如果超级Niven Number不仅可以被其数字的总和而且可以被其(非零)数字的任何子集的总和整除，则它是数字N。
例如：
68040 is a Super Niven Number because it is divisible by 6, 8, 4, 6+8, 6+4, 4+8 and 6+4+8.
检查N是否为超级数
给定数字N ，任务是检查N是否为超级尼文数。如果N是超级尼文数，则打印“是”，否则打印“否” 。
例子：
Input: N = 68040
Output: Yes
Explanation:
68040 is divisible by 6, 8, 4, 6+8, 6+4, 4+8 and 6+4+8.
and N begins also with ’25’.
Input: N = 72
Output: No
方法：
- 我们将数字N的所有数字存储在数组arr中
- 现在我们将找到数组每个子集的总和，并检查数字N是否可被所有子集整除
- 如果N无法被任何子集整除，则返回false，否则最后返回true。
下面是上述方法的实现：
C++
// C++ implementagtion to check if a number
// is Super Niven Number or not.
#include
using namespace std;
// Checks if sums of all subsets of digits array
// divides the number N
bool isDivBySubsetSums(vector arr, int num)
{
// to calculate length of array arr
int n = arr.size();
// There are totoal 2^n subsets
long long total = 1 << n;
// Consider all numbers from 0 to 2^n - 1
for (long long i = 0; i < total; i++) {
long long sum = 0;
// Consider binary reprsentation of
// current i to decide which elements
// to pick.
for (int j = 0; j < n; j++)
if (i & (1 << j))
sum += arr[j];
// check sum of picked elements.
if (sum != 0 && num % sum != 0)
return false;
}
return true;
}
// Function to check if a number is
// a super-niven number
bool isSuperNivenNum(int n)
{
int temp = n;
// to stor digits of N
vector digits;
while (n != 0) {
int digit = n % 10;
digits.push_back(digit);
n = n / 10;
}
return isDivBySubsetSums(digits, temp);
}
// Driver code
int main()
{
int n = 500;
if (isSuperNivenNum(n))
cout << "yes";
else
cout << "No";
return 0;
} Java
// Java implementagtion to check if a number
// is Super Niven Number or not.
import java.util.*;
class GFG{
// Checks if sums of all subsets of digits array
// divides the number N
static boolean isDivBySubsetSums(Vector arr,
int num)
{
// to calculate length of array arr
int n = arr.size();
// There are totoal 2^n subsets
long total = 1 << n;
// Consider all numbers from 0 to 2^n - 1
for (long i = 0; i < total; i++)
{
long sum = 0;
// Consider binary reprsentation of
// current i to decide which elements
// to pick.
for (int j = 0; j < n; j++)
if ((i & (1 << j)) > 0)
sum += arr.get(j);
// check sum of picked elements.
if (sum != 0 && num % sum != 0)
return false;
}
return true;
}
// Function to check if a number is
// a super-niven number
static boolean isSuperNivenNum(int n)
{
int temp = n;
// to stor digits of N
Vector digits = new Vector();
while (n != 0)
{
int digit = n % 10;
digits.add(digit);
n = n / 10;
}
return isDivBySubsetSums(digits, temp);
}
// Driver code
public static void main(String[] args)
{
int n = 500;
if (isSuperNivenNum(n))
System.out.print("yes");
else
System.out.print("No");
}
}
// This code is contributed by Amit Katiyar Python3
# Python3 implementagtion to check if a
# number is Super Niven Number or not.
# Checks if sums of all subsets of digits
# array divides the number N
def isDivBySubsetSums(arr, num):
# To calculate length of array arr
n = len(arr)
# There are totoal 2^n subsets
total = 1 << n
# Consider all numbers from 0 to 2^n - 1
i = 0
while i < total:
sum = 0
# Consider binary reprsentation of
# current i to decide which elements
# to pick.
j = 0
while j < n:
if (i & (1 << j)):
sum += arr[j]
j += 1
# Check sum of picked elements.
if (sum != 0) and (num % sum != 0):
return False
i += 1
return True
# Function to check if a number is
# a super-niven number
def isSuperNivenNum(n):
temp = n
# To store digits of N
digits = []
while (n > 1):
digit = int(n) % 10
digits.append(digit)
n = n / 10
return isDivBySubsetSums(digits, temp)
# Driver code
if __name__ == '__main__':
n = 500
if isSuperNivenNum(n):
print("Yes")
else:
print("No")
# This code is contributed by jana_sayantanC#
// C# implementagtion to check if a number
// is Super Niven Number or not.
using System;
using System.Collections.Generic;
class GFG{
// Checks if sums of all subsets of digits array
// divides the number N
static bool isDivBySubsetSums(List arr,
int num)
{
// to calculate length of array arr
int n = arr.Count;
// There are totoal 2^n subsets
long total = 1 << n;
// Consider all numbers from 0 to 2^n - 1
for (long i = 0; i < total; i++)
{
long sum = 0;
// Consider binary reprsentation of
// current i to decide which elements
// to pick.
for (int j = 0; j < n; j++)
if ((i & (1 << j)) > 0)
sum += arr[j];
// check sum of picked elements.
if (sum != 0 && num % sum != 0)
return false;
}
return true;
}
// Function to check if a number is
// a super-niven number
static bool isSuperNivenNum(int n)
{
int temp = n;
// to stor digits of N
List digits = new List();
while (n != 0)
{
int digit = n % 10;
digits.Add(digit);
n = n / 10;
}
return isDivBySubsetSums(digits, temp);
}
// Driver code
public static void Main(String[] args)
{
int n = 500;
if (isSuperNivenNum(n))
Console.Write("yes");
else
Console.Write("No");
}
}
// This code is contributed by gauravrajput1 输出：
yes
参考：http://www.numbersaplenty.com/set/super_Niven_number/