3或7的倍数

给定正整数n，请找到小于或等于n的3或7的所有倍数的计数。
例子 ：

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遍历从1到n的所有数字，并在数字为3或7或两者的倍数时递增计数。

C++

// A Simple C++ program to find count of all
// numbers that multiples
#include
using namespace std;
 
// Returns count of all numbers smaller than
// or equal to n and multples 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

// 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 multples 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

Python3

# A Simple Python3 program to
# find count of all numbers
# that multiples
 
# Returns count of all numbers
# smaller than or equal to n
# and multples 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 multples 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

Javascript

 
// A Simple JavaScript program to find count
// of all numbers that multiples
 
// Returns count of all numbers
// smaller than or equal to n
// and multples 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
 

输出 ：

Count = 10

时间复杂度： O(n)
一个有效的解决方案可以在O(1)时间内解决上述问题。这个想法是计算3的倍数，再加上7的倍数，然后减去21的倍数，因为它们被计数两次。

count = n/3 + n/7 - n/21

C++

// A better C++ program to find count of all
// numbers that multiples
#include
using namespace std;
 
// Returns count of all numbers smaller than
// or equal to n and multples 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

// 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 multples 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

的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
# multples 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 multples 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

Java脚本

 
// JavaScript program to find count
// of all numbers that multiples
 
// Returns count of all numbers
// smaller than or equal to n
// and multples 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
 

输出 ：

Count = 10