给定整数N ，任务是为N的给定值打印以下图案。
For N = 5 Below is the given pattern:
例子：
Input: N = 4
Output:
1020304017018019020
**50607014015016
****809012013
******10011
Input: N = 3
Output:
10203010011012
**4050809
****607
方法：想法是要了解给定模式背后的逻辑，如下所述：
通过仔细观察，我们发现通过用空格替换零之间的模式，可以更清楚地看到图案。该模式进一步分为三个不同的模式。
- 情况1：星号(*)字符模式遵循一个从0开始的序列，并在每行中添加另外两个星号，其中该行等于N。
- 情况2：在这一部分中，模式很容易理解，即列和行的数量将等于N，并遵循类似于1、2、3、4、5…的序列
- 情况3：后续或自下而上的顺序是它有趣的部分，其中数字从下到上表示。
下面是上述方法的实现：
C++
// C++ implementation to print
// the given pattern
#include
using namespace std;
// Function to find the sum of
// N integers from 1 to N
int sum(int n)
{
return n * (n - 1) / 2;
}
// Function to print the given
// pattern
void BSpattern(int N)
{
int Val = 0, Pthree = 0,
cnt = 0, initial;
string s = "**";
// Iterate over [0, N - 1]
for (int i = 0; i < N; i++) {
cnt = 0;
// Sub-Pattern - 1
if (i > 0) {
cout << s;
s += "**";
}
// Sub-Pattern - 2
for (int j = i; j < N; j++) {
// Count the number of element
// in rows and sub-pattern 2 and 3
// will have same rows
if (i > 0) {
cnt++;
}
// Increment Val to print the
// series 1, 2, 3, 4, 5 ...
cout << ++Val;
cout << 0;
}
// To get the first element of sub
// pattern 3 find the sum of first N-1
// elements first N-1 elements in row1
// previous of Sub-Pattern 2
// Finally, add the (N-1)th element
// i.e., 5 and increment it by 1
if (i == 0) {
int Sumbeforelast = sum(Val) * 2;
Pthree = Val + Sumbeforelast + 1;
initial = Pthree;
}
// Initial is used to give the initial
// value of the row in Sub-Pattern 3
initial = initial - cnt;
Pthree = initial;
// Sub-Pattern 3
for (int k = i; k < N; k++) {
cout << Pthree++;
// Skip printing zero at the last
if (k != N - 1) {
cout << 0;
}
}
cout << "\n";
}
}
// Driver Code
int main()
{
// Given N
int N = 5;
// Function Call
BSpattern(N);
return 0;
} Java
// Java implementation to print
// the given pattern
import java.util.*;
class GFG{
// Function to find the sum of
// N integers from 1 to N
static int sum(int n)
{
return n * (n - 1) / 2;
}
// Function to print the given
// pattern
static void BSpattern(int N)
{
int Val = 0, Pthree = 0,
cnt = 0, initial = -1;
String s = "**";
// Iterate over [0, N - 1]
for(int i = 0; i < N; i++)
{
cnt = 0;
// Sub-Pattern - 1
if (i > 0)
{
System.out.print(s);
s += "**";
}
// Sub-Pattern - 2
for(int j = i; j < N; j++)
{
// Count the number of element
// in rows and sub-pattern 2
// and 3 will have same rows
if (i > 0)
{
cnt++;
}
// Increment Val to print the
// series 1, 2, 3, 4, 5 ...
System.out.print(++Val);
System.out.print("0");
}
// To get the first element of sub
// pattern 3 find the sum of first N-1
// elements first N-1 elements in row1
// previous of Sub-Pattern 2
// Finally, add the (N-1)th element
// i.e., 5 and increment it by 1
if (i == 0)
{
int Sumbeforelast = sum(Val) * 2;
Pthree = Val + Sumbeforelast + 1;
initial = Pthree;
}
// Initial is used to give the initial
// value of the row in Sub-Pattern 3
initial = initial - cnt;
Pthree = initial;
// Sub-Pattern 3
for(int k = i; k < N; k++)
{
System.out.print(Pthree++);
// Skip printing zero at the last
if (k != N - 1)
{
System.out.print("0");
}
}
System.out.println();
}
}
// Driver code
public static void main(String[] args)
{
// Given N
int N = 5;
// Function call
BSpattern(N);
}
}
// This code is contributed by offbeatPython3
# Python3 implementation to print
# the given pattern
# Function to find the sum of
# N integers from 1 to N
def sum(n):
return n * (n - 1) // 2
# Function to print the given
# pattern
def BSpattern(N):
Val = 0
Pthree = 0,
cnt = 0
initial = -1
s = "**"
# Iterate over [0, N - 1]
for i in range(N):
cnt = 0
# Sub-Pattern - 1
if (i > 0):
print(s, end = "")
s += "**"
# Sub-Pattern - 2
for j in range(i, N):
# Count the number of element
# in rows and sub-pattern 2 and 3
# will have same rows
if (i > 0):
cnt += 1
# Increment Val to print the
# series 1, 2, 3, 4, 5 ...
Val += 1
print(Val, end = "")
print(0, end = "")
# To get the first element of sub
# pattern 3 find the sum of first N-1
# elements first N-1 elements in row1
# previous of Sub-Pattern 2
# Finally, add the (N-1)th element
# i.e., 5 and increment it by 1
if (i == 0):
Sumbeforelast = sum(Val) * 2
Pthree = Val + Sumbeforelast + 1
initial = Pthree
# Initial is used to give the initial
# value of the row in Sub-Pattern 3
initial = initial - cnt
Pthree = initial
# Sub-Pattern 3
for k in range(i, N):
print(Pthree, end = "")
Pthree += 1
# Skip printing zero at the last
if (k != N - 1):
print(0, end = "")
print()
# Driver Code
# Given N
N = 5
# Function call
BSpattern(N)
# This code is contributed by sanjoy_62C#
// C# implementation to print
// the given pattern
using System;
class GFG{
// Function to find the sum of
// N integers from 1 to N
static int sum(int n)
{
return n * (n - 1) / 2;
}
// Function to print the given
// pattern
static void BSpattern(int N)
{
int Val = 0, Pthree = 0,
cnt = 0, initial = -1;
String s = "**";
// Iterate over [0, N - 1]
for(int i = 0; i < N; i++)
{
cnt = 0;
// Sub-Pattern - 1
if (i > 0)
{
Console.Write(s);
s += "**";
}
// Sub-Pattern - 2
for(int j = i; j < N; j++)
{
// Count the number of element
// in rows and sub-pattern 2
// and 3 will have same rows
if (i > 0)
{
cnt++;
}
// Increment Val to print the
// series 1, 2, 3, 4, 5 ...
Console.Write(++Val);
Console.Write("0");
}
// To get the first element of sub
// pattern 3 find the sum of first N-1
// elements first N-1 elements in row1
// previous of Sub-Pattern 2
// Finally, add the (N-1)th element
// i.e., 5 and increment it by 1
if (i == 0)
{
int Sumbeforelast = sum(Val) * 2;
Pthree = Val + Sumbeforelast + 1;
initial = Pthree;
}
// Initial is used to give the initial
// value of the row in Sub-Pattern 3
initial = initial - cnt;
Pthree = initial;
// Sub-Pattern 3
for(int k = i; k < N; k++)
{
Console.Write(Pthree++);
// Skip printing zero at the last
if (k != N - 1)
{
Console.Write("0");
}
}
Console.WriteLine();
}
}
// Driver code
public static void Main(String[] args)
{
// Given N
int N = 5;
// Function call
BSpattern(N);
}
}
// This code is contributed by shikhasingrajput输出：
102030405026027028029030
**6070809022023024025
****10011012019020021
******13014017018
********15016时间复杂度： O(N 2 )
辅助空间： O(1)