Coins and Queries（CodeForces 1003D）-优快云博客

本文链接：https://blog.youkuaiyun.com/Baiyi_destroyer/article/details/81239976

本文介绍了一个关于硬币组合的问题及解决思路。问题的核心在于利用数量有限的2的次幂价值硬币来组合出特定数值，并尽可能减少硬币使用数量。通过从大到小枚举硬币面值的方式实现最优解。

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Polycarp has nn coins, the value of the ii-th coin is aiai. It is guaranteed that all the values are integer powers of 22 (i.e. ai=2dai=2d for some non-negative integer number dd).

Polycarp wants to know answers on qq queries. The jj-th query is described as integer number bjbj. The answer to the query is the minimum number of coins that is necessary to obtain the value bjbj using some subset of coins (Polycarp can use only coins he has). If Polycarp can't obtain the value bjbj, the answer to the jj-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 nn and qq (1≤n,q≤2⋅1051≤n,q≤2⋅105) — the number of coins and the number of queries.

The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an — values of coins (1≤ai≤2⋅1091≤ai≤2⋅109). It is guaranteed that all aiai are integer powers of 22 (i.e. ai=2dai=2d for some non-negative integer number dd).

The next qq lines contain one integer each. The jj-th line contains one integer bjbj — the value of the jj-th query (1≤bj≤1091≤bj≤109).

Output

Print qq integers ansjansj. The jj-th integer must be equal to the answer on the jj-th query. If Polycarp can't obtain the value bjbj the answer to the jj-th query is -1.

Example

Input

Output

题解：由二进制表示一个数可知，任何一个数均能由若干个数（均为2的次幂）的和构成，要想用最少的硬币凑成x，则应该尽量用面值比较大的硬币，所以，从大到小枚举硬币面值。

代码如下：

#include <iostream>
#include <cstdio>
#include <stdlib.h>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string.h>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <ctime>
#define maxn 1007
//#define N 100005
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define lowbit(x) (x&(-x))
#define eps 0.000000001
#define read(x) scanf("%d",&x)
#define put(x) printf("%d\n",x)
#define Debug(x) cout<<x<<" "<<endl
using namespace std;
typedef long long ll;


const int N = 2e5+5;
map<int,int>mp;
int main()
{
    int n,q;
    cin>>n>>q;
    for(int i=0; i<n; i++)
    {
        int x;
        cin>>x;
        mp[x]++;
    }
    while(q--)
    {
        int x;
        cin>>x;
        int ans=0;
        for(int i=1<<30; i>=1; i/=2)
        {
            int k=min(mp[i],x/i);
            ans+=k;
            x-=k*i;
        }
        if(x)
            cout<<"-1"<<endl;
        else
            cout<<ans<<endl;
    }
    return 0;
}