计算范围内的整数，这些范围可被其欧拉上位值整除

给定2个整数L和R ，任务是找出[L，R]范围内的整数数量，以使它们可以完全被其Euler拉伸值整除。
例子：

Input: L = 2, R = 3 
Output: 1 

(2) = 2 => 2 % 

(2) = 0 

(3) = 2 => 3 % 

(3) = 1 
Hence 2 satisfies the condition.
Input: L = 12, R = 21 
Output: 3 
Only 12, 16 and 18 satisfy the condition. 
 

方法：我们知道一个数字的欧拉上位函数如下：

$\phi(n) = n * (1 - \frac{1}{p_1}) * (1 - \frac{1}{p_2}) * ... * (1 - \frac{1}{p_k})$

重新排列条款，我们得到：

$\frac{n}{\phi(n)} = \frac{p_1 * p_2 * ... * p_k}{(p_1 - 1) * (p_2 -1) * ... * (p_k -1)}$

如果我们仔细研究RHS，我们会发现只有2和3是满足n％的素数

*** QuickLaTeX cannot compile formula:
 

*** Error message:
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= 0 。这是因为对于素数p ₁ = 2和p ₂ = 3，p ₁ – 1 = 1和p ₂ – 1 = 2 。因此，当处于范围[L，R]时，仅需要计数2 ^p 3 ^q形式的数字，其中p> = 1且q> = 0 。
下面是上述方法的实现：

C++

*** QuickLaTeX cannot compile formula:
 

*** Error message:
Error: Nothing to show, formula is empty

Java

*** QuickLaTeX cannot compile formula:
 

*** Error message:
Error: Nothing to show, formula is empty

Python3

*** QuickLaTeX cannot compile formula:
 

*** Error message:
Error: Nothing to show, formula is empty

C#

*** QuickLaTeX cannot compile formula:
 

*** Error message:
Error: Nothing to show, formula is empty

PHP

// C++ implementation of the above approach.
#include 
 
#define ll long long
using namespace std;
 
// Function to return a^n
ll power(ll a, ll n)
{
    if (n == 0)
        return 1;
 
    ll p = power(a, n / 2);
    p = p * p;
 
    if (n & 1)
        p = p * a;
 
    return p;
}
 
// Function to return count of integers
// that satisfy n % phi(n) = 0
int countIntegers(ll l, ll r)
{
 
    ll ans = 0, i = 1;
    ll v = power(2, i);
 
    while (v <= r) {
 
        while (v <= r) {
 
            if (v >= l)
                ans++;
            v = v * 3;
        }
 
        i++;
        v = power(2, i);
    }
 
    if (l == 1)
        ans++;
 
    return ans;
}
 
// Driver Code
int main()
{
    ll l = 12, r = 21;
    cout << countIntegers(l, r);
 
    return 0;
}

Javascript

// Java implementation of the above approach.
class GFG
{
 
// Function to return a^n
static long power(long a, long n)
{
    if (n == 0)
        return 1;
 
    long p = power(a, n / 2);
    p = p * p;
 
    if (n%2== 1)
        p = p * a;
 
    return p;
}
 
// Function to return count of integers
// that satisfy n % phi(n) = 0
static int countIntegers(long l, long r)
{
 
    long ans = 0, i = 1;
    long v = power(2, i);
 
    while (v <= r)
    {
        while (v <= r)
        {
 
            if (v >= l)
                ans++;
            v = v * 3;
        }
 
        i++;
        v = power(2, i);
    }
 
    if (l == 1)
        ans++;
 
    return (int) ans;
}
 
// Driver Code
public static void main(String[] args)
{
    long l = 12, r = 21;
    System.out.println(countIntegers(l, r));
}
}
 
// This code contributed by Rajput-Ji

输出：

# Python3 implementation of the approach
 
# Function to return a^n
def power(a, n):
 
    if n == 0:
        return 1
 
    p = power(a, n // 2)
    p = p * p
 
    if n & 1:
        p = p * a
 
    return p
 
# Function to return count of integers
# that satisfy n % phi(n) = 0
def countIntegers(l, r):
 
    ans, i = 0, 1
    v = power(2, i)
 
    while v <= r:
 
        while v <= r:
 
            if v >= l:
                ans += 1
             
            v = v * 3
 
        i += 1
        v = power(2, i)
     
    if l == 1:
        ans += 1
 
    return ans
 
# Driver Code
if __name__ == "__main__":
 
    l, r = 12, 21
    print(countIntegers(l, r))
     
# This code is contributed
# by Rituraj Jain