可被整数 n 整除的 m 位整数的计数

原文:https://www . geesforgeks . org/count-m-digits-integers-整除-integer-n/

给定两个数字 m 和 n，计算可被 n 整除的 m 位数的个数 示例:

Input : m = 2 
        n = 6
Output : 15
Two digit numbers that are divisible by 6
are 12, 18, 24, 30, 36, ....., 96.

Input : m = 3 
        n = 5
Output :180

一个简单的解决方法就是两次尝试所有 m 位数。对于每个数字，检查它是否可以被 n 整除。如果可以，我们增加计数。 一个高效的解决方案包括以下步骤。 这个想法是基于这样一个事实:从第一个可除数开始，每第 n 个数都可以被 n 整除。

  1. 找到最大的 m 位数。
  2. 找到最大的 m-1 位数。
  3. 将两个数除以 n，然后从前面减去。

下面是以上步骤的实现。

C++

// C++ program to count m digit numbers having
// n as divisor.
#include<bits/stdc++.h>
using namespace std;

// Returns count of m digit numbers having n
// as divisor
int findCount(int m, int n)
{   
    // generating largest number of m digit
    int num1 = 0;
    for (int i = 0; i < m; i++)
        num1 = (num1 * 10) + 9;

    // generating largest number of m-1 digit
    int num2 = 0;
    for (int i = 0; i < (m - 1); i++)
        num2 = (num2 * 10) + 9;

    // returning number of dividend
    return ((num1 / n) - (num2 / n));
}

// Driver code
int main()
{
    int m = 2, n = 6;
    printf("%d\n", findCount(m, n));
    return 0;
}

Java 语言(一种计算机语言，尤用于创建网站)

// Java program to count m digit numbers having
// n as divisor.

class Main
{
    // Returns count of m digit numbers having n
    // as divisor
    static int findCount(int m, int n)
    {   
        // generating largest number of m digit
        int num1 = 0;
        for (int i = 0; i < m; i++)
            num1 = (num1 * 10) + 9;

        // generating largest number of m-1 digit
        int num2 = 0;
        for (int i = 0; i < (m - 1); i++)
            num2 = (num2 * 10) + 9;

        // returning number of dividend
        return ((num1 / n) - (num2 / n));
    }

    // main function
    public static void main (String[] args)
    {
        int m = 2, n = 6;
        System.out.println(findCount(m, n));
    }
}

/* This code is contributed by Harsh Agarwal */

Python 3

# Python3 program to count m digit
# numbers having n as divisor.

# Returns count of m digit
# numbers having n as divisor
def findCount(m, n):

    # Generating largest number of m digit
    num1 = 0

    for i in range(0, m):
        num1 = (num1 * 10) + 9

    # Generating largest number of m-1 digit
    num2 = 0

    for i in range(0, (m - 1)):
        num2 = (num2 * 10) + 9

    # returning number of dividend
    return int((num1 / n) - (num2 / n))

# Driver code
m = 2; n = 6
print(findCount(m, n))

# This code is contributed by Smitha Dinesh Semwal

C

// C# program to count m digit numbers
// having n as divisor.
using System;

class GfG {

    // Returns count of m digit numbers
    // having n as divisor
    static int findCount(int m, int n)
    {

        // generating largest number
        // of m digit
        int num1 = 0;
        for (int i = 0; i < m; i++)
            num1 = (num1 * 10) + 9;

        // generating largest number
        // of m-1 digit
        int num2 = 0;
        for (int i = 0; i < (m - 1); i++)
            num2 = (num2 * 10) + 9;

        // returning number of dividend
        return ((num1 / n) - (num2 / n));
    }

    // main function
    public static void Main ()
    {
        int m = 2, n = 6;

        Console.Write(findCount(m, n));
    }
}

// This code is contributed by parashar.

服务器端编程语言（Professional Hypertext Preprocessor 的缩写）

<?php
// PHP program to count m digit
// numbers having n as divisor.

// Returns count of m digit numbers
// having n as divisor
function findCount($m, $n)
{
    // generating largest number
    // of m digit
    $num1 = 0;
    for ($i = 0; $i < $m; $i++)
        $num1 = ($num1 * 10) + 9;

    // generating largest number
    // of m-1 digit
    $num2 = 0;
    for ($i = 0; $i < ($m - 1); $i++)
        $num2 = ($num2 * 10) + 9;

    // returning number of dividend
    return (($num1 / $n) - ($num2 / $n));
}

// Driver code
$m = 2; $n = 6;
echo findCount($m, $n), "\n";

// This code is contributed by ajit
?>

java 描述语言

<script>
// Javascript program to count m digit
// numbers having n as divisor.

// Returns count of m digit numbers
// having n as divisor
function findCount(m, n)
{

    // generating largest number
    // of m digit
    let num1 = 0;
    for (let i = 0; i < m; i++)
        num1 = (num1 * 10) + 9;

    // generating largest number
    // of m-1 digit
    let num2 = 0;
    for (let i = 0; i < (m - 1); i++)
        num2 = (num2 * 10) + 9;

    // returning number of dividend
    return ((num1 / n) - (num2 / n));
}

// Driver code
let m = 2; n = 6;
document.write(findCount(m, n) + "<br>");

// This code is contributed by gfgking
</script>

输出:

15

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