可被给定三个数整除的最大 N 位数
给定四个整数 x，y，z 和 n ，任务是找到最大的 n 位数，该位数可被 x，y 和 z 整除。 举例:
输入: x = 2，y = 3，z = 5，n = 4 输出: 9990 9990 是可被 2、3、5 整除的最大 4 位数。 输入: x = 3，y = 23，z = 6，n = 2 输出:不可能
进场:
- 找到最大的 n 位数，即幂(10，n)–1，并将其存储在变量 largestN 中。
- 找出给定的三个数字 x，y，z 的 LCM，说 LCM 。
- 计算最大值除以 LCM 即最大值% LCM 时的余数，并将其存储在变量余数中。
- 从大中减去余数。如果结果仍然是 n 数字，则打印结果。
- 否则打印不可能。
以下是上述方法的实现:
C++
// C++ program to find largest n digit number
// which is divisible by x, y and z.
#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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");
}
}
Python 3
# 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
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find largest n digit number
// which is divisible by x, y and z.
// Recursive function to return gcd of a and b
function 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 - $a);
}
// Function to return the LCM
// of three numbers
function LCM($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
java 描述语言
<script>
// Javascript program to find largest n
// digit number which is divisible
// by x, y and z.
// Recursive function to return
// gcd of a and b
function 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
function LCM(x, y, z)
{
var ans = parseInt((x * y) / (gcd(x, y)));
return parseInt((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
var lcm = LCM(x, y, z);
// find largest n-digit number
var largestNDigitNum = Math.pow(10, n) - 1;
var 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 >= Math.pow(10, n - 1))
return largestNDigitNum;
else
return 0;
}
// Driver code
var n = 2, x = 3, y = 4, z = 6;
var res = findDivisible(n, x, y, z);
// if the number is found
if (res != 0)
document.write(res);
else
document.write("Not possible");
</script>
Output:
96
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