计算可被给定数字除尽的 n 位数
给定 n 位数和一个数，任务是计算所有能被该数整除的 n 位数。 例:
Input : n = 2, number = 7
Output : 9
There are nine n digit numbers that
are divisible by 7\. Numbers are 14,
21, 28, 35, 42, 49, .... 70.
Input : n = 3, number = 7
Output : 128
Input : n = 4, number = 4
Output : 2250
原生方式:遍历所有 n 位数。对于每个数字，检查可除性，
C++
// Simple CPP program to count n digit
// divisible numbers.
#include <cmath>
#include <iostream>
using namespace std;
// Returns count of n digit numbers
// divisible by 'number'
int numberofterm(int n, int number)
{
// compute the first and last term
int firstnum = pow(10, n - 1);
int lastnum = pow(10, n);
// count total number of which having
// n digit and divisible by number
int count = 0;
for (int i = firstnum; i < lastnum; i++)
if (i % number == 0)
count++;
return count;
}
// Driver code
int main()
{
int n = 3, num = 7;
cout << numberofterm(n, num) << "\n";
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Simple Java program to count n digit
// divisible numbers.
import java.io.*;
class GFG {
// Returns count of n digit numbers
// divisible by 'number'
static int numberofterm(int n, int number)
{
// compute the first and last term
int firstnum = (int)Math.pow(10, n - 1);
int lastnum = (int)Math.pow(10, n);
// count total number of which having
// n digit and divisible by number
int count = 0;
for (int i = firstnum; i < lastnum; i++)
if (i % number == 0)
count++;
return count;
}
// Driver code
public static void main (String[] args)
{
int n = 3, num = 7;
System.out.println(numberofterm(n, num));
}
}
// This code is contributed by Ajit.
Python 3
# Simple Python 3 program to count n digit
# divisible numbers
import math
# Returns count of n digit
# numbers divisible by number
def numberofterm(n, number):
# compute the first and last term
firstnum = math.pow(10, n - 1)
lastnum = math.pow(10, n)
# count total number of which having
# n digit and divisible by number
count = 0
for i in range(int(firstnum), int(lastnum)):
if (i % number == 0):
count += 1
return count
# Driver code
n = 3
num = 7
print(numberofterm(n, num))
# This article is contributed
# by Smitha Dinesh Semwal
C
// Simple C# program to count n digit
// divisible numbers.
using System;
class GFG
{
// Returns count of n digit numbers
// divisible by 'number'
static int numberofterm(int n, int number)
{
// compute the first and last term
int firstnum = (int)Math.Pow(10, n - 1);
int lastnum = (int)Math.Pow(10, n);
// count total number of which having
// n digit and divisible by number
int count = 0;
for (int i = firstnum; i < lastnum; i++)
if (i % number == 0)
count++;
return count;
}
// Driver code
public static void Main ()
{
int n = 3, num = 7;
Console.Write(numberofterm(n, num));
}
}
// This code is contributed by nitin mittal
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// Simple php program to count n digit
// divisible numbers.
// Returns count of n digit numbers
// divisible by 'number'
function numberofterm($n, $number)
{
// compute the first and last term
$firstnum = pow(10, $n - 1);
$lastnum = pow(10, $n);
// count total number of which having
// n digit and divisible by number
$count = 0;
for ($i = $firstnum; $i < $lastnum; $i++)
if ($i % $number == 0)
$count++;
return $count;
}
// Driver code
$n = 3;
$num = 7;
echo numberofterm($n, $num);
// This code is contributed by mits
?>
java 描述语言
<script>
// JavaScript program to count n digit
// divisible numbers.
// Returns count of n digit numbers
// divisible by 'number'
function numberofterm(n, number)
{
// compute the first and last term
let firstnum = Math.pow(10, n - 1);
let lastnum = Math.pow(10, n);
// count total number of which having
// n digit and divisible by number
let count = 0;
for (let i = firstnum; i < lastnum; i++)
if (i % number == 0)
count++;
return count;
}
// Driver Code
let n = 3, num = 7;
document.write(numberofterm(n, num));
// This code is contributed by code_hunt.
</script>
Output:
128
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