给定一个由N个自然数组成的数组arr [] ，任务是从不包含任何不足数的数组中找到最长子序列的长度。
例子：
Input: arr[] = {13, 55, 240, 32, 24, 27, 56, 80, 100, 330, 89}
Output: 6
Explanation:
Array elements which are not deficient numbers are {240, 24, 56, 80, 100, 330}
Divisors of 13 are {1, 13}. Therefore, sum of divisors = 14, which is less than 26 ( = 2 * 13)
Divisors of 55 are {1, 5, 11, 55}. Therefore, sum of divisors = 72, which is less than 110 ( = 2 * 55).
Divisors of 32 are {1, 2, 4, 8, 16, 32}. Therefore, sum of divisors = 63, which is less than 64 ( = 2 * 32).
Divisors of 27 are {1, 3, 9, 27}. Therefore, sum of divisors = 40, which is less than 54 ( = 2 * 27).
Divisors of 89 are {1, 89}. Therefore, sum of divisors = 90, which is less than 178 ( = 2 * 89).
Therefore, the required count is 6.
Input: arr[] = {1, 2, 3, 4, 5, 6}
Output: 6
Explanation: Array elements which are non-deficient numbers are {1, 2, 3, 4, 5, 6}
方法：解决问题的想法是简单地计算数组中存在的不足数字的数量。所有剩余的数组元素的计数将是所需的最长子序列的长度。请按照以下步骤解决问题：
- 初始化一个变量，例如res ，以存储非不足的数组元素的数量。
- 遍历数组arr []。对于每个数组元素，请检查其是否为不足的数字。
- 计算当前数组元素所有除数的总和，如果总和小于2 * arr [i] ，则返回true 。否则，返回false 。
- 如果发现数组元素为假，则增加res 。
- 完成上述步骤后，打印 res的值作为必需的答案。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to check if n is
// a deficient number or not
bool isNonDeficient(int n)
{
// Stores sum of divisors
int sum = 0;
// Iterate over the range [1, sqrt(N)]
for (int i = 1; i <= sqrt(n); i++) {
// If n is divisible by i
if (n % i == 0) {
// If divisors are equal,
// add only one of them
if (n / i == i) {
sum = sum + i;
}
// Otherwise add both
else {
sum = sum + i;
sum = sum + (n / i);
}
}
}
return sum >= 2 * n;
}
// Function to print the longest
// subsequence which does not
// contain any deficient numbers
int LongestNonDeficientSubsequence(int arr[], int n)
{
// Stores the count of array
// elements which are non-deficient
int res = 0;
// Traverse the array
for (int i = 0; i < n; i++) {
// If element is non-deficient
if (isNonDeficient(arr[i])) {
res += 1;
}
}
// Return the answer
return res;
}
// Driver Code
int main()
{
int arr[]
= { 13, 55, 240, 32, 24, 27,
56, 80, 100, 330, 89 };
int N = sizeof(arr) / sizeof(arr[0]);
cout << LongestNonDeficientSubsequence(arr, N);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to check if n is
// a deficient number or not
static boolean isNonDeficient(int n)
{
// Stores sum of divisors
int sum = 0;
// Iterate over the range [1, sqrt(N)]
for(int i = 1; i <= Math.sqrt(n); i++)
{
// If n is divisible by i
if (n % i == 0)
{
// If divisors are equal,
// add only one of them
if (n / i == i)
{
sum = sum + i;
}
// Otherwise add both
else
{
sum = sum + i;
sum = sum + (n / i);
}
}
}
return sum >= 2 * n;
}
// Function to print the longest
// subsequence which does not
// contain any deficient numbers
static int LongestNonDeficientSubsequence(int arr[],
int n)
{
// Stores the count of array
// elements which are non-deficient
int res = 0;
// Traverse the array
for(int i = 0; i < n; i++)
{
// If element is non-deficient
if (isNonDeficient(arr[i]))
{
res += 1;
}
}
// Return the answer
return res;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 13, 55, 240, 32, 24, 27,
56, 80, 100, 330, 89 };
int N = arr.length;
System.out.print(
LongestNonDeficientSubsequence(arr, N));
}
}
// This code is contributed by splevel62Python3
# Python3 program for the above approach
import math
# Function to check if n is
# a deficient number or not
def isNonDeficient(n):
# Stores sum of divisors
sum = 0
# Iterate over the range [1, sqrt(N)]
for i in range(1, int(math.sqrt(n)) + 1):
# If n is divisible by i
if (n % i == 0):
# If divisors are equal,
# add only one of them
if (n // i == i):
sum = sum + i
# Otherwise add both
else:
sum = sum + i
sum = sum + (n // i)
return sum >= 2 * n
# Function to print the longest
# subsequence which does not
# contain any deficient numbers
def LongestNonDeficientSubsequence(arr, n):
# Stores the count of array
# elements which are non-deficient
res = 0
# Traverse the array
for i in range(n):
# If element is non-deficient
if (isNonDeficient(arr[i])):
res += 1
# Return the answer
return res
# Driver Code
if __name__ == "__main__":
arr = [ 13, 55, 240, 32, 24, 27,
56, 80, 100, 330, 89 ]
N = len(arr)
print(LongestNonDeficientSubsequence(arr, N))
# This code is contributed by chitranayalC#
// C# program for above approach
using System;
public class GFG
{
// Function to check if n is
// a deficient number or not
static bool isNonDeficient(int n)
{
// Stores sum of divisors
int sum = 0;
// Iterate over the range [1, sqrt(N)]
for(int i = 1; i <= Math.Sqrt(n); i++)
{
// If n is divisible by i
if (n % i == 0)
{
// If divisors are equal,
// add only one of them
if (n / i == i)
{
sum = sum + i;
}
// Otherwise add both
else
{
sum = sum + i;
sum = sum + (n / i);
}
}
}
return sum >= 2 * n;
}
// Function to print the longest
// subsequence which does not
// contain any deficient numbers
static int LongestNonDeficientSubsequence(int[] arr,
int n)
{
// Stores the count of array
// elements which are non-deficient
int res = 0;
// Traverse the array
for(int i = 0; i < n; i++)
{
// If element is non-deficient
if (isNonDeficient(arr[i]))
{
res += 1;
}
}
// Return the answer
return res;
}
// Driver code
public static void Main(String[] args)
{
int[] arr = { 13, 55, 240, 32, 24, 27,
56, 80, 100, 330, 89 };
int N = arr.Length;
Console.WriteLine(
LongestNonDeficientSubsequence(arr, N));
}
}
// This code is contributed by splevel62.输出：
6时间复杂度： O(N 3/2 )
辅助空间： O(N)