Educational Codeforces Round 21 C Tea Party

原創

2020-02-20 19:47

Polycarp invited all his friends to the tea party to celebrate the holiday. He hasn cups, one for each of hisn friends, with volumes a₁, a₂, ..., a_n. His teapot stores w milliliters of tea (w ≤ a₁ + a₂ + ... + a_n). Polycarp wants to pour tea in cups in such a way that:

Every cup will contain tea for at least half of its volume
Every cup will contain integer number of milliliters of tea
All the tea from the teapot will be poured into cups
All friends will be satisfied.

Friend with cup i won't be satisfied, if there exists such cup j that cupi contains less tea than cupj but a_i > a_j.

For each cup output how many milliliters of tea should be poured in it. If it's impossible to pour all the tea and satisfy all conditions then output-1.

Input

The first line contains two integer numbers n andw (1 ≤ n ≤ 100, $\text{[math]}$ ).

The second line contains n numbers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 100).

Output

Output how many milliliters of tea every cup should contain. If there are multiple answers, print any of them.

If it's impossible to pour all the tea and satisfy all conditions then output -1.

Examples

Input

2 10
8 7

Output

6 4

Input

4 4
1 1 1 1

Output

1 1 1 1

Input

3 10
9 8 10

Output

-1

Note

In the third example you should pour to the first cup at least 5 milliliters, to the second one at least 4, to the third one at least 5. It sums up to 14, which is greater than 10 milliliters available.

題意：現在有n個茶杯，水壺有int k的水，每個茶杯的水至少要倒容量的一半且不能爲分數，而且水壺必須要倒完，而且容量大的茶杯的茶不能比容量小的少。不滿足這些條件的就輸出-1，否則輸出沒個茶杯裝多少水。

分析：簡單談心法，先看看所有杯子能否倒滿，如果能的話從大的杯子開始倒直到全部倒滿或者茶壺空。

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;
struct node{
    int va;
    int ke;
}a[128],b[128];
bool cmp1(node q,node w){
    return q.va<w.va;
}
bool cmp2(node q,node w){
    return q.ke<w.ke;
}
int main()
{

    int n,w;
    while(cin>>n>>w){
        int tem; memset(b,0,sizeof(b));
        for(int i=0;i<n;i++){
            cin>>tem;
            a[i].ke=i;
            b[i].ke=i;
            a[i].va=tem;
        }
        sort(a,a+n,cmp1);
        for(int i=n-1;i>=0;i--){
            w-=(a[i].va+1)/2;
            b[a[i].ke].va+=(a[i].va+1)/2;
        }
        if(w<0)
            cout<<-1<<endl;
        else{
            for(int i=n-1;i>=0;i--){
                if(a[i].va-b[a[i].ke].va<=w){
                    w-=a[i].va-b[a[i].ke].va;
                    b[a[i].ke].va=a[i].va;
                }
                else
                {
                    b[a[i].ke].va+=w;
                    w=0;
                    break;
                }
            }
            if(w>0) cout<<-1;
            else
                for(int i=0;i<n;i++)
                    printf("%d ",b[i].va);
            cout<<endl;
        }
    }
    return 0;
}