完美数字不变量数
原文:https://www . geesforgeks . org/perfect-digital-不变量-number/
如果一个正整数的数字的某个固定幂的和等于数字本身，则称之为完美数字不变数。 对于任意数， abcd… =幂(a，n) +幂(b，n) +幂(c，n) +幂(d，n) + …。、 其中 n 可以是大于 0 的任意整数。
检查 N 是否是完美数字不变数
给定一个数字 N ，任务是检查给定的数字 N 是否为完美数字不变数。如果 N 是完美数字不变数，则打印“是”否则打印“否”。 例:
输入: N = 153 输出:是 解释: 153 是一个完美的数字不变量数至于 n = 3 我们有 13+53+33= 153 输入: 4150 输出:是
逼近:对于数字 N 中的每一位数字，从固定的数字 1 开始计算其数位幂的和，直到 N 的数位幂之和超过 N 。如果 N 是完美数字不变数，则打印“是”否则打印“否”。
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to calculate x raised
// to the power y
int power(int x, unsigned int y)
{
if (y == 0) {
return 1;
}
if (y % 2 == 0) {
return (power(x, y / 2)
* power(x, y / 2));
}
return (x
* power(x, y / 2)
* power(x, y / 2));
}
// Function to check whether the given
// number is Perfect Digital Invariant
// number or not
bool isPerfectDigitalInvariant(int x)
{
for (int fixed_power = 1;; fixed_power++) {
int temp = x, sum = 0;
// For each digit in temp
while (temp) {
int r = temp % 10;
sum += power(r, fixed_power);
temp = temp / 10;
}
// If satisfies Perfect Digital
// Invariant condition
if (sum == x) {
return true;
}
// If sum exceeds n, then not possible
if (sum > x) {
return false;
}
}
}
// Driver Code
int main()
{
// Given Number N
int N = 4150;
// Function Call
if (isPerfectDigitalInvariant(N))
cout << "Yes";
else
cout << "No";
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.util.*;
class GFG{
// Function to calculate x raised
// to the power y
static int power(int x, int y)
{
if (y == 0)
{
return 1;
}
if (y % 2 == 0)
{
return (power(x, y / 2) *
power(x, y / 2));
}
return (x * power(x, y / 2) *
power(x, y / 2));
}
// Function to check whether the given
// number is Perfect Digital Invariant
// number or not
static boolean isPerfectDigitalInvariant(int x)
{
for (int fixed_power = 1;; fixed_power++)
{
int temp = x, sum = 0;
// For each digit in temp
while (temp > 0)
{
int r = temp % 10;
sum += power(r, fixed_power);
temp = temp / 10;
}
// If satisfies Perfect Digital
// Invariant condition
if (sum == x)
{
return true;
}
// If sum exceeds n, then not possible
if (sum > x)
{
return false;
}
}
}
// Driver Code
public static void main(String[] args)
{
// Given Number N
int N = 4150;
// Function Call
if (isPerfectDigitalInvariant(N))
System.out.print("Yes");
else
System.out.print("No");
}
}
// This code is contributed by gauravrajput1
Python 3
# Python3 implementation of the above approach
# Function to find the
# sum of divisors
def power(x, y):
if (y == 0):
return 1
if (y % 2 == 0):
return (power(x, y//2) * power(x, y//2))
return (x * power(x, y//2) * power(x, y//2))
# Function to check whether the given
# number is Perfect Digital Invariant
# number or not
def isPerfectDigitalInvariant(x):
fixed_power = 0
while True:
fixed_power += 1
temp = x
summ = 0
# For each digit in temp
while (temp):
r = temp % 10
summ = summ + power(r, fixed_power)
temp = temp//10
# If satisfies Perfect Digital
# Invariant condition
if (summ == x):
return (True)
# If sum exceeds n, then not possible
if (summ > x):
return (False)
# Driver code
# Given Number N
N = 4150
# Function Call
if (isPerfectDigitalInvariant(N)):
print("Yes")
else:
print("No")
# This code is contributed by vikas_g
C
// C# program for the above approach
using System;
class GFG{
// Function to calculate x raised
// to the power y
static int power(int x, int y)
{
if (y == 0)
{
return 1;
}
if (y % 2 == 0)
{
return (power(x, y / 2) *
power(x, y / 2));
}
return (x * power(x, y / 2) *
power(x, y / 2));
}
// Function to check whether the given
// number is Perfect Digital Invariant
// number or not
static bool isPerfectDigitalInvariant(int x)
{
for(int fixed_power = 1;; fixed_power++)
{
int temp = x, sum = 0;
// For each digit in temp
while (temp > 0)
{
int r = temp % 10;
sum += power(r, fixed_power);
temp = temp / 10;
}
// If satisfies Perfect Digital
// Invariant condition
if (sum == x)
{
return true;
}
// If sum exceeds n, then not possible
if (sum > x)
{
return false;
}
}
}
// Driver Code
public static void Main(String[] args)
{
// Given number N
int N = 4150;
// Function call
if (isPerfectDigitalInvariant(N))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by PrinciRaj1992
java 描述语言
<script>
// Javascript implementation
// Function to calculate x raised
// to the power y
function power(x, y)
{
if (y == 0) {
return 1;
}
if (y % 2 == 0) {
return (power(x, Math.floor(y / 2))
* power(x, Math.floor(y / 2)));
}
return (x
* power(x, Math.floor(y / 2))
* power(x, Math.floor(y / 2)));
}
// Function to check whether the given
// number is Perfect Digital Invariant
// number or not
function isPerfectDigitalInvariant(x)
{
for (var fixed_power = 1;; fixed_power++) {
var temp = x, sum = 0;
// For each digit in temp
while (temp) {
var r = temp % 10;
sum += power(r, fixed_power);
temp = Math.floor(temp / 10);
}
// If satisfies Perfect Digital
// Invariant condition
if (sum == x) {
return true;
}
// If sum exceeds n, then not possible
if (sum > x) {
return false;
}
}
}
// Driver Code
// Given Number N
var N = 4150;
// Function Call
if (isPerfectDigitalInvariant(N))
document.write("Yes");
else
document.write("No");
// This code is contributed by shubhamsingh10
</script>
Output:
Yes
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