打印所有小于或等于N的强数字

原文:https://www . geesforgeks . org/print-all-strong-numbers-小于或等于 n/

给定一个数字 N ，打印所有小于或等于 N 的强数字。

强数是一个特殊数，其位数的阶乘之和等于原数。
比如:145 是强号。从，1！+ 4!+ 5!= 145.

示例:

输入: N = 100
输出: 1 2
说明:
只有 1 和 2 是 1 到 100 的强号，因为
1！= 1，
2！= 2

输入: N = 1000
输出: 1 2 145
说明:
只有 1、2、145 是 1 到 1000 的强号，因为
1！= 1，
2！= 2，
(1！+ 4!+ 5!) = 145

方法:思路是从【1，N】迭代，检查范围内是否有强数。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     {
        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 <        }
    }
}
// Driver Code
int main()
{
    // Given number
    int N = 1000;
    // Function call
    printStrongNumbers(N);
    return 0;
}
// This code is contributed by rutvik_56


Java 语言(一种计算机语言，尤用于创建网站)

// 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             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);
    }
}


Python 3

# 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     {
        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


java 描述语言

Output: 

1 2 145


时间复杂度: O(N)