高托宾数k是对方程φ(x)= k有更多解的整数，其中φ是欧拉的托宾函数
序列 ：
1, 2, 4, 8, 12, 24, 48, 72, 144, 240, 432, 480, 576, 720, 1152, 1440, 2880, 4320, 5760, 8640
解释 ：
- 1个有2个解决方案
- 2有3个解决方案
- 4有4个解决方案
- 8有5个解决方案
- 12有6个解决方案
- 24有10个解决方案
对于给定的N ，任务是打印第一个N个高强度数字。
例子：
Input : N = 10
Output : 1, 2, 4, 8, 12, 24, 48, 72, 144, 240
Input : N = 20
Output : 1, 2, 4, 8, 12, 24, 48, 72, 144, 240, 432, 480, 576, 720, 1152, 1440, 2880, 4320, 5760, 8640
方法：
一种有效的方法是将最多10 5的所有φ(x)值及其频率存储在映射中，然后从1开始循环运行，直到高度托宾数的计数小于N为止。对于每一个我，我们会检查我的频率比以前更大的元素，如果是，那么打印数量，并增加了其他数增量的数量。
下面是上述方法的实现：
C++
// CPP program to find highly totient numbers
#include
using namespace std;
// Function to find euler totient number
int phi(int n)
{
int result = n; // Initialize result as n
// Consider all prime factors of n and
// subtract their multiples from result
for (int p = 2; p * p <= n; ++p) {
// Check if p is a prime factor.
if (n % p == 0) {
// If yes, then update n and result
while (n % p == 0)
n /= p;
result -= result / p;
}
}
// If n has a prime factor greater than sqrt(n)
// (There can be at-most one such prime factor)
if (n > 1)
result -= result / n;
return result;
}
// Function to find first n highly totient numbers
void Highly_Totient(int n)
{
// Count of Highly totient numbers
// and value of count of phi of previous numbers
int count = 0, p_count = -1, i = 1;
// Store all the values of phi(x) upto
// 10^5 with frequencies
map mp;
for (int i = 1; i < 100000; i++)
mp[phi(i)]++;
while (count < n) {
// If count is greater than count of
// previous element
if (mp[i] > p_count)
{
// Display the number
cout << i;
if(count < n-1)
cout << ", ";
// Store the value of phi
p_count = mp[i];
count++;
}
i++;
}
}
// Driver code
int main()
{
int n = 20;
// Function call
Highly_Totient(n);
return 0;
} Java
// Java program to find highly totient numbers
import java.util.*;
class GFG
{
// Function to find euler totient number
static int phi(int n)
{
int result = n; // Initialize result as n
// Consider all prime factors of n and
// subtract their multiples from result
for (int p = 2; p * p <= n; ++p)
{
// Check if p is a prime factor.
if (n % p == 0)
{
// If yes, then update n and result
while (n % p == 0)
n /= p;
result -= result / p;
}
}
// If n has a prime factor greater than sqrt(n)
// (There can be at-most one such prime factor)
if (n > 1)
result -= result / n;
return result;
}
// Function to find first n highly totient numbers
static void Highly_Totient(int n)
{
// Count of Highly totient numbers
// and value of count of phi of previous numbers
int count = 0, p_count = -1;
// Store all the values of phi(x) upto
// 10^5 with frequencies
HashMap mp = new HashMap();
for (int i = 1; i < 100000; i++)
{
if(mp.containsKey(phi(i)))
{
mp.put(phi(i), mp.get(phi(i)) + 1);
}
else
{
mp.put(phi(i), 1);
}
}
int i = 1;
while (count < n)
{
// If count is greater than count of
// previous element
if (mp.containsKey(i)&&mp.get(i) > p_count)
{
// Display the number
System.out.print(i);
if(count < n - 1)
System.out.print(", ");
// Store the value of phi
p_count = mp.get(i);
count++;
}
i++;
}
}
// Driver code
public static void main(String[] args)
{
int n = 20;
// Function call
Highly_Totient(n);
}
}
// This code is contributed by Rajput-Ji Python3
# Python3 program to find highly totient numbers
# Function to find euler totient number
def phi(n):
result = n; # Initialize result as n
# Consider all prime factors of n and
# subtract their multiples from result
p = 2
while(p*p <= n):
# Check if p is a prime factor.
if (n % p == 0):
# If yes, then update n and result
while (n % p == 0):
n //= p;
result -= (result // p);
p += 1
# If n has a prime factor greater than sqrt(n)
# (There can be at-most one such prime factor)
if (n > 1):
result -= (result // n);
return result;
# Function to find first n highly totient numbers
def Highly_Totient(n):
# Count of Highly totient numbers
# and value of count of phi of previous numbers
count = 0
p_count = -1
# Store all the values of phi(x) upto
# 10^5 with frequencies
mp = dict()
i = 1
while i < 100000:
tmp = phi(i)
if tmp not in mp:
mp[tmp] = 0
mp[tmp] += 1;
i += 1
i = 1
while (count < n):
# If count is greater than count of
# previous element
if ((i in mp) and mp[i] > p_count):
# Display the number
print(i, end = '');
if(count < n - 1):
print(", ", end = '');
# Store the value of phi
p_count = mp[i];
count += 1
i += 1
# Driver code
if __name__=='__main__':
n = 20;
# Function call
Highly_Totient(n);
# This code is contributed by rutvik_56C#
// C# program to find highly totient numbers
using System;
using System.Collections.Generic;
class GFG
{
// Function to find euler totient number
static int phi(int n)
{
int result = n; // Initialize result as n
// Consider all prime factors of n and
// subtract their multiples from result
for (int p = 2; p * p <= n; ++p)
{
// Check if p is a prime factor.
if (n % p == 0)
{
// If yes, then update n and result
while (n % p == 0)
n /= p;
result -= result / p;
}
}
// If n has a prime factor greater than sqrt(n)
// (There can be at-most one such prime factor)
if (n > 1)
result -= result / n;
return result;
}
// Function to find first n highly totient numbers
static void Highly_Totient(int n)
{
// Count of Highly totient numbers
// and value of count of phi of
// previous numbers
int count = 0, p_count = -1, i;
// Store all the values of phi(x) upto
// 10^5 with frequencies
Dictionary mp = new Dictionary();
for (i = 1; i < 100000; i++)
{
if(mp.ContainsKey(phi(i)))
{
mp[phi(i)] = mp[phi(i)] + 1;
}
else
{
mp.Add(phi(i), 1);
}
}
i = 1;
while (count < n)
{
// If count is greater than count of
// previous element
if (mp.ContainsKey(i)&&mp[i] > p_count)
{
// Display the number
Console.Write(i);
if(count < n - 1)
Console.Write(", ");
// Store the value of phi
p_count = mp[i];
count++;
}
i++;
}
}
// Driver code
public static void Main(String[] args)
{
int n = 20;
// Function call
Highly_Totient(n);
}
}
// This code is contributed by Rajput-Ji 输出：
1, 2, 4, 8, 12, 24, 48, 72, 144, 240, 432, 480, 576, 720, 1152, 1440, 2880, 4320, 5760, 8640
此方法不能用于查找超过1000个“高姿态”号码。