Codeforces 1003D（贪心）

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题面：

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

题目描述：

    给你n个硬币，每个硬币的面值都为2的幂次方。先有q个询问，每个询问给出一个数b，问你至少用多少个硬币才能使得他们的面值凑成b。若不能凑出，则输出-1。

题目分析：

    因为题目中所给的硬币的面值都是2的倍数，因此，若我们要使得用最少的硬币凑出一个数，我们则应该贪心的从最大的2的倍数进行枚举，优先选取大的硬币，并将该数除以该面值。而若将这个过程做完后还不能使得原来的数为0，则输出-1。

代码：

#include <bits/stdc++.h>
using namespace std;
map<int,int>mp;//用map来存储硬币的面额
int main()
{
    int n,q;
    cin>>n>>q;
    for(int i=0;i<n;i++){
        int num;
        scanf("%d",&num);
        mp[num]++;
    }
    int ans=0;
    while(q--){
        int num;
        ans=0;
        scanf("%d",&num);
        for(int i=1<<30;i>=1;i>>=1){
            if(mp.count(i)==0) continue;//如果不存在i的面额
            else{
                int x=min(mp[i],num/i);//取面额i和num/i中的最小值
                num-=x*i;
                ans+=x;
            }
        }
        if(num){
            puts("-1");
            continue;
        }
        else cout<<ans<<endl;
    }
    return 0;
}