最大N位数字可被给定的三个数字整除

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📜 最大N位数字可被给定的三个数字整除

📅 最后修改于: 2021-05-04 13:29:51 🧑 作者: Mango

给定四个整数x，y，z和n ，任务是找到可被x，y和z整除的最大n位数字。

例子：

Input: x = 2, y = 3, z = 5, n = 4
Output: 9990
9990 is the largest 4-digit number which is divisible by 2, 3 and 5.

Input: x = 3, y = 23, z = 6, n = 2
Output: Not possible

为什么编程需要懂一点英语

方法：

找到最大的n位数字，即pow(10，n)– 1并将其存储在变量maximumN中。
找到给定的三个数字x，y和z的LCM表示LCM 。
当最大N除以LCM即最大N％LCM时，计算余数并将其存储在可变余数中。
从maximumN减去余数。如果结果仍然是n位数字，则打印结果。
其他打印不可能。

下面是上述方法的实现：

C++

// C++ program to find largest n digit number
// which is divisible by x, y and z.
#include 
using namespace std;
  
// Function to return the LCM of three numbers
int LCM(int x, int y, int z)
{
    int ans = ((x * y) / (__gcd(x, y)));
    return ((z * ans) / (__gcd(ans, z)));
}
  
// Function to return the largest n-digit
// number which is divisible by x, y and z
int findDivisible(int n, int x, int y, int z)
{
  
    // find the LCM
    int lcm = LCM(x, y, z);
  
    // find largest n-digit number
    int largestNDigitNum = pow(10, n) - 1;
  
    int remainder = largestNDigitNum % lcm;
  
    // If largest number is the answer
    if (remainder == 0)
        return largestNDigitNum ;
  
    // find closest smaller number
    // divisible by LCM
    largestNDigitNum -= remainder;
  
    // if result is an n-digit number
    if (largestNDigitNum >= pow(10, n - 1))
        return largestNDigitNum;
    else
        return 0;
}
  
// Driver code
int main()
{
    int n = 2, x = 3, y = 4, z = 6;
    int res = findDivisible(n, x, y, z);
  
    // if the number is found
    if (res != 0)
        cout << res;
    else
        cout << "Not possible";
  
    return 0;
}

Java

// Java program to find largest n digit number
// which is divisible by x, y and z.
import java.math.*;
class GFG {
      
// Recursive function to return gcd of a and b 
    static int gcd(int a, int b) 
    { 
        // Everything divides 0  
        if (a == 0) 
          return b; 
        if (b == 0) 
          return a; 
         
        // base case 
        if (a == b) 
            return a; 
         
        // a is greater 
        if (a > b) 
            return gcd(a-b, b); 
        return gcd(a, b-a); 
    } 
      
// Function to return the LCM of three numbers
static int LCM(int x, int y, int z)
{
    int ans = ((x * y) / (gcd(x, y)));
    return ((z * ans) / (gcd(ans, z)));
}
  
// Function to return the largest n-digit
// number which is divisible by x, y and z
static int findDivisible(int n, int x, int y, int z)
{
  
    // find the LCM
    int lcm = LCM(x, y, z);
  
    // find largest n-digit number
    int largestNDigitNum = (int)Math.pow(10, n) - 1;
  
    int remainder = largestNDigitNum % lcm;
  
    // If largest number is the answer
    if (remainder == 0)
        return largestNDigitNum ;
  
    // find closest smaller number
    // divisible by LCM
    largestNDigitNum -= remainder;
  
    // if result is an n-digit number
    if (largestNDigitNum >= (int)Math.pow(10, n - 1))
        return largestNDigitNum;
    else
        return 0;
}
  
// Driver code
public static void main(String args[])
{
    int n = 2, x = 3, y = 4, z = 6;
    int res = findDivisible(n, x, y, z);
  
    // if the number is found
    if (res != 0)
        System.out.println(res);
    else
        System.out.println("Not possible");
  
}
}

Python3

# Python3 program to find largest n digit
# number which is divisible by x, y and z.
  
# Recursive function to return 
# gcd of a and b 
def gcd(a, b):
  
    # Everything divides 0 
    if (a == 0): 
        return b; 
    if (b == 0): 
        return a; 
      
    # base case 
    if (a == b): 
        return a; 
      
    # a is greater 
    if (a > b): 
        return gcd(a - b, b); 
    return gcd(a, b - a); 
  
# Function to return the LCM 
# of three numbers
def LCM(x, y, z):
    ans = ((x * y) / (gcd(x, y)));
    return ((z * ans) / (gcd(ans, z)));
  
# Function to return the largest n-digit
# number which is divisible by x, y and z
def findDivisible(n, x, y, z):
      
    # find the LCM
    lcm = LCM(x, y, z);
      
    # find largest n-digit number
    largestNDigitNum = int(pow(10, n)) - 1;
      
    remainder = largestNDigitNum % lcm;
      
    # If largest number is the answer
    if (remainder == 0):
        return largestNDigitNum ;
      
    # find closest smaller number
    # divisible by LCM
    largestNDigitNum -= remainder;
      
    # if result is an n-digit number
    if (largestNDigitNum >= int(pow(10, n - 1))):
        return largestNDigitNum;
    else:
        return 0;
  
# Driver code
n = 2; x = 3; 
y = 4; z = 6;
res = int(findDivisible(n, x, y, z));
  
# if the number is found
if (res != 0):
    print(res);
else:
    print("Not possible");
  
# This code is contributed 
# by mits

C#

// C# program to find largest n 
// digit number which is divisible 
// by x, y and z.
using System;
  
class GFG 
{
// Recursive function to return
// gcd of a and b 
static int gcd(int a, int b) 
{ 
    // Everything divides 0 
    if (a == 0) 
        return b; 
    if (b == 0) 
        return a; 
      
    // base case 
    if (a == b) 
        return a; 
      
    // a is greater 
    if (a > b) 
        return gcd(a - b, b); 
    return gcd(a, b - a); 
} 
  
// Function to return the 
// LCM of three numbers
static int LCM(int x, int y, int z)
{
    int ans = ((x * y) / (gcd(x, y)));
    return ((z * ans) / (gcd(ans, z)));
}
  
// Function to return the largest 
// n-digit number which is divisible
// by x, y and z
static int findDivisible(int n, int x, 
                         int y, int z)
{
  
    // find the LCM
    int lcm = LCM(x, y, z);
  
    // find largest n-digit number
    int largestNDigitNum = (int)Math.Pow(10, n) - 1;
  
    int remainder = largestNDigitNum % lcm;
  
    // If largest number is the answer
    if (remainder == 0)
        return largestNDigitNum ;
  
    // find closest smaller number
    // divisible by LCM
    largestNDigitNum -= remainder;
  
    // if result is an n-digit number
    if (largestNDigitNum >= (int)Math.Pow(10, n - 1))
        return largestNDigitNum;
    else
        return 0;
}
  
// Driver code
static void Main()
{
    int n = 2, x = 3, y = 4, z = 6;
    int res = findDivisible(n, x, y, z);
  
    // if the number is found
    if (res != 0)
        Console.WriteLine(res);
    else
        Console.WriteLine("Not possible");
}
}
  
// This code is contributed by ANKITRAI1

PHP

 $b) 
        return gcd($a - $b, $b); 
    return gcd($a, $b - $a); 
} 
  
// Function to return the LCM 
// of three numbers
function LCM($x, $y, $z)
{
$ans = (($x * $y) / (gcd($x, $y)));
return (($z * $ans) / (gcd($ans, $z)));
}
  
// Function to return the largest n-digit
// number which is divisible by x, y and z
function findDivisible($n, $x, $y, $z)
{
      
    // find the LCM
    $lcm = LCM($x, $y, $z);
      
    // find largest n-digit number
    $largestNDigitNum = (int)pow(10, $n) - 1;
      
    $remainder = $largestNDigitNum % $lcm;
      
    // If largest number is the answer
    if ($remainder == 0)
        return $largestNDigitNum ;
      
    // find closest smaller number
    // divisible by LCM
    $largestNDigitNum -= $remainder;
      
    // if result is an n-digit number
    if ($largestNDigitNum >= (int)pow(10, $n - 1))
        return $largestNDigitNum;
    else
        return 0;
}
  
// Driver code
$n = 2; $x = 3; $y = 4; $z = 6;
$res = findDivisible($n, $x, $y, $z);
  
// if the number is found
if ($res != 0)
    echo $res;
else
    echo "Not possible";
  
// This code is contributed 
// by Akanksha Rai

输出：