hdoj-1027-Ignatius and the Princess II(逆康拓展开)

题目链接

 /*
     Name:
     Copyright:
     Author:
     Date: 2018/5/2 11:07:16
     Description:输出第m小的序列
 */
 #include <iostream>
 #include <cstdio>
 #include <vector>
 #include <algorithm>
 #include <cstring>
 using namespace std;
 int  fac[] = {,,,,,,,,};//阶乘
 //康托展开的逆运算,{1...n}的全排列，中的第k个数为s[]
 void reverse_kangtuo(int n,int k,int s[])
 {
     int i, j, t, vst[]={};
     --k;
     for (i=; i<n; i++)
     {
         if (n-i- > ) {
             t = k/fac[];
         } else {
             t = k/fac[n-i-];
         }
         for (j=; j<=n; j++)
             if (!vst[j])
             {
                 if (t == ) break;
                 --t;
             }
         s[i] = j;
         vst[j] = ;
         if (n-i- > ) {
             k %= fac[];
         } else {
             k %= fac[n-i-];
         }
     }
 }

 int main()
 {
     int s[] ;
     int m, n;
     while (cin>>m>>n) {
         memset(s, , sizeof(s)) ;
         reverse_kangtuo(m, n, s);
         cout<<s[];
         for (int i=; i<m; i++) {
             cout<<" "<<s[i];
         }
         cout<<endl;
     }
     return ;
 }