Coins and Queries(思维+贪心)-优快云博客

本文链接：https://blog.youkuaiyun.com/codeswarrior/article/details/81459392

本篇博客介绍了一道关于硬币组合的算法问题，通过给定的n个2的幂次数值的硬币，来回答q次如何使用最少数量的硬币达到目标值bj的查询。通过对2的幂次特性的利用，实现高效查询解答。

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Coins and Queries

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

Polycarp wants to know answers on q
queries. The j-th query is described as integer number bj. The answer to the query is the minimum number of coins that is necessary to obtain the value bj using some subset of coins (Polycarp can use only coins he has). If Polycarp can’t obtain the value bj, 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≤n,q≤2⋅105

) — the number of coins and the number of queries.

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

The next q
lines contain one integer each. The j-th line contains one integer bj — the value of the j-th query (1≤bj≤109

).

Output

Print q

integers ansj. The j-th integer must be equal to the answer on the j-th query. If Polycarp can’t obtain the value bj the answer to the j

-th query is -1.

Example
Input

题意：

n个硬币，q次询问。第二行给你n个硬币的值（保证都是2的次幂），每次询问组成b块钱至少需要多少硬币

分析：

因为每个数都是2的幂次，那么可一个想象一个二进制数，任何数可以通过若干个2的幂次的数构成，而题目既然要求用最小的硬币，那么就先选择大的，因为数据范围1e9，所以最多枚举30次即可
从大到小枚举

code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
using namespace std;
map<int,int>mp;
int main(){
    int n,q;
    scanf("%d%d",&n,&q);
    for(int i = 0; i < n; i++){
        int a;
        scanf("%d",&a);
        if(!mp.count(a)) mp[a] = 1;
        else mp[a]++;
    }
    while(q--){
        int b;
        scanf("%d",&b);
        int ans = 0;
        for(int i = 1 << 30; i >= 1; i /= 2){//从大到小枚举
            int times = min(mp[i],b/i);//b/i表示应该需要的这个幂次的个数，mp[i]表示实际有多个这个样的，那么最终取的就是两者较小的
            ans += times;//答案加上需要的这个2的幂次的个数
            b -= times * i;//然后减去总共拿的这个2的幂次
        }
        if(b) ans = -1;
        printf("%d\n",ans);
    }
    return 0;
}