C.Subsequence Counting cf474 [构造]

Pikachu had an array with him. He wrote down all the non-empty subsequences of the array on paper. Note that an array of size n has 2n - 1 non-empty subsequences in it.

Pikachu being mischievous as he always is, removed all the subsequences in which Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ d

Pikachu was finally left with X subsequences.

However, he lost the initial array he had, and now is in serious trouble. He still remembers the numbers X and d. He now wants you to construct any such array which will satisfy the above conditions. All the numbers in the final array should be positive integers less than 1018.

Note the number of elements in the output array should not be more than 104. If no answer is possible, print - 1.

Input

The only line of input consists of two space separated integers X and d (1 ≤ X, d ≤ 109).

Output

Output should consist of two lines.

First line should contain a single integer n (1 ≤ n ≤ 10 000)— the number of integers in the final array.

Second line should consist of n space separated integers — a1, a2, ... , an (1 ≤ ai < 1018).

If there is no answer, print a single integer -1. If there are multiple answers, print any of them.

         Examples       
           input                       Copy          
10 5
           output         
6
5 50 7 15 6 100
           input                       Copy          
4 2
           output         
4
10 100 1000 10000

Note

In the output of the first example case, the remaining subsequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 5 are [5], [5, 7], [5, 6], [5, 7, 6], [50], [7], [7, 6], [15], [6], [100]. There are 10 of them. Hence, the array [5, 50, 7, 15, 6, 100] is valid.

Similarly, in the output of the second example case, the remaining sub-sequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 2 are [10], [100], [1000], [10000]. There are 4 of them. Hence, the array [10, 100, 1000, 10000] is valid.

思路:把x进行分解,构造出可满足的 2^k-1个集合,每个集合互不干扰(+d+1),直至x=0

#include<bits/stdc++.h>
#define PI acos(-1.0)
using namespace std;
typedef long long ll;

const int N=1e5+50;
const int MOD=1e9+7;
const int INF=0x3f3f3f3f;

vector <ll> ans;

int main(void){
    ll x,d;
    cin >>x >> d;
    ll now=1;
    for(int i=31;i>=1;i--){
        while(x>=(1<<i)-1){//显然while次数很少
            for(int j=1;j<=i;j++)   ans.push_back(now);
            x-=(1<<i)-1;
            now+=d+1;
        }
    }
//    cout <<"x="<<x<<endl;
    cout <<  ans.size()<<endl;
    for(int i=0;i<(int)ans.size();++i)
        printf("%lld ",ans[i]);
    return 0;
}
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