山地序列模式
给定一个数字 N ，任务是生成一个接一个包含 N 金字塔的金字塔序列模式，如下例所示。 例:
Input: N = 3
Output:
* * *
*** *** ***
***************
Input: N = 4
Output:
* * * *
*** *** *** ***
********************
迭代方法:为给定编号 N 打印山脉序列模式的迭代方法步骤:
- 运行两个嵌套循环。
- 外部循环将关注图案的行。
- 内部循环将会关心列的模式。
- 取三个变量 k1、k2 和间隙，这有助于生成模式。
- 打印完图案行后，将 k1 和 k2 的值更新为:
- k1 = k1 +间隙
- k2 = k2 +间隙
下面是迭代方法的实现:
C++
// C++ program for the above approach
#include <iostream>
using namespace std;
// Function to create the mountain
// sequence pattern
void printPatt(int n)
{
int k1 = 3;
int k2 = 3;
int gap = 5;
// Outer loop to handle the row
for (int i = 1; i <= 3; i++) {
// Inner loop to handle the
// Column
for (int j = 1;
j <= (5 * n); j++) {
if (j > k2 && i < 3) {
k2 += gap;
k1 += gap;
}
// Condition to print the
// star in mountain pattern
if (j >= k1 && j <= k2) {
cout << "*";
}
else {
cout << " ";
}
}
// Condition to adjust the value of
// K1 and K2 for printing desire
// Pattern
if (i + 1 == 3) {
k1 = 1;
k2 = (5 * n);
}
else {
k1 = 3;
k2 = 3;
k1--;
k2++;
}
cout << endl;
}
}
// Driver Code
int main()
{
// Given Number N
int N = 5;
// Function call
printPatt(N);
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the above approach
class GFG{
// Function to create the mountain
// sequence pattern
static void printPatt(int n)
{
int k1 = 3;
int k2 = 3;
int gap = 5;
// Outer loop to handle the row
for(int i = 1; i <= 3; i++)
{
// Inner loop to handle the
// Column
for(int j = 1; j <= (5 * n); j++)
{
if (j > k2 && i < 3)
{
k2 += gap;
k1 += gap;
}
// Condition to print the
// star in mountain pattern
if (j >= k1 && j <= k2)
{
System.out.print("*");
}
else
{
System.out.print(" ");
}
}
// Condition to adjust the value of
// K1 and K2 for printing desire
// Pattern
if (i + 1 == 3)
{
k1 = 1;
k2 = (5 * n);
}
else
{
k1 = 3;
k2 = 3;
k1--;
k2++;
}
System.out.println();
}
}
// Driver code
public static void main (String[] args)
{
// Given Number N
int N = 5;
// Function call
printPatt(N);
}
}
// This code is contributed by Pratima Pandey
Python 3
# Python3 program for the above approach
# Function to create the mountain
# sequence pattern
def printPatt(n):
k1 = 3; k2 = 3; gap = 5;
# Outer loop to handle the row
for i in range(1, 4):
# Inner loop to handle the
# Column
for j in range(1, (5 * n) + 1):
if (j > k2 and i < 3):
k2 += gap;
k1 += gap;
# Condition to print the
# star in mountain pattern
if (j >= k1 and j <= k2):
print("*", end = "");
else:
print(" ", end = "");
print("\n", end = "");
# Condition to adjust the value of
# K1 and K2 for printing desire
# Pattern
if (i + 1 == 3):
k1 = 1;
k2 = (5 * n);
else:
k1 = 3;
k2 = 3;
k1 -= 1;
k2 += 1;
print(end = "");
# Driver Code
# Given Number N
N = 5;
# Function call
printPatt(N);
# This code is contributed by Code_Mech
C
// C# implementation of the above approach
using System;
class GFG{
// Function to create the mountain
// sequence pattern
static void printPatt(int n)
{
int k1 = 3;
int k2 = 3;
int gap = 5;
// Outer loop to handle the row
for(int i = 1; i <= 3; i++)
{
// Inner loop to handle the
// Column
for(int j = 1; j <= (5 * n); j++)
{
if (j > k2 && i < 3)
{
k2 += gap;
k1 += gap;
}
// Condition to print the
// star in mountain pattern
if (j >= k1 && j <= k2)
{
Console.Write("*");
}
else
{
Console.Write(" ");
}
}
// Condition to adjust the value of
// K1 and K2 for printing desire
// Pattern
if (i + 1 == 3)
{
k1 = 1;
k2 = (5 * n);
}
else
{
k1 = 3;
k2 = 3;
k1--;
k2++;
}
Console.WriteLine();
}
}
// Driver code
public static void Main (String[] args)
{
// Given Number N
int N = 5;
// Function call
printPatt(N);
}
}
// This code is contributed by shivanisinghss2110
java 描述语言
<!-- Javascript program for the above approach. -->
<script>
// Function to create the mountain
// sequence pattern
function printPatt( n)
{
var k1 = 3;
var k2 = 3;
var gap = 5;
// Outer loop to handle the row
for(let i = 1; i <= 3; i++)
{
// Inner loop to handle the
// Column
for(let j = 1; j <= (5 * n); j++)
{
if (j > k2 && i < 3)
{
k2 += gap;
k1 += gap;
}
// Condition to print the
// star in mountain pattern
if (j >= k1 && j <= k2)
{ document.write("*");
}
else
{
document.write(" ");
}
}
// Condition to adjust the value of
// K1 and K2 for printing desire
// Pattern
if (i + 1 == 3)
{
k1 = 1;
k2 = (5 * n);
}
else
{
k1 = 3;
k2 = 3;
k1--;
k2++;
}
document.write("<br>");
}
}
//Driver Code
var N=3;
printPatt(N);
</script>
<!-- This code in contributed by nirajgusain5 -->
Output:
* * * * *
*** *** *** *** ***
*************************
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