codeforces1003D（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 $\mathrm{nn coins,\; the\; value\; of\; the ii-th\; coin\; is aiai.\; It\; is\; guaranteed\; that\; all\; the\; values\; are\; integer\; powers\; of 22 (i.e. ai=2dai=2d for\; some non-negative integer\; number dd).}$

Polycarp wants to know answers on $\mathrm{qq queries.\; The jj-th\; query\; is\; described\; as\; integer\; number bjbj.\; The\; answer\; to\; the\; query\; is\; the\; minimum\; number\; of\; coins\; that\; is\; necessary\; to\; obtain\; the\; value bjbj using\; some\; subset\; of\; coins\; (Polycarp\; can\; use\; only\; coins\; he\; has).\; If\; Polycarp\; can\text{'}t\; obtain\; the\; value bjbj,\; the\; answer\; to\; the jj-th\; query\; is -1.}$

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 $\mathrm{nn and qq (1\le n,q\le 2\cdot 1051\le n,q\le 2\cdot 105)\; —\; the\; number\; of\; coins\; and\; the\; number\; of\; queries.}$

The second line of the input contains $\mathrm{nn integers a1,a2,\dots ,ana1,a2,\dots ,an —\; values\; of\; coins\; (1\le ai\le 2\cdot 1091\le ai\le 2\cdot 109).\; It\; is\; guaranteed\; that\; all aiai are\; integer\; powers\; of 22 (i.e. ai=2dai=2d for\; some non-negative integer\; number dd).}$

The next $\mathrm{qq lines\; contain\; one\; integer\; each.\; The jj-th\; line\; contains\; one\; integer bjbj —\; the\; value\; of\; the jj-th\; query\; (1\le bj\le 1091\le bj\le 109).}$

Output

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

Example

input

Copy

5 4
2 4 8 2 4
8
5
14
10

output

Copy

1
-1
3
2

 题意：给你n个面值为2指数次方的硬币，然后有q个问题，每次给出一个数值，问是否能用最少的所给出的硬币组成这个面值，能组成则输出最小值，不能输出-1.

思路：能用大面值的就用大面值的

代码：

#include<stdio.h>
#include<map>
using namespace std;
map<int,int> mp;
struct{
	int sm;
}st[35];
int mn;
int s[35];
void init(){
	int i,a;
	a=1;
	mp[1]=0;
	st[0].sm=1;
	for(i=1;i<=31;i++){
		a=a*2;
		mp[a]=i;
		st[i].sm=a;
	}
}
int min(int a,int b){
	if(a<b)
	return a;
	return b;
}
int dfs(int k,int j,int cnt){
	int i;
	if(j<0){
		if(k==0)
		mn=cnt;
		return 0;
	}
	i=min(k/st[j].sm,s[mp[st[j].sm]]);
		dfs(k-i*st[j].sm,j-1,cnt+i);
	return 0;
}
int main(){
	init();
	int n,q;
	int i,j,a;
	scanf("%d%d",&n,&q);
	for(i=0;i<n;i++){
		scanf("%d",&a);
		s[mp[a]]++;
	}
	while(q--){
		mn=0;
		scanf("%d",&a);
		dfs(a,31,0);
		if(mn==0)
		printf("-1
");
		else
		printf("%d
",mn);
	} 
	return 0;
}