Pascal's Triangle in C++

Program to print Pascal's triangle

AdvancedPattern ProgramsExample 6 of 6

pascal-s-triangle.cpp

1#include <iostream>
2using namespace std;
3
4int main() {
5    int rows;
6    
7    cout << "Enter number of rows: ";
8    cin >> rows;
9    
10    for (int i = 0; i < rows; i++) {
11        int num = 1;
12        // Print spaces
13        for (int j = 0; j < rows - i - 1; j++) {
14            cout << " ";
15        }
16        // Print numbers
17        for (int j = 0; j <= i; j++) {
18            cout << num << " ";
19            num = num * (i - j) / (j + 1);
20        }
21        cout << endl;
22    }
23    
24    return 0;
25}

Output

Enter number of rows: 5
    1
   1 1
  1 2 1
 1 3 3 1
1 4 6 4 1

What's going on

Pascal's triangle: each entry is a binomial coefficient. Row i (counting from 0) is C(i,0) C(i,1) … C(i,i). 5 rows: 1 / 1 1 / 1 2 1 / 1 3 3 1 / 1 4 6 4 1.

This program does not add the two numbers above. It walks the row with:

num = num * (i - j) / (j + 1)

for (int i = 0; i < rows; i++) — 0-based. i < rows not <=.

Pad: j < rows - i - 1 spaces, then the numbers.

int num = 1; at the start of each row. Always print 1 first.

Then for j from 0 to i: print num, then update. Multiply before divide so integer math stays exact: num * (i - j) / (j + 1). Reverse that (num / (j+1) * (i-j)) and 4/2 is fine but 5/2 truncates and the row is wrong.

Walk i = 4:

j=0: print 1, num = 1*(4-0)/1 = 4

j=1: print 4, num = 4*(4-1)/2 = 6

j=2: print 6, num = 6*(4-2)/3 = 4

j=3: print 4, num = 4*(4-3)/4 = 1

j=4: print 1

That's 1 4 6 4 1.

The visual "sum of the two parents" is true. This code never looks at the row above.

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