Coins and Queries（map函数初体验）

最新推荐文章于 2020-03-26 09:19:14 发布

不想写代码了啊啊

最新推荐文章于 2020-03-26 09:19:14 发布

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本文链接：https://blog.youkuaiyun.com/shezjoe/article/details/81234457

本文介绍了一个关于硬币组合的问题：给定一系列值为2的幂次的硬币和若干查询值，目标是最小化硬币数量以达到指定的查询值。通过使用map存储不同面值硬币的数量并迭代减去每个查询值的最大可能面额来解决该问题。

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map函数map<键，键值>实现一对一映射关系，以后再写博客详细说。

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

#include <bits/stdc++.h>
using namespace std;
int a[200010];
map<int,int> mp;
int main()
{
    int n,q,t,i,d,ans;
    scanf("%d%d", &n, &q);
    for(int i = 0; i < n; i++)
    {
        scanf("%d",&a[i]);
        mp[a[i]]++;
    }
    while(q--)
    {
        scanf("%d", &d);
        ans=0;
        for(i=1<<30; i >= 1; i/=2)
        {
            t=min(d/i,mp[i]);
            ans+=t;
            d-=t*i;
        }
        if(d!=0)
            puts("-1\n");
        else
            printf("%d\n",ans);
    }
    return 0;
}