codeforces 1003 D Coins and Queries （贪心+暴力）

D. Coins and Queries

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarp has $n$ coins, the value of the $i$-th coin is $a_i$. It is guaranteed that all the values are integer powers of $2$ (i.e. $a_i=2^d$ for some non-negative integer number $d$).

Polycarp wants to know answers on $q$ queries. The $j$-th query is described as integer number $b_j$. The answer to the query is the minimum number of coins that is necessary to obtain the value $b_j$ using some subset of coins (Polycarp can use only coins he has). If Polycarp can't obtain the value $b_j$, the answer to the $j$-th query is -1.

The queries are independent (the answer on the query doesn't affect Polycarp's coins).

Input

The first line of the input contains two integers $n$ and $q$ ($1\le n,q\le 2\cdot {10}^5$) — the number of coins and the number of queries.

The second line of the input contains $n$ integers $a_1,a_2,\dots ,a_n$ — values of coins ($1\le a_i\le 2\cdot {10}^9$). It is guaranteed that all $a_i$ are integer powers of $2$ (i.e. $a_i=2^d$ for some non-negative integer number $d$).

The next $q$ lines contain one integer each. The $j$-th line contains one integer $b_j$ — the value of the $j$-th query ($1\le b_j\le {10}^9$).

Output

Print $q$ integers $an{s}_j$. The $j$-th integer must be equal to the answer on the $j$-th query. If Polycarp can't obtain the value $b_j$ the answer to the $j$-th query is -1.

Example

input

Copy

5 4
2 4 8 2 4
8
5
14
10

output

Copy

1
-1
3
2

题意： 有n个硬币q次询问，每个硬币为2^d， 每次询问给你一个数，问能否用硬币凑成这个数。

思路：因为都是倍数关系，所以有大的用大的，最多就32个不同的数，所以可以桶装，这里用的反向迭代器

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <map>
#include <algorithm>
using namespace std;
map<int, int> mp;
int main()
{
    int n, q, x, cnt = 0;
    scanf("%d%d", &n, &q);
    for(int i = 1; i <= n; i++)
    {
        scanf("%d", &x);
        mp[x]++;
    }
    while(q--)
    {
        scanf("%d", &x);
        int ans = 0, flag = 0;
        for(map<int,int>::reverse_iterator it = mp.rbegin(); it != mp.rend(); it++)
        {
            int tmp = min(x/it->first, it->second);
            ans += tmp;
            x -= tmp * it->first;
            if(!x) break;
        }
        if(!x) printf("%d\n", ans);
        else puts("-1");
    }
    return 0;
}