cf437B The Child and Set

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在儿童节的玩笑之后，Picks的家被孩子捣乱，丢失了心爱的集合S。幸运的是，Picks记得集合的一些条件：元素是1到限制之间的唯一整数，且特定元素的值等于其二进制表示中最低位1的个数之和。本文将指导您如何找到满足所有条件的集合S。

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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

input
4 3

output
3
2 3 1

input
5 1

output
-1

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.

贪心……从第一位向上贪心，然后每次往上合并。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
int bh[20][100010],fa[20][100010],gs[20];
int pre[100010];
int n,limt,sum,ans[100010];
int getw(int x)
{
    int res=0;
    while(x)
    {
       x=x>>1;res++;
    }
    return res-1;
}
void solve()
{
    int i,j;
    for(i=0;i<=18;i++) 
    {
        if((1<<i) & sum) 
        {
            if(gs[i]>0)
            {
                for(j=bh[i][gs[i]];j;j=pre[j]) ans[++ans[0]]=j;
                gs[i]--;
            }
            else {printf("-1\n");return;}
        }
        for(j=1;j<=gs[i];j+=2)
        {
            if(j+1<=gs[i])
            {
                pre[fa[i][j+1]]=bh[i][j];
                gs[i+1]++;bh[i+1][gs[i+1]]=bh[i][j+1];fa[i+1][gs[i+1]]=fa[i][j];
            }
        }
    }
    printf("%d\n",ans[0]);
    for(i=1;i<=ans[0];i++) printf("%d ",ans[i]);
}
int main()
{
    int i,x,y;
    cin>>sum>>limt;    
    for(i=1;i<=limt;i++)
    {
       x=(i & (-i));
       y=getw(x);
       gs[y]++;
       fa[y][gs[y]]=i;bh[y][gs[y]]=i;
    }
    solve();
    return 0;
}