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

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 $\text{[math]}$ 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

///1085422276
#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=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){
        if(ch=='-')f=-1;ch=getchar();
    }
    while(ch>='0'&&ch<='9'){
        x=x*10+ch-'0';ch=getchar();
    }return x*f;
}
//************************************************
const int maxn=100000+5;

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=0;
     int sum=0,mn=0;
    for(int i=1;i<=m;i++){
        if(i%2){
            a[i]=1;
        }
        else {
                int tmp=i;
                int ans=1;
            while(tmp){
                tmp/=2;

                ans*=2;
                if(tmp%2)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=0;
    if(sum<n){
        cout<<-1<<endl;return 0;
    }
    int flag;
    for(int i=mn;i>=1;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=0;j<G[i].size();j++){
            cout<<G[i][j]<<" ";
        }
    }
   for(int i=0;i<n;i++){
    cout<<G[1][i]<<" ";
   }
   return 0;
}

代码