codeforces 1003d Coins and Queries

http://www.elijahqi.win/archives/3922
題意 讓你用最少的2^x的金幣湊出想要的數字那麼我直接貪心即可 就儘量用大的去滿足我的要求 簡單dfs即可

#include<bits/stdc++.h>
using namespace std;
inline char gc(){
    static char now[1<<16],*S,*T;
    if (T==S){T=(S=now)+fread(now,1,1<<16,stdin);if (T==S) return EOF;}
    return *S++;
}
inline int read(){
    int x=0,f=1;char ch=gc();
    while(!isdigit(ch)) {if (ch=='-') f=-1;ch=gc();}
    while(isdigit(ch)) {x=x*10+ch-'0',ch=gc();}
    return x*f;
}
const int N=2e5+10;
int a[N],n,q,cnt[33],cnt1[33];
inline void gao(int x){
    int tot=0;while(x!=1) ++tot,x>>=1;
    ++cnt[tot];
}
inline int dfs(int x,int nm){
    if (!x) return 0;
    if (nm<0) return -1;int res=0;
    if((1<<nm)<=x) {
        int t=x/(1<<nm);
        t=min(t,cnt[nm]);res=dfs(x-t*(1<<nm),nm-1);
        if (res==-1) return -1;
        else return t+res;
    }
    else return dfs(x,nm-1);
}
int main(){
//  freopen("d.in","r",stdin);
    n=read();q=read();
    for (int i=1;i<=n;++i) a[i]=read();
    for (int i=1;i<=n;++i) gao(a[i]);
    for (int i=1;i<=q;++i){
        int x=read();memcpy(cnt1,cnt,sizeof(cnt));
        printf("%d\n",dfs(x,31));
    }
    return 0;
}