already written, p is the maximum possible mark for a test, x is the maximum total number of points so that the classmates don't yet disturb Vova, y is the minimum median point so that mom still lets him play computer games.

The second line contains k space-separated integers: a1, ..., ak (1 ≤ ai ≤ p) — the marks that Vova got for the tests he has already written.

Output

If Vova cannot achieve the desired result, print "-1".

Otherwise, print n - k space-separated integers — the marks that Vova should get for the remaining tests. If there are multiple possible solutions, print any of them.

Sample test(s)

input

5 3 5 18 4
3 5 4

output

4 1

input

5 3 5 16 4
5 5 5

output

-1

Note

The median of sequence a1, ..., an where n is odd (in this problem n is always odd) is the element staying on (n + 1) / 2 position in the sorted list of ai.

In the first sample the sum of marks equals 3 + 5 + 4 + 4 + 1 = 17, what doesn't exceed 18, that means that Vova won't be disturbed by his classmates. And the median point of the sequence {1, 3, 4, 4, 5} equals to 4, that isn't less than 4, so his mom lets him play computer games.

Please note that you do not have to maximize the sum of marks or the median mark. Any of the answers: "4 2", "2 4", "5 1", "1 5", "4 1", "1 4" for the first test is correct.

In the second sample Vova got three '5' marks, so even if he gets two '1' marks, the sum of marks will be 17, that is more than the required value of 16. So, the answer to this test is "-1".

 

题意：一共n个数，给出其中k个数，要求这n个数的中位数为y,这n个数的和不超过x,补全剩下的n-k个数

 

先统计给出的k个数里面比中位数小的数，

如果cnt<=n/2,说明中位数还没有出现，把这n/2-cnt个数都补上1，剩下的都补上y

如果cnt>n/2,说明中位数不存在

 

（把这n/2-cnt个数都补上1，剩下的都补上y）比赛的时候， 我思路很乱， 没有理清思路。 后来搜题解的时候， 刚开始有点不敢相信， 这么短的代码能把所以的情况考虑清楚吗？？？  再调试一遍后， 突然醒悟， 补 1 和补 y ， 是它能得到的最少分数， 如果他的最少分数都大于 x ， 那么肯定是不行的 ， 唉， 还是So young

 

 

 

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

#define N 1100

int main()
{
    int n, k, p, x, y;

    while(scanf("%d%d%d%d%d", &n, &k, &p, &x, &y)!=EOF)
    {
        int i, sum=0, ans=0, number;

        for(i=0; i<k; i++)
        {
            scanf("%d", &number);
            if(number<y) ans++;
            sum += number;
        }

        if(ans<=n/2)
        {
            int l, r;
            l = min(n/2-ans, n-k);
            r = n-l-k;
            sum += l + r*y;

            if(sum>x) printf("-1
");
            else
            {
                for(i=0; i<l; i++) printf("1 ");
                for(i=0; i<r; i++) printf("%d ", y);
                printf("
");
            }
        }
        else  printf("-1
");

    }
    return 0;
}