Codeforces Round #250 (Div. 2)B. The Child and Set 暴力

B. The Child and Set

 

At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important things were lost, in particular the favorite set of Picks.

Fortunately, Picks remembers something about his set S:

• its elements were distinct integers from 1 to limit;
• the value of  was equal to sum; here lowbit(x) equals 2k where k is the position of the first one in the binary representation of x. For example, lowbit(100102) = 102, lowbit(100012) = 12, lowbit(100002) = 100002 (binary representation).

Can you help Picks and find any set S, that satisfies all the above conditions?

Input

The first line contains two integers: sum, limit (1 ≤ sum, limit ≤ 105).

Output

In the first line print an integer n (1 ≤ n ≤ 105), denoting the size of S. Then print the elements of set S in any order. If there are multiple answers, print any of them.

If it's impossible to find a suitable set, print -1.

Sample test(s)

input

5 5

output

2
4 5

Note

In sample test 1: lowbit(4) = 4, lowbit(5) = 1, 4 + 1 = 5.

In sample test 2: lowbit(1) = 1, lowbit(2) = 2, lowbit(3) = 1, 1 + 2 + 1 = 4.

题意：就是给一个lowbit(x)  x在二进制下  从左想右边数第一个为1的数的大小

x属于1到m   问你是否让着m中的某几个数的lowbit和为sum

 题解:

我是预处理 lowbit值，然后从大的找，暴力跑就是了

///
#include<bits/stdc++.h>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,127,sizeof(a));
#define inf 1000000007
#define mod 1000000007
inline ll read()
{
    ll x=,f=;char ch=getchar();
    while(ch<''||ch>''){
        if(ch=='-')f=-;ch=getchar();
    }
    while(ch>=''&&ch<=''){
        x=x*+ch-'';ch=getchar();
    }return x*f;
}
//************************************************
const int maxn=+;

struct ss
{
    int s,i;
}b[maxn];
int cmp(ss s1,ss s2)
{
    return s1.s<s2.s;
}
vector<int >G[maxn];
int a[maxn];
int main(){

   int n=read();
   int m=read();
     set<int >s;
     int k=;
     int sum=,mn=;
    for(int i=;i<=m;i++){
        if(i%){
            a[i]=;
        }
        else {
                int tmp=i;
                int ans=;
            while(tmp){
                tmp/=;

                ans*=;
                if(tmp%)break;

            }
            a[i]=ans;
        }
        sum+=a[i];
        b[++k].i=i;
        b[k].s=a[i];
        G[a[i]].push_back(i);
        mn=max(a[i],mn);
    }   int A=;
    if(sum<n){
        cout<<-<<endl;return ;
    }
    int flag;
    for(int i=mn;i>=;i--){
        if(n>G[i].size()*i){
            n-=G[i].size()*i;
            A+=G[i].size();
        }
        else {
            flag=i;
            A+=n;
            break;
        }
    }

    cout<<A<<endl;
    for(int i=mn;i>flag;i--){
        for(int j=;j<G[i].size();j++){
            cout<<G[i][j]<<" ";
        }
    }
   for(int i=;i<n;i++){
    cout<<G[][i]<<" ";
   }
   return ;
}

代码