B. Shashlik Cooking
time limit per test1 second
memory limit per test512 megabytes
inputstandard input
outputstandard output
Long story short, shashlik is Miroslav’s favorite food. Shashlik is prepared on several skewers simultaneously. There are two states for each skewer: initial and turned over.
This time Miroslav laid out n skewers parallel to each other, and enumerated them with consecutive integers from 1 to n in order from left to right. For better cooking, he puts them quite close to each other, so when he turns skewer number i, it leads to turning k closest skewers from each side of the skewer i, that is, skewers number i−k, i−k+1, …, i−1, i+1, …, i+k−1, i+k (if they exist).
For example, let n=6 and k=1. When Miroslav turns skewer number 3, then skewers with numbers 2, 3, and 4 will come up turned over. If after that he turns skewer number 1, then skewers number 1, 3, and 4 will be turned over, while skewer number 2 will be in the initial position (because it is turned again).
As we said before, the art of cooking requires perfect timing, so Miroslav wants to turn over all n skewers with the minimal possible number of actions. For example, for the above example n=6 and k=1, two turnings are sufficient: he can turn over skewers number 2 and 5.
Help Miroslav turn over all n skewers.
Input
The first line contains two integers n and k (1≤n≤1000, 0≤k≤1000) — the number of skewers and the number of skewers from each side that are turned in one step.
Output
The first line should contain integer l — the minimum number of actions needed by Miroslav to turn over all n skewers. After than print l integers from 1 to n denoting the number of the skewer that is to be turned over at the corresponding step.
Examples
inputCopy
7 2
outputCopy
2
1 6
inputCopy
5 1
outputCopy
2
1 4
Note
In the first example the first operation turns over skewers 1, 2 and 3, the second operation turns over skewers 4, 5, 6 and 7.
In the second example it is also correct to turn over skewers 2 and 5, but turning skewers 2 and 4, or 1 and 5 are incorrect solutions because the skewer 3 is in the initial state after these operations.
#include<iostream>
using namespace std;
int main()
{
int n,k,d,i,q,r;
cin>>n>>k;
d=2*k+1;
q=n/d;
r=n%d;
if(r!=0)
q++;
else
r=2*k;
cout<<q<<endl;
for(i=1+r/2;i<=n;i+=d)
cout<<i<<" ";
return 0;
}