Leetcode-1006 Clumsy Factorial(笨阶乘)

 1 #define pb push_back
 2 #define _for(i,a,b) for(int i = (a);i < (b);i ++)
 3 const int maxn = 50003;
 4 
 5 class Solution
 6 {
 7     public:
 8         int clumsy(int N)
 9         {
10             if(N==1)
11                 return 1;
12             else if(N==2)
13                 return 2;
14             else if(N==3)
15                 return 6;
16                 
17             int tt = N*(N-1)/(N-2);
18             N -= 3;
19             int rnt = tt;
20             int flag = 1;
21             while(flag || N>3)
22             {
23                 if(flag)
24                 {
25                     rnt += N;
26                     if(N>4&&(N-1)*(N-2)%(N-3)==0)
27                         tt -= 3;
28                     else
29                         tt -= 4;
30                     N --;
31 
32                     flag = 0;
33                     continue;
34                 }
35                 rnt -= tt;
36                 
37                 N -= 3;
38                 if(N>4&&(N-1)*(N-2)%(N-3)==0)
39                     tt -= 3;
40                 else
41                     tt -= 4;
42                 rnt += N;
43                 N --;
44             } 
45             if(N==1)
46                 rnt -= 1;
47             else if(N==2)
48                 rnt -= 2;
49             else if(N==3)
50                 rnt -= 6;
51             return rnt;
52         }
53 };

数学题，A*(A-1)/(A-2)一定比(A-4)*(A-5)/(A-6)大3或4，具体是3还是4，主要看(A-4)*(A-5)是否能除尽(A-6)，能除尽，那就差3，否则差4

标签：return,Factorial,tt,else,int,flag,阶乘,1006,rnt
来源： https://www.cnblogs.com/Asurudo/p/10504871.html