如果由divisorsSum(n)表示的数字的所有除数的总和小于数字n的值的两倍，则将数字n称为不足数。这两个值之间的差异称为不足。
从数学上讲，如果满足以下条件，则该数字称为“不足”：
divisorsSum(n) < 2 * n
deficiency = (2 * n) - divisorsSum(n)前几个不足编号是：
1、2、3、4、5、7、8、9、10、11、13、14、15、16、17、19…..
给定数字n，我们的任务是查找该数字是否为不足数字。
例子 ：
Input: 21
Output: YES
Divisors are 1, 3, 7 and 21. Sum of divisors is 32.
This sum is less than 2*21 or 42.
Input: 12
Output: NO
Input: 17
Output: YES一个简单的解决方案是将所有数字从1迭代到n，然后检查数字是否除以n并计算总和。检查此总和是否小于2 * n。
这种方法的时间复杂度：O(n)
优化的解决方案：
如果我们仔细观察，数字n的除数是成对出现的。例如，如果n = 100，则所有对数除数分别为：(1，100)，(2，50)，(4，25)，(5，20)，(10，10)
利用这一事实，我们可以加快程序速度。
在检查除数时，我们必须注意是否存在两个相等的除数，例如(10，10)。在这种情况下，我们在计算总和时将只采用其中之一。
实施优化方法
C++
// C++ program to implement an Optimized Solution
// to check Deficient Number
#include
using namespace std;
// Function to calculate sum of divisors
int divisorsSum(int n)
{
int sum = 0; // Initialize sum of prime factors
// 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);
}
}
}
return sum;
}
// Function to check Deficient Number
bool isDeficient(int n)
{
// Check if sum(n) < 2 * n
return (divisorsSum(n) < (2 * n));
}
/* Driver program to test above function */
int main()
{
isDeficient(12) ? cout << "YES\n" : cout << "NO\n";
isDeficient(15) ? cout << "YES\n" : cout << "NO\n";
return 0;
} Java
// Java program to check Deficient Number
import java.io.*;
class GFG {
// Function to calculate sum of divisors
static int divisorsSum(int n)
{
int sum = 0; // Initialize sum of prime factors
// 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);
}
}
}
return sum;
}
// Function to check Deficient Number
static boolean isDeficient(int n)
{
// Check if sum(n) < 2 * n
return (divisorsSum(n) < (2 * n));
}
/* Driver program to test above function */
public static void main(String args[])
{
if (isDeficient(12))
System.out.println("YES");
else
System.out.println("NO");
if (isDeficient(15))
System.out.println("YES");
else
System.out.println("NO");
}
}
// This code is contributed by Nikita TiwariPython
# Python program to implement an Optimized
# Solution to check Deficient Number
import math
# Function to calculate sum of divisors
def divisorsSum(n) :
sum = 0 # Initialize sum of prime factors
# Note that this loop runs till square
# root of n
i = 1
while i<= math.sqrt(n) :
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)
i = i + 1
return sum
# Function to check Deficient Number
def isDeficient(n) :
# Check if sum(n) < 2 * n
return (divisorsSum(n) < (2 * n))
# Driver program to test above function
if ( isDeficient(12) ):
print "YES"
else :
print "NO"
if ( isDeficient(15) ) :
print "YES"
else :
print "NO"
# This Code is contributed by Nikita TiwariC#
// C# program to implement an Optimized Solution
// to check Deficient Number
using System;
class GFG {
// Function to calculate sum of
// divisors
static int divisorsSum(int n)
{
// Initialize sum of prime factors
int sum = 0;
// 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);
}
}
}
return sum;
}
// Function to check Deficient Number
static bool isDeficient(int n)
{
// Check if sum(n) < 2 * n
return (divisorsSum(n) < (2 * n));
}
/* Driver program to test above function */
public static void Main()
{
string var = isDeficient(12) ? "YES" : "NO";
Console.WriteLine(var);
string var1 = isDeficient(15) ? "YES" : "NO";
Console.WriteLine(var1);
}
}
// This code is contributed by vt_mPHP
Javascript
输出 ：
NO
YES