C++：帕斯卡三角

Question

I am struggling on how to interpret and/or relate between comb(i, j) to long comb(int n, int k) . May you please explain to me how the for loop of the long comb(int n, int k) works? Thank you very much.

#include<iostream>
#include<iomanip>
using namespace std;

long comb(int, int);

int main()
{
 int m;
 cout << "Type a number and then press ENTER: ";
 cin >> m;

 for(int i = 0; i < m; i++)
 {
  for(int j = 1; j < (m - i); j++)
  {
   cout << setw(2) << "";
  }
  for(int j = 0; j <= i; j++)
  {
   cout << setw(4) << comb(i, j);
  }
 cout << endl;
 }
}

long comb(int n, int k)
{
 if (n < 0 || k < 0 || n < k)
 {
  return 0;
 }
 long c= 1;
 for(int i = 1; i <= k; i++, n--)
 {
  c = c*(n/i);
 }
 return c;
}

Edit: Thank Yunnosch for the correction.

Answer 1

The for loop in the long comb(int n, int k) is calculating the Binomial Coefficients using the Multiplicative Formula .

You could also write that for loop like below (which looks more like the multiplicative formula):

for(int i = 1; i <= k; i++) {
    c = c * ((n - i + 1) / i); 
}

The

long comb(int, int);

before int main() is the declaration of the long comb(int n, int k) function. You have to do it, so that the code compiles. Another way would be to define the long comb(int n, int k) function completely before main() instead of after it. You should understand the difference between declaration and definition in C++.

By the way, your code seems to have a bug, and does not print the correct Pascal triangle.

Answer 2

These are the meanings of the different occurrences of comb() in your code.

long comb(int, int);

This means "Dear compiler. Later on I will provide code for a function of this name with these parameter types. Please accept that I call it before you have seen the code."
It is called a "prototype".

cout << setw(4) << comb(i, j);

Means "Dear compiler, remember the function I told you about but have not shown you the code for? Call it here please, as part of making output; the code is below somewhere."

long comb(int n, int k)
{
/* ... */
}

Means "Dear compiler, with the prototype above I promised to provide code for a function of this name and parameters. I even asked you to call it from other code. Here I finally provide the code for it. Please use this code for the execution of the call from main()." (Some of this last part actually addresses more the linker than the compiler, but I think that level of detail is out of scope here.)

Answer 3

I have the impression that you don't understand how to learn how to program:
An experienced programmer looks at a piece of code and understands what it means, but for a starting programmer most of the times this does not work: you need to try the piece of code, debug it, try to understand it and mostly, try it out .

Let's have a look at comb(n,k) : apparently n must be larger than k, so let's have a look at what happens for the values n=8 and k=3:

for(int i = 1; i <= k; i++, n--)
 {
  c = c*(n/i);
 }

What happens with i and c?

   i      k      n      c
   1      3      8      1*8/1
   2      3      7      1*8/1*7/2
   3      3      6      1*8/1*7/2*6/3

So, at the end: c becomes 8*7*6/(1*2*3) .

I admit: two variables get changed during the loop ( i++ and n-- ), which is confusing indeed, but by trying out you'll get the hang of it.