不溶物编号
不可分解数是一个数 N ，如果它可以被和整除，也可以被其数字平方的乘积整除。 极少不溶物编号为:
111、11112、1122112、111111111、122121216、11111112112……
检查一个号码是否为不可溶号码
给定一个数字 N ，任务是检查 N 是否为曝光量数字。如果 N 是不溶物号，则打印“是”否则打印“否”。 示例:
输入: N = 1122112 输出: Yes 解释: 1122112 是一个不可分解数，因为 其位数的平方和 1^2+1^2+2^2+2^2+1^2+1^2+2^2 = 16 其位数的平方和(1122112)^2 = 64 和 1122112 可同时被 16 和 66 整除 输入: N = 11 输出:否 T21【解释:
方法:不溶物数是一个数 N 如果它能被其数字的平方和和乘积整除。所以我们会找到 N 位数的平方和，N 位数的平方和的乘积，然后检查 N 是否能被和和乘积整除。如果可分，则打印“是”否则打印“否”。 以下是上述方法的实施:
C++
// C++ implementation for the
// above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check if a number
// is an Insolite numbers
bool isInsolite(int n)
{
int N = n;
// To store sum of squares of digits
int sum = 0;
// To store product of
// squares of digits
int product = 1;
while (n != 0) {
// extracting digit
int r = n % 10;
sum = sum + r * r;
product = product * r * r;
n = n / 10;
}
return (N % sum == 0)
&& (N % product == 0);
}
// Driver Code
int main()
{
int N = 111;
// Function Call
if (isInsolite(N))
cout << "Yes";
else
cout << "No";
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation for the
// above approach
class GFG{
// Function to check if a number
// is an Insolite numbers
static boolean isInsolite(int n)
{
int N = n;
// To store sum of squares of digits
int sum = 0;
// To store product of
// squares of digits
int product = 1;
while (n != 0)
{
// extracting digit
int r = n % 10;
sum = sum + r * r;
product = product * r * r;
n = n / 10;
}
return (N % sum == 0) &&
(N % product == 0);
}
// Driver Code
public static void main (String[] args)
{
int N = 111;
// Function Call
if (isInsolite(N))
System.out.print("Yes");
else
System.out.print("No");
}
}
// This code is contributed by rock_cool
Python 3
# Python3 implementation for the
# above approach
# Function to check if a number
# is an Insolite numbers
def isInsolite(n):
N = n;
# To store sum of squares of digits
sum = 0;
# To store product of
# squares of digits
product = 1;
while (n != 0):
# extracting digit
r = n % 10;
sum = sum + r * r;
product = product * r * r;
n = n // 10;
return ((N % sum == 0) and
(N % product == 0));
# Driver Code
if __name__ == '__main__':
N = 111;
# Function Call
if (isInsolite(N)):
print("Yes");
else:
print("No");
# This code is contributed by 29AjayKumar
C
// C# implementation for the
// above approach
using System;
class GFG{
// Function to check if a number
// is an Insolite numbers
static bool isInsolite(int n)
{
int N = n;
// To store sum of squares of digits
int sum = 0;
// To store product of
// squares of digits
int product = 1;
while (n != 0)
{
// extracting digit
int r = n % 10;
sum = sum + r * r;
product = product * r * r;
n = n / 10;
}
return (N % sum == 0) &&
(N % product == 0);
}
// Driver Code
public static void Main()
{
int N = 111;
// Function Call
if (isInsolite(N))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by Code_Mech
java 描述语言
<script>
// Javascript implementation for the
// above approach
// Function to check if a number
// is an Insolite numbers
function isInsolite( n) {
let N = n;
// To store sum of squares of digits
let sum = 0;
// To store product of
// squares of digits
let product = 1;
while (n != 0) {
// extracting digit
let r = n % 10;
sum = sum + r * r;
product = product * r * r;
n = parseInt(n / 10);
}
return (N % sum == 0) && (N % product == 0);
}
// Driver Code
let N = 111;
// Function Call
if (isInsolite(N))
document.write("Yes");
else
document.write("No");
// This code contributed by Rajput-Ji
</script>
Output:
Yes
时间复杂度:O(n) T5】参考:http://www.numbersaplenty.com/set/insolite_number/
版权属于：月萌API www.moonapi.com，转载请注明出处
