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

题目链接

 1 /*
 2     Name:
 3     Copyright:
 4     Author:
 5     Date: 2018/5/2 11:07:16
 6     Description:输出第m小的序列
 7 */
 8 #include <iostream>
 9 #include <cstdio>
10 #include <vector> 
11 #include <algorithm>
12 #include <cstring>
13 using namespace std;
14 int  fac[] = {1,1,2,6,24,120,720,5040,40320};//阶乘
15 //康托展开的逆运算,{1...n}的全排列,中的第k个数为s[]
16 void reverse_kangtuo(int n,int k,int s[])
17 {
18     int i, j, t, vst[1001]={0};
19     --k;
20     for (i=0; i<n; i++)
21     {
22         if (n-i-1 > 8) {
23             t = k/fac[8];    
24         } else {
25             t = k/fac[n-i-1];
26         }
27         for (j=1; j<=n; j++)
28             if (!vst[j])
29             {
30                 if (t == 0) break;
31                 --t;
32             }
33         s[i] = j;
34         vst[j] = 1;
35         if (n-i-1 > 8) {
36             k %= fac[8];
37         } else {
38             k %= fac[n-i-1];
39         }
40     }
41 }
42 
43 int main()
44 {
45     int s[1005] ;
46     int m, n;
47     while (cin>>m>>n) {
48         memset(s, 0, sizeof(s)) ;
49         reverse_kangtuo(m, n, s);
50         cout<<s[0];
51         for (int i=1; i<m; i++) {
52             cout<<" "<<s[i];
53         }
54         cout<<endl;
55     }
56     return 0;
57 }

 

posted @ 2018-05-02 15:16  朤尧  阅读(263)  评论(0编辑  收藏  举报