1 second
512 megabytes
standard input
standard 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 nn skewers parallel to each other, and enumerated them with consecutive integers from 11 to nn in order from left to right. For better cooking, he puts them quite close to each other, so when he turns skewer number ii, it leads to turning kk closest skewers from each side of the skewer ii, that is, skewers number i−ki−k, i−k+1i−k+1, ..., i−1i−1, i+1i+1, ..., i+k−1i+k−1, i+ki+k (if they exist).
For example, let n=6n=6 and k=1k=1. When Miroslav turns skewer number 33, then skewers with numbers 22, 33, and 44 will come up turned over. If after that he turns skewer number 11, then skewers number 11, 33, and 44 will be turned over, while skewer number 22 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 nn skewers with the minimal possible number of actions. For example, for the above example n=6n=6 and k=1k=1, two turnings are sufficient: he can turn over skewers number 22 and 55.
Help Miroslav turn over all nn skewers.
The first line contains two integers nn and kk (1≤n≤10001≤n≤1000, 0≤k≤10000≤k≤1000) — the number of skewers and the number of skewers from each side that are turned in one step.
The first line should contain integer ll — the minimum number of actions needed by Miroslav to turn over all nn skewers. After than print llintegers from 11 to nn denoting the number of the skewer that is to be turned over at the corresponding step.
7 2
2
1 6
5 1
2
1 4
In the first example the first operation turns over skewers 11, 22 and 33, the second operation turns over skewers 44, 55, 66 and 77.
In the second example it is also correct to turn over skewers 22 and 55, but turning skewers 22 and 44, or 11 and 55 are incorrect solutions because the skewer 33 is in the initial state after these operations.
题目大意：
翻烤串问题，有N个烤串需要被翻面，每次翻第i个烤串可以使得第i-k到第i+k个烤串也同时翻转，问需要至少多少次的翻转使得翻转烤串的数量最大，输出次数和每次翻转烤串的位置。
思路：
首先，每一次翻转烤串的最大影响个数为2*k+1（本身和左右都影响到的烤串）。那么我们可以先将（2*k+1）的倍数烤串翻转，余数进行判断。
如果余数大于k，那么翻转次数加一；如果余数为零，那么翻转次数就为n/（2*k+1）；如果余数小于k，那么翻转次数加一，并使第一个翻转烤串的位置从余数开始，保证答案正确。
AC代码如下：
#include<stdio.h> int main()
{
int n,k;
scanf("%d%d",&n,&k);
int mi=n%(*k+);
if(mi>k) mi=k+,printf("%d\n",n/(*k+)+);
else if(mi==) mi=k+,printf("%d\n",n/(*k+));
else {
printf("%d\n",n/(*k+)+);
}
for(;mi<=n;mi+=*k+) printf("%d ",mi);
return ;
}