CF#808 C. Tea Party（贪心）

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本文介绍了一个有趣的算法问题：如何在确保公平的前提下将有限的茶水合理分配到不同容量的茶杯中。问题需要考虑茶杯的容量、茶水总量及公平分配原则。

Polycarp invited all his friends to the tea party to celebrate the holiday. He has n cups, one for each of his n friends, with volumesa1, a2, ..., an. His teapot stores w milliliters of tea (w ≤ a1 + a2 + ... + an). 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 cup i contains less tea than cup j but ai > aj.

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 and w (1 ≤ n ≤ 100, $\text{[math]}$ ).

The second line contains n numbers a1, a2, ..., an (1 ≤ ai ≤ 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.

题意：有 n 个茶杯，茶壶里有 m 毫升茶，每个茶杯大小不等，我们要满足以下三个条件：1. 每个茶杯中的茶要超过茶杯容量的一半 2. 每个茶杯中倒的茶必须是整数毫升 3. 大茶杯中的茶的容量不能少于任意一个小茶杯中的茶的容量，若 n ，m 不满足条件，输出 -1 ，否则输出一种满足条件的情况

#include<bits/stdc++.h>
using namespace std;

const int N = 100 + 10;
int n, w;
struct xx{
    int num, pos;
    int ans;
} a[N];

bool cmp1(xx a, xx b){
    return a.num > b.num;
}

bool cmp2(xx a, xx b){
    return a.pos < b.pos;
}

int main()
{
    while(scanf("%d%d", &n, &w) == 2){
        int ww = w;
        int Maxsum = 0, Minsum = 0;
        for(int i = 1; i <= n; i++){
            scanf("%d", &a[i].num);
            a[i].pos = i;
            a[i].ans = a[i].num/2;
            if(a[i].num&1) a[i].ans++;
            Maxsum += a[i].num;
            Minsum += a[i].ans;
            w -= a[i].ans;
        }
        if(ww > Maxsum || ww < Minsum){
            printf("-1\n");
        }
        else
        {
            sort(a+1, a+1+n, cmp1);
            for(int i = 1; i <= n; i++){
                if(w + a[i].ans >= a[i].num){
                    w -= a[i].num-a[i].ans;
                    a[i].ans = a[i].num;
                }
                else{
                    a[i].ans += w;
                    w = 0;
                }

                if(!w) break;
            }
            sort(a+1, a+1+n, cmp2);
            for(int i = 1; i <= n; i++){
                printf("%d%c", a[i].ans, i<n? ' ':'\n');
            }
        }
    }
}