Polycarp has nn coins, the value of the i-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 j -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
5 4
2 4 8 2 4
8
5
14
10
1
-1
3
2
题解：有n个数,q次询问,对于要询问的数x，找出数组中能组成x的最小个数，若不能输出-1.这道题很明显是贪心，每一次从最大的值开始判断是否选择。由于数组中的每一个数都是2的幂，所以可以二进制优化的思想进行遍历判断，还需要使用
map容器进行一个映射，来判断当前值是否存在，然后再判断当前值与x的大小即可。
1 #include <iostream> 2 #include<cmath> 3 #include<cstdio> 4 #include<cstring> 5 #include<stack> 6 #include<queue> 7 #include<deque> 8 #include<map> 9 #include<algorithm> 10 #define PI acos(-1.0) 11 using namespace std; 12 typedef long long ll; 13 ll m,n,k; 14 int dp[1999]; 15 map<int,int>::iterator it; 16 ll fac(ll x) 17 { 18 ll ans=1; 19 for(ll i=1;i<=x;i++) 20 { 21 ans*=2; 22 } 23 return ans; 24 } 25 int main() 26 { 27 cin>>m>>n; 28 map<ll,ll>mp; 29 mp.clear(); 30 ll i; 31 for(i=0;i<m;i++) 32 { 33 ll p; 34 cin>>p; 35 mp[p]++; 36 } 37 int str[32]={1}; 38 for(i=1;i<=31;i++) 39 { 40 str[i]=str[i-1]*2; 41 } 42 while(n--) 43 { 44 ll m,k,p,s; 45 cin>>m; 46 ll sum=0,l=31; 47 while(l>=0) 48 { 49 50 k=str[l--]; 51 if(mp.count(k)==0) 52 continue; 53 p=min(mp[k],m/k); 54 sum+=p; 55 m-=p*k; 56 57 } 58 if(m) 59 sum=-1; 60 printf("%lld ",sum); 61 } 62 return 0; 63 }