给定一个数字n，我们需要检查n是否是一个有抱负的数字。如果数字n的等分序列以一个完美数字结尾，并且它本身不是一个完美数字，则它被称为有抱负数字。前几个有抱负的数字是：25、95、119、143、417、445、565、608、650、652…。
例子 ：
Input : 25
Output : Yes.
Explanation : Terminating number of
aliquot sequence of 25 is 6 which is
perfect number.
Input : 24
Output : No.
Explanation : Terminating number of
aliquot sequence of 24 is 0 which is
not a perfect number.方法：首先，找到给定输入的等分序列的终止号，然后检查其是否为理想数(根据定义)。下面给出的是检查有抱负数字的实现。
C++
// C++ implementation to check whether
// a number is aspiring or not
#include
using namespace std;
// Function to calculate sum of all proper
// divisors
int getSum(int n)
{
int sum = 0; // 1 is a proper divisor
// Note that this loop runs till square root
// of n
for (int i = 1; i <= sqrt(n); i++) {
if (n % i == 0) {
// If divisors are equal, take only one
// of them
if (n / i == i)
sum = sum + i;
else // Otherwise take both
{
sum = sum + i;
sum = sum + (n / i);
}
}
}
// calculate sum of all proper divisors only
return sum - n;
}
// Function to get last number of Aliquot Sequence.
int getAliquot(int n)
{
unordered_set s;
s.insert(n);
int next = 0;
while (n > 0) {
// Calculate next term from previous term
n = getSum(n);
if (s.find(n) != s.end())
return n;
s.insert(n);
}
return 0;
}
// Returns true if n is perfect
bool isPerfect(int n)
{
// To store sum of divisors
long long int sum = 1;
// Find all divisors and add them
for (long long int i = 2; i * i <= n; i++)
if (n % i == 0)
sum = sum + i + n / i;
// If sum of divisors is equal to
// n, then n is a perfect number
if (sum == n && n != 1)
return true;
return false;
}
// Returns true if n is aspiring
// else returns false
bool isAspiring(int n)
{
// checking condition for aspiring
int alq = getAliquot(n);
if (isPerfect(alq) && !isPerfect(n))
return true;
else
return false;
}
// Driver program
int main()
{
int n = 25;
if (isAspiring(n))
cout << "Aspiring" << endl;
else
cout << "Not Aspiring" << endl;
return 0;
} Java
// Java implementation to check whether
// a number is aspiring or not
import java.util.*;
class GFG
{
// Function to calculate sum of
// all proper divisors
static int getSum(int n)
{
int sum = 0; // 1 is a proper divisor
// Note that this loop runs till
// square root of n
for (int i = 1; i <= Math.sqrt(n); i++)
{
if (n % i == 0)
{
// If divisors are equal,
// take only one of them
if (n / i == i)
{
sum = sum + i;
}
else // Otherwise take both
{
sum = sum + i;
sum = sum + (n / i);
}
}
}
// calculate sum of all
// proper divisors only
return sum - n;
}
// Function to get last number
// of Aliquot Sequence.
static int getAliquot(int n)
{
TreeSet s = new TreeSet();
s.add(n);
int next = 0;
while (n > 0)
{
// Calculate next term from previous term
n = getSum(n);
if (s.contains(n) & n != s.last())
{
return n;
}
s.add(n);
}
return 0;
}
// Returns true if n is perfect
static boolean isPerfect(int n)
{
// To store sum of divisors
int sum = 1;
// Find all divisors and add them
for (int i = 2; i * i <= n; i++)
{
if (n % i == 0)
{
sum = sum + i + n / i;
}
}
// If sum of divisors is equal to
// n, then n is a perfect number
if (sum == n && n != 1)
{
return true;
}
return false;
}
// Returns true if n is aspiring
// else returns false
static boolean isAspiring(int n)
{
// checking condition for aspiring
int alq = getAliquot(n);
if (isPerfect(alq) && !isPerfect(n))
{
return true;
}
else
{
return false;
}
}
// Driver code
public static void main(String[] args)
{
int n = 25;
if (isAspiring(n))
{
System.out.println("Aspiring");
}
else
{
System.out.println("Not Aspiring");
}
}
}
/* This code has been contributed
by PrinciRaj1992*/ Python3
# Python3 implementation to check whether
# a number is aspiring or not
# Function to calculate sum of all proper
# divisors
def getSum(n):
# 1 is a proper divisor
sum = 0
# Note that this loop runs till
# square root of n
for i in range(1, int((n) ** (1 / 2)) + 1):
if not n % i:
# If divisors are equal, take
# only one of them
if n // i == i:
sum += i
# Otherwise take both
else:
sum += i
sum += (n // i)
# Calculate sum of all proper
# divisors only
return sum - n
# Function to get last number
# of Aliquot Sequence.
def getAliquot(n):
s = set()
s.add(n)
next = 0
while (n > 0):
# Calculate next term from
# previous term
n = getSum(n)
if n not in s:
return n
s.add(n)
return 0
# Returns true if n is perfect
def isPerfect(n):
# To store sum of divisors
sum = 1
# Find all divisors and add them
for i in range(2, int((n ** (1 / 2))) + 1):
if not n % i:
sum += (i + n // i)
# If sum of divisors is equal to
# n, then n is a perfect number
if sum == n and n != 1:
return True
return False
# Returns true if n is aspiring
# else returns false
def isAspiring(n):
# Checking condition for aspiring
alq = getAliquot(n)
if (isPerfect(alq) and not isPerfect(n)):
return True
else:
return False
# Driver Code
n = 25
if (isAspiring(n)):
print("Aspiring")
else:
print("Not Aspiring")
# This code is contributed by rohitsingh07052C#
// C# implementation to check whether
// a number is aspiring or not
using System;
using System.Collections.Generic;
class GFG
{
// Function to calculate sum of
// all proper divisors
static int getSum(int n)
{
int sum = 0; // 1 is a proper divisor
// Note that this loop runs till
// square root of n
for (int i = 1; i <= (int)Math.Sqrt(n); i++)
{
if (n % i == 0)
{
// If divisors are equal,
// take only one of them
if (n / i == i)
{
sum = sum + i;
}
else // Otherwise take both
{
sum = sum + i;
sum = sum + (n / i);
}
}
}
// calculate sum of all
// proper divisors only
return sum - n;
}
// Function to get last number
// of Aliquot Sequence.
static int getAliquot(int n)
{
HashSet s = new HashSet();
s.Add(n);
while (n > 0)
{
// Calculate next term from previous term
n = getSum(n);
if (s.Contains(n))
{
return n;
}
s.Add(n);
}
return 0;
}
// Returns true if n is perfect
static bool isPerfect(int n)
{
// To store sum of divisors
int sum = 1;
// Find all divisors and add them
for (int i = 2; i * i <= n; i++)
{
if (n % i == 0)
{
sum = sum + i + n / i;
}
}
// If sum of divisors is equal to
// n, then n is a perfect number
if (sum == n && n != 1)
{
return true;
}
return false;
}
// Returns true if n is aspiring
// else returns false
static bool isAspiring(int n)
{
// checking condition for aspiring
int alq = getAliquot(n);
if (isPerfect(alq) && !isPerfect(n))
{
return true;
}
else
{
return false;
}
}
// Driver code
public static void Main(String[] args)
{
int n = 25;
if (isAspiring(n))
{
Console.WriteLine("Aspiring");
}
else
{
Console.WriteLine("Not Aspiring");
}
}
}
// This code is contributed by subhammahato348 输出：
Aspiring时间复杂度： O(n)