题解 | #K-skipPermutation#

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;
}