题解在代码里~

#include 
     
      #include 
      
       #include 
       
        using namespace std;int main(){    int n, k, f[100];    n = 12; cin>>k;    //链表做法,复杂度O(n*k)    list 
        
          L;    for(int i = 1; i <= n; i++) f[i] = i, L.push_back(i);    list
         
          ::iterator pos = L.begin();    while(L.size() > 1)    {        for(int i = 1; i < k; i++)        {            ++pos;            if(pos == L.end()) pos = L.begin();        }        f[*pos] = 0; pos = L.erase(pos); if(pos == L.end()) pos = L.begin();        for(int i = 1; i <= n; i++) cout<
          
           <