codeforces B. School Marks

B. School Marks

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Little Vova studies programming in an elite school. Vova and his classmates are supposed to writen progress tests, for each test they will get a mark from 1 top. 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 valuex, 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 thany 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 marksa₁, ..., a_k. 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, ..., a_k (1 ≤ a_i ≤ 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 43 5 4

Output

4 1

Input

5 3 5 16 45 5 5

Output

-1

Note

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

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".

WA在15：

#include<iostream>#include<string>#include <stdio.h>#include<algorithm>using namespace std;int main(){    int n,k,p,x,y,sum=0;    int a[1002],b[1002];    cin>>n>>k>>p>>x>>y;    for(int i=0;i<k;i++)    {         cin>>a[i];         sum+=a[i];    }    int c=x-sum; int flag=0;     if(c<n-k){cout<<"-1"<<endl;return 0;}    //sort(a,a+k);    if(k==0){         for(int i=k;i<n;i++)    {        if(c>=y+n-i-1)        {a[i]=y;c-=y;b[i]=y;}        else {b[i]=1;a[i]=1;}    }    }    else    {        for(int i=k;i<n;i++)    {        if(c>=y+n-i&&flag<=n/2)        {a[i]=y;c-=y;b[i]=y;flag++;}        else {b[i]=1;a[i]=1;}    }    }    sort(a,a+n);    if(a[(n-1)/2]<y){cout<<"-1"<<endl;return 0;}    for(int i=k;i<n-1;i++)        cout<<b[i]<<' ';    cout<<b[n-1]<<endl;   return 0;}

方法是一样的，关键是怎么简便的实现

First count the number of marks that are less than y. If there are more than such marks, we can't satisfy the second condition (about the median), and the answer is -1. Otherwise we can get exactly such number of y marks so that the total number of marks greater than or equal toy is at least (maybe it's already satisfied). This is the required action for satisfying the second condition.

Now, in order not to break the first condition, get the remaining marks as lower as possible — all ones — and check the sum of the marks. If it is greater thanx, the answer is -1, otherwise the correct answer is found.

AC：

#include<iostream>#include<string>#include <stdio.h>#include<algorithm>using namespace std;int main(){    int n,k,p,x,y;    int a[1002];    cin>>n>>k>>p>>x>>y;    int ans=0;    int sum=0;    for(int i=1;i<=k;i++)    {        cin>>a[i];        sum+=a[i];        if(a[i]<y) ans++;    }    int l,r;    if(ans<=n/2){        l=min(n/2-ans,n-k);        r=n-l-k;        sum+=l+r*y;        if(sum>x) printf("-1\n");        else{            for(int i=1;i<=l;i++) printf("1 ");            for(int i=1;i<=r;i++) printf("%d ",y);            printf("\n");        }    }    else printf("-1\n");    return 0;}