题解 | #K-skipPermutation#
K-skipPermutation
https://ac.nowcoder.com/acm/problem/222763
显然，当枚举的值等于k+1时，结束循环即可
#include <bits stdc++.h>
using namespace std;
int n,k;
const int N = 1e6 + 10;
bool st[N];
int main(){
cin >> n >> k;
int a = 1;
while(1){
if(a == k + 1) break;
for(int i=a;i<=n;i+=k){
if(!st[i]) cout << i << " ";
st[i] = true;
}
a++;
}
for(int i=1;i<=n;i++)
if(!st[i]) cout << i << " ";
return 0;
}