程序从给定的孔中查找编号

给定一个数字H ，该数字代表孔的总数。任务是找到具有那么多孔的最小数量。
笔记：

1. 0、4、6、9每个都有1个孔，而8个则有2个孔。
2. 该数字不应包含前导零。

例子：

Input: H = 1 
Output: 0
Input: H = 5 
Output: 488 
Explanation: 
Number which has 5 holes in it is 488. i.e (1 + 2 + 2) 
 

参考：计算整数中的孔数

方法：

1. 首先，检查给定的孔数是0还是1，如果为0，则打印1；如果为1，则打印0。
2. 如果给定的孔数大于1，则将孔数除以2，并将剩余的数存储在‘rem’变量中。和‘quo’变量中的商。
3. 现在，如果rem变量的值等于1，则首先打印4次，然后打印8次。
4. 否则仅打印8次。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
 
// Function that will find out
// the number
void printNumber(int holes)
{
 
    // If number of holes
    // equal 0 then return 1
    if (holes == 0)
        cout << "1";
 
    // If number of holes
    // equal 0 then return 0
    else if (holes == 1)
        cout << "0";
 
    // If number of holes
    // is more than 0 or 1.
    else {
        int rem = 0, quo = 0;
 
        rem = holes % 2;
        quo = holes / 2;
 
        // If number of holes is
        // odd
        if (rem == 1)
            cout << "4";
 
        for (int i = 0; i < quo; i++)
            cout << "8";
    }
}
 
// Driver code
int main()
{
    int holes = 3;
 
    // Calling Function
    printNumber(holes);
 
    return 0;
}

Java

// Java implementation of the above approach
import java.io.*;
 
class GFG
{
         
// Function that will find out
// the number
static void printNumber(int holes)
{
 
    // If number of holes
    // equal 0 then return 1
    if (holes == 0)
        System.out.print("1");
 
    // If number of holes
    // equal 0 then return 0
    else if (holes == 1)
        System.out.print("0");
 
    // If number of holes
    // is more than 0 or 1.
    else
    {
        int rem = 0, quo = 0;
 
        rem = holes % 2;
        quo = holes / 2;
 
        // If number of holes is
        // odd
        if (rem == 1)
            System.out.print("4");
 
        for (int i = 0; i < quo; i++)
                System.out.print("8");
    }
}
 
// Driver code
public static void main (String[] args)
{
    int holes = 3;
     
    // Calling Function
    printNumber(holes);
}
}
 
// This code is contributed by Sachin.

Python3

# Python3 implementation of
# the above approach
 
# Function that will find out
# the number
def printNumber(holes):
 
    # If number of holes
    # equal 0 then return 1
    if (holes == 0):
        print("1")
 
    # If number of holes
    # equal 0 then return 0
    elif (holes == 1):
        print("0", end = "")
 
    # If number of holes
    # is more than 0 or 1.
    else:
        rem = 0
        quo = 0
 
        rem = holes % 2
        quo = holes // 2
 
        # If number of holes is
        # odd
        if (rem == 1):
            print("4", end = "")
 
        for i in range(quo):
            print("8", end = "")
 
# Driver code
holes = 3
 
# Calling Function
printNumber(holes)
 
# This code is contributed by Mohit kumar

C#

// C# implementation of the above approach
using System;
 
class GFG
{
     
// Function that will find out
// the number
static void printNumber(int holes)
{
 
    // If number of holes
    // equal 0 then return 1
    if (holes == 0)
        Console.Write ("1");
 
    // If number of holes
    // equal 0 then return 0
    else if (holes == 1)
        Console.Write ("0");
 
    // If number of holes
    // is more than 0 or 1.
    else
    {
        int rem = 0, quo = 0;
 
        rem = holes % 2;
        quo = holes / 2;
 
        // If number of holes is
        // odd
        if (rem == 1)
        Console.Write ("4");
 
        for (int i = 0; i < quo; i++)
            Console.Write ("8");
    }
}
 
// Driver code
static public void Main ()
{
    int holes = 3;
     
    // Calling Function
    printNumber(holes);
}
}
 
// This code is contributed by jit_t

Javascript

    // Javascript implementation of the above approach
     
    // Function that will find out
    // the number
    function printNumber(holes)
    {
 
        // If number of holes
        // equal 0 then return 1
        if (holes == 0)
            document.write("1");
 
        // If number of holes
        // equal 0 then return 0
        else if (holes == 1)
            document.write("0");
 
        // If number of holes
        // is more than 0 or 1.
        else {
            let rem = 0, quo = 0;
 
            rem = holes % 2;
            quo = parseInt(holes / 2, 10);
 
            // If number of holes is
            // odd
            if (rem == 1)
                document.write("4");
 
            for (let i = 0; i < quo; i++)
                document.write("8");
        }
    }
     
    let holes = 3;
   
    // Calling Function
    printNumber(holes);
     
    // This code is contributed by divyeshrabadiya07.

输出：

48