3 或 7 的倍数
给定一个正整数 n，求小于等于 n 的 3 或 7 的所有倍数的计数 例:
Input : n = 10
Output : Count = 4
The multiples are 3, 6, 7 and 9
Input : n = 25
Output : Count = 10
The multiples are 3, 6, 7, 9, 12, 14, 15, 18, 21 and 24
一个简单的解决方案是迭代从 1 到 n 的所有数字，并且每当一个数字是 3 或 7 的倍数或者两者都是时递增计数。
C++
// A Simple C++ program to find count of all
// numbers that multiples
#include<iostream>
using namespace std;
// Returns count of all numbers smaller than
// or equal to n and multiples of 3 or 7 or both
int countMultiples(int n)
{
int res = 0;
for (int i=1; i<=n; i++)
if (i%3==0 || i%7 == 0)
res++;
return res;
}
// Driver code
int main()
{
cout << "Count = " << countMultiples(25);
}
Java 语言(一种计算机语言，尤用于创建网站)
// A Simple Java program to
// find count of all numbers
// that multiples
import java.io.*;
class GFG
{
// Returns count of all numbers
// smaller than or equal to n
// and multiples of 3 or 7 or both
static int countMultiples(int n)
{
int res = 0;
for (int i = 1; i <= n; i++)
if (i % 3 == 0 || i % 7 == 0)
res++;
return res;
}
// Driver Code
public static void main (String[] args)
{
System.out.print("Count = ");
System.out.println(countMultiples(25));
}
}
// This code is contributed by m_kit
Python 3
# A Simple Python3 program to
# find count of all numbers
# that multiples
# Returns count of all numbers
# smaller than or equal to n
# and multiples of 3 or 7 or both
def countMultiples(n):
res = 0;
for i in range(1, n + 1):
if (i % 3 == 0 or i % 7 == 0):
res += 1;
return res;
# Driver code
print("Count =", countMultiples(25));
# This code is contributed by mits
C
// A Simple C# program to
// find count of all numbers
// that are multiples of 3 or 7
using System;
class GFG
{
// Returns count of all
// numbers smaller than
// or equal to n and
// are multiples of 3 or
// 7 or both
static int countMultiples(int n)
{
int res = 0;
for (int i = 1; i <= n; i++)
if (i % 3 == 0 || i % 7 == 0)
res++;
return res;
}
// Driver Code
static public void Main ()
{
Console.Write("Count = ");
Console.WriteLine(countMultiples(25));
}
}
// This code is contributed by ajit
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// A Simple PHP program to find count
// of all numbers that multiples
// Returns count of all numbers
// smaller than or equal to n
// and multiples of 3 or 7 or both
function countMultiples($n)
{
$res = 0;
for ($i = 1; $i <= $n; $i++)
if ($i % 3 == 0 || $i % 7 == 0)
$res++;
return $res;
}
// Driver code
echo "Count = " ,countMultiples(25);
// This code is contributed by aj_36
?>
java 描述语言
<script>
// A Simple JavaScript program to find count
// of all numbers that multiples
// Returns count of all numbers
// smaller than or equal to n
// and multiples of 3 or 7 or both
function countMultiples(n)
{
let res = 0;
for (let i = 1; i <= n; i++)
if (i % 3 == 0 || i % 7 == 0)
res++;
return res;
}
// Driver code
document.write( "Count = " +countMultiples(25));
// This code is contributed by bobby
</script>
输出:
Count = 10
时间复杂度: O(n) 一个高效的解决方案可以在 O(1)时间内解决上述问题。这个想法是数 3 的倍数，加上 7 的倍数，然后减去 21 的倍数，因为这些是数两次。
count = n/3 + n/7 - n/21
C++
// A better C++ program to find count of all
// numbers that multiples
#include<iostream>
using namespace std;
// Returns count of all numbers smaller than
// or equal to n and multiples of 3 or 7 or both
int countMultiples(int n)
{
return n/3 + n/7 -n/21;
}
// Driver code
int main()
{
cout << "Count = " << countMultiples(25);
}
Java 语言(一种计算机语言，尤用于创建网站)
// A better Java program to
// find count of all numbers
// that multiples
import java.io.*;
class GFG
{
// Returns count of all numbers
// smaller than or equal to n
// and multiples of 3 or 7 or both
static int countMultiples(int n)
{
return n / 3 + n / 7 - n / 21;
}
// Driver code
public static void main (String args [] )
{
System.out.println("Count = " +
countMultiples(25));
}
}
// This code is contributed by aj_36
Python 3
# Python 3 program to find count of
# all numbers that multiples
# Returns count of all numbers
# smaller than or equal to n and
# multiples of 3 or 7 or both
def countMultiples(n):
return n / 3 + n / 7 - n / 21;
# Driver code
n = ((int)(countMultiples(25)));
print("Count =", n);
# This code is contributed
# by Shivi_Aggarwal
C
// A better Java program to
// find count of all numbers
// that multiples
using System;
class GFG
{
// Returns count of all numbers
// smaller than or equal to n
// and multiples of 3 or 7 or both
static int countMultiples(int n)
{
return n / 3 + n / 7 - n / 21;
}
// Driver Code
static public void Main ()
{
Console.WriteLine("Count = " +
countMultiples(25));
}
}
// This code is contributed by m_kit
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// A better PHP program to find count
// of all numbers that multiples
// Returns count of all numbers
// smaller than or equal to n
// and multiples of 3 or 7 or both
function countMultiples($n)
{
return floor($n / 3 + $n / 7 - $n / 21);
}
// Driver code
echo "Count = " , countMultiples(25);
// This code is contributed by ajit
?>
java 描述语言
<script>
// JavaScript program to find count
// of all numbers that multiples
// Returns count of all numbers
// smaller than or equal to n
// and multiples of 3 or 7 or both
function countMultiples(n)
{
return Math.floor(n / 3 + n / 7 - n / 21);
}
// Driver code
document.write( "Count = " +countMultiples(25));
// This code is contributed by bobby
</script>
输出:
Count = 10
时间复杂度: O(1) 练习: 现在试一下在 O(1)时间内求所有小于或等于 n 的数的和以及 3 或 7 的倍数或两者的问题。 本文由绍拉布·古普塔供稿。如果你喜欢极客博客并想投稿，你也可以写一篇文章并把你的文章邮寄到 review-team@geeksforgeeks.org。看到你的文章出现在极客博客主页上，帮助其他极客。 如果发现有不正确的地方，或者想分享更多关于上述话题的信息，请写评论。
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