On the calculation of recursion number

There is such a topic:

Recursive functions:

1 intXintN)2 {3     if(n<=3)4     {5         return 1;6     }7     Else8     {9         returnX (n2) +x (n4)+1;Ten     } One}

Calculates the number of recursive invocations of X (x (8)).

Most probably think it's a very simple topic, and it's really simple.

But to correctly calculate this recursion without a compiler

The number of calls actually requires a bit of patience.

X (x (8)) We first calculate X (8), we count the number of recursive calls with count=0

1.x (8) =x (6) +x (4) +1 count=1;

2.x (6) =x (4) +x (2) +1,x (4) =x (2) +x (0) +1 x (8) =x (4) +2*x (2) +x (0) +3 count=3;

3.x (4) =x (2) +x (0) +1 x (8) =3*x (2) +2*x (0) +4 count=4

4.x (2) =1,x (0) = 1; X (8) =9 count=9

Recalculate X (9)

1.x (9) =x (7) +x (5) +1 count=10

2.x (7) =x (5) +x (3) +1,x (5) =x (3) +x (1) +1 x (9) =x (5) +2*x (3) +x (1) +3 count=12

3.x (5) =x (3) +x (1) +1 x (9) =3*x (3) +2*x (1) +4 count=13

4.x (3) =1 x (1) =1 x (9) =3+2+4=9 count=18

Next we'll use the program to verify:

1#include <iostream>2 using namespacestd;3 4 Static intCount=0;5 6 intXintN)7 {8     if(n<=3)9     {Tencount++; One         return 1; A     } -     Else -     { thecount++; -         returnX (n2) +x (n4)+1; -     } - } +  - intMainvoid) + {  Acout<<"x (x8) ="<<x (X (8)) <<Endl; atcout<<"count="<<count<<Endl; -System"Pause"); -     return 0; -}

Run:

The validation is correct.

For this calculation recursive calls must be clear-minded, it is best to recursive all recursive calls to the recursive exit

Local re-unification of the call for the export of the recursive, so not prone to confusion, personal opinion, for reference only.

On the calculation of recursion number