📜  打印偶数奇数金字塔的程序

📅  最后修改于: 2022-05-13 01:57:07.220000             🧑  作者: Mango

打印偶数奇数金字塔的程序

给定总行数为 n,任务是打印给定的模式。

例子:

Input: n = 5
Output:
*
1*
*2*
1*3*
*2*4*

Input: n = 10
Output:
*
1*
*2*
1*3*
*2*4*
1*3*5*
*2*4*6*
1*3*5*7*
*2*4*6*8*
1*3*5*7*9*

以下是上述问题的解决方案:

C++
// CPP program to print Even Odd Number Pyramid
 
#include 
using namespace std;
 
// function for creating pattern
void Pattern(int n)
{
    // Initialization
    int i, j, k;
    for (i = 1; i <= n; i++) {
 
        for (j = 1, k = i; j <= i; j++, k--) {
 
            if (k % 2 == 0) {
 
                // displaying the numbers
                cout << j;
            }
            else {
 
                // displaying the stars
                cout << "*";
            }
        }
        cout << "\n";
    }
}
 
// driver code
int main()
{
 
    // Get n
    int n = 5;
 
    // Print the pattern
    Pattern(n);
 
    return 0;
}


C
// C program to print Even Odd Number Pyramid
 
#include 
 
// function for creating pattern
void Pattern(int n)
{
    // Initialization
    int i, j, k;
    for (i = 1; i <= n; i++) {
        for (j = 1, k = i; j <= i; j++, k--) {
            if (k % 2 == 0) {
 
                // displaying the numbers
                printf("%d", j);
            }
            else {
 
                // displaying the stars
                printf("*");
            }
        }
        printf("\n");
    }
}
 
// driver code
int main()
{
 
    // Get n
    int n = 5;
 
    // Print the pattern
    Pattern(n);
 
    return 0;
}


Java
// Java program to print above pattern
 
import java.util.Scanner;
 
class Pattern {
    static void display(int n)
    {
 
        int i, j, k;
        for (i = 1; i <= n; i++) {
 
            for (j = 1, k = i; j <= i; j++, k--) {
 
                if (k % 2 == 0) {
                    // displaying the numbers
                    System.out.print(j);
                }
                else {
                    // displaying the stars
                    System.out.print("*");
                }
            }
            System.out.print("\n");
        }
    }
    // Driver Code
    public static void main(String[] args)
    {
 
        // Get n
        int n = 5;
 
        // Print the pattern
        display(n);
    }
}


Python3
# Python3 program to print above pattern
def display(n):
    for i in range(1, n + 1):
        k = i
        for j in range(1, i + 1):
            if k % 2 == 0:
                 
                # Displaying the numbers
                print(j, end = '')
            else:
                 
                # Displaying the stars
                print('*', end = '')
            k -= 1
        print()
 
# Driver Code
 
# Get n
n = 5
 
# Print the pattern
display(n)
 
# This code is contributed by SamyuktaSHegde


C#
// C# program to print above pattern
using System;
 
class GFG
{
static void display(int n)
{
 
    int i, j, k;
    for (i = 1; i <= n; i++)
    {
 
        for (j = 1, k = i; j <= i; j++, k--)
        {
 
            if (k % 2 == 0)
            {
                // displaying the numbers
                Console.Write(j);
            }
            else
            {
                // displaying the stars
                Console.Write("*");
            }
        }
        Console.Write("\n");
    }
}
 
// Driver Code
public static void Main()
{
 
    // Get n
    int n = 5;
 
    // Print the pattern
    display(n);
}
}
 
// This code is contributed by anuj_67


PHP


Javascript


输出:
*
1*
*2*
1*3*
*2*4*