Coins and Queries CodeForces - 1003D （贪心）

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

Codeforces (c) Copyright 2010-2018 Mike Mirzayanov

题意：给出n张钱ai（ai是2的幂，也就是说，ai=2^x,x是任意正整数），给出m次询问，每次给出一个数，问，是否可以用最少的纸币拼出这个数，可以，则输出最少的纸币数，否则，输出-1

思路：由于ai是2的幂，那优先取面值最大的纸币就好了。

#include "iostream"
#include "map"
using namespace std;
typedef long long ll;
map<ll,int> mp;
const int Max=2e5+10;
int main()
{
    ios::sync_with_stdio(false);
    mp.clear();
    int n,m,x;
    cin>>n>>m;
    for(int i=0;i<n;i++) {cin>>x;mp[x]++;}
    while(m--){
        int ans=0;
        cin>>x;
        for(int i=1<<30;i>0;i>>=1){
            if(mp[i]){
                int t=mp[i]<(x/i)?mp[i]:x/i;
                x-=t*i;
                ans+=t;
            }
            if(x==0) break;
        }
        if(!x) cout<<ans<<endl;
        else cout<<"-1"<<endl;
    }
    return 0;
}