给定数字N ,请打印所有小于或等于N的强数。
Strong number is a special number whose sum of the factorial of digits is equal to the original number.
For Example: 145 is strong number. Since, 1! + 4! + 5! = 145.
例子:
Input: N = 100
Output: 1 2
Explanation:
Only 1 and 2 are the strong numbers from 1 to 100 because
1! = 1, and
2! = 2
Input: N = 1000
Output: 1 2 145
Explanation:
Only 1, 2 and 145 are the strong numbers from 1 to 1000 because
1! = 1,
2! = 2, and
(1! + 4! + 5!) = 145
方法:想法是从[1,N]进行迭代,并检查范围之间的任何数字是否为强数。如果是,则打印相应的号码,否则检查下一个号码。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Store the factorial of all the
// digits from [0, 9]
int factorial[] = { 1, 1, 2, 6, 24, 120,
720, 5040, 40320, 362880 };
// Function to return true
// if number is strong or not
bool isStrong(int N)
{
// Converting N to String so that
// can easily access all it's digit
string num = to_string(N);
// sum will store summation of
// factorial of all digits
// of a number N
int sum = 0;
for(int i = 0; i < num.length(); i++)
{
sum += factorial[num[i] - '0'];
}
// Returns true of N is strong number
return sum == N;
}
// Function to print all
// strong number till N
void printStrongNumbers(int N)
{
// Iterating from 1 to N
for(int i = 1; i <= N; i++)
{
// Checking if a number is
// strong then print it
if (isStrong(i))
{
cout << i << " ";
}
}
}
// Driver Code
int main()
{
// Given number
int N = 1000;
// Function call
printStrongNumbers(N);
return 0;
}
// This code is contributed by rutvik_56
Java
// Java program for the above approach
class GFG {
// Store the factorial of all the
// digits from [0, 9]
static int[] factorial = { 1, 1, 2, 6, 24, 120,
720, 5040, 40320, 362880 };
// Function to return true
// if number is strong or not
public static boolean isStrong(int N)
{
// Converting N to String so that
// can easily access all it's digit
String num = Integer.toString(N);
// sum will store summation of
// factorial of all digits
// of a number N
int sum = 0;
for (int i = 0;
i < num.length(); i++) {
sum += factorial[Integer
.parseInt(num
.charAt(i)
+ "")];
}
// Returns true of N is strong number
return sum == N;
}
// Function to print all
// strong number till N
public static void
printStrongNumbers(int N)
{
// Iterating from 1 to N
for (int i = 1; i <= N; i++) {
// Checking if a number is
// strong then print it
if (isStrong(i)) {
System.out.print(i + " ");
}
}
}
// Driver Code
public static void
main(String[] args)
throws java.lang.Exception
{
// Given Number
int N = 1000;
// Function Call
printStrongNumbers(N);
}
}
Python3
# Python3 program for the
# above approach
# Store the factorial of
# all the digits from [0, 9]
factorial = [1, 1, 2, 6, 24, 120,
720, 5040, 40320, 362880]
# Function to return true
# if number is strong or not
def isStrong(N):
# Converting N to String
# so that can easily access
# all it's digit
num = str(N)
# sum will store summation
# of factorial of all
# digits of a number N
sum = 0
for i in range (len(num)):
sum += factorial[ord(num[i]) -
ord('0')]
# Returns true of N
# is strong number
if sum == N:
return True
else:
return False
# Function to print all
# strong number till N
def printStrongNumbers(N):
# Iterating from 1 to N
for i in range (1, N + 1):
# Checking if a number is
# strong then print it
if (isStrong(i)):
print (i, end = " ")
# Driver Code
if __name__ == "__main__":
# Given number
N = 1000
# Function call
printStrongNumbers(N)
# This code is contributed by Chitranayal
C#
// C# program for the above approach
using System;
class GFG{
// Store the factorial of all the
// digits from [0, 9]
static int[] factorial = { 1, 1, 2, 6, 24, 120,
720, 5040, 40320, 362880 };
// Function to return true
// if number is strong or not
public static bool isStrong(int N)
{
// Converting N to String so that
// can easily access all it's digit
String num = N.ToString();
// sum will store summation of
// factorial of all digits
// of a number N
int sum = 0;
for (int i = 0; i < num.Length; i++)
{
sum += factorial[int.Parse(num[i] + "")];
}
// Returns true of N is strong number
return sum == N;
}
// Function to print all
// strong number till N
public static void printStrongNumbers(int N)
{
// Iterating from 1 to N
for (int i = 1; i <= N; i++)
{
// Checking if a number is
// strong then print it
if (isStrong(i))
{
Console.Write(i + " ");
}
}
}
// Driver Code
public static void Main(String[] args)
{
// Given Number
int N = 1000;
// Function Call
printStrongNumbers(N);
}
}
// This code is contributed by sapnasingh4991
Javascript
输出:
1 2 145
时间复杂度: O(N)