C++ Program to Print Floyd Triangle

You can print Floyd's triangle in C++ using nested for loops and a counter variable, for example: int num = 1; for (int i = 1; i <= n; i++) { for (int j = 1; j <= i; j++) { std::cout << num++ << " "; } std::cout << std::endl; }.

Examples

Input3

Output1 2 3 4 5 6

Input5

Output1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Input1

Output1

How to Think About It

To print Floyd's triangle, think of printing numbers in rows where the first row has 1 number, the second has 2, and so on. Use a counter starting at 1 and increase it after printing each number. Use two loops: the outer loop for rows and the inner loop for numbers in each row.

Algorithm

1

Get the number of rows (n) from the user.

2

Initialize a counter variable to 1.

3

For each row from 1 to n:

4

Print numbers equal to the current row number, increasing the counter after each print.

5

Move to the next line after each row.

Code

cpp

#include <iostream>

int main() {
    int n, num = 1;
    std::cout << "Enter number of rows: ";
    std::cin >> n;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= i; j++) {
            std::cout << num++ << " ";
        }
        std::cout << std::endl;
    }
    return 0;
}

Output

Enter number of rows: 5
1 
2 3 
4 5 6 
7 8 9 10 
11 12 13 14 15 

Dry Run

Let's trace the program for input n=3 through the code.

1

Initialize variables

n = 3, num = 1

2

First row (i=1)

Print 1 number: print num=1, then num=2

3

Second row (i=2)

Print 2 numbers: print num=2, num=3, then num=4

4

Third row (i=3)

Print 3 numbers: print num=4, num=5, num=6, then num=7

Row (i)	Numbers Printed	num after row
1	1	2
2	2 3	4
3	4 5 6	7

Why This Works

Step 1: Outer loop controls rows

The for loop with variable i runs from 1 to n, controlling how many rows to print.

Step 2: Inner loop prints numbers per row

The inner for loop runs i times to print the correct count of numbers in each row.

Step 3: Counter increments after each print

The variable num starts at 1 and increases by 1 after printing each number to continue the sequence.

Alternative Approaches

Using while loops

cpp

#include <iostream>

int main() {
    int n, num = 1, i = 1;
    std::cout << "Enter number of rows: ";
    std::cin >> n;
    while (i <= n) {
        int j = 1;
        while (j <= i) {
            std::cout << num++ << " ";
            j++;
        }
        std::cout << std::endl;
        i++;
    }
    return 0;
}

This uses <code>while</code> loops instead of <code>for</code> loops, which some beginners find easier to understand.

Using recursion

cpp

#include <iostream>

void printRow(int count, int &num) {
    if (count == 0) return;
    std::cout << num++ << " ";
    printRow(count - 1, num);
}

void printFloyd(int row, int n, int &num) {
    if (row > n) return;
    printRow(row, num);
    std::cout << std::endl;
    printFloyd(row + 1, n, num);
}

int main() {
    int n, num = 1;
    std::cout << "Enter number of rows: ";
    std::cin >> n;
    printFloyd(1, n, num);
    return 0;
}

This recursive approach is elegant but less intuitive for beginners and uses function calls instead of loops.

Complexity: O(n^2) time, O(1) space

Time Complexity

The program uses nested loops where the outer loop runs n times and the inner loop runs up to n times cumulatively, resulting in O(n^2) time.

Space Complexity

Only a few integer variables are used, so the space complexity is O(1), constant space.

Which Approach is Fastest?

The loop-based approach is fastest and simplest. The recursive method adds overhead due to function calls.

Approach	Time	Space	Best For
Nested for loops	O(n^2)	O(1)	Simple and efficient
Nested while loops	O(n^2)	O(1)	Beginners preferring while loops
Recursion	O(n^2)	O(n)	Learning recursion, less efficient