查找数字根为X的N个正数

给定数字X(1 <= X <= 9)和正数N，找到数字根为X的第N个正数。
数字根：正数的数字根是通过对数字进行迭代求和而获得的。在每次迭代中，该数字都用其数字的总和替换，并且当该数字减少为一个数字时，迭代将停止。这个单位数字称为数字根。例如，数字根数65是2，因为6 + 5 = 11和1 +1 = 2。

例子：

Input : X = 3, N = 100 
Output : 894 
The N-th Number whose digit root is X is 894

Input : X = 7, N = 43 
Output : 385

简单方法：简单的方法是从1开始计算每个数字的数字根，每当遇到一个数字根等于X的数字时，我们将计数器加1。当计数器为1时，我们将停止搜索。等于N

下面是上述方法的实现：

C++

// C++ program to find the N-th number whose
// digital root is X
 
#include 
using namespace std;
 
// Function to find the digital root of
// a number
int findDigitalRoot(int num)
{
    int sum = INT_MAX, tempNum = num;
 
    while (sum >= 10) {
        sum = 0;
 
        while (tempNum > 0) {
            sum += tempNum % 10;
            tempNum /= 10;
        }
 
        tempNum = sum;
    }
 
    return sum;
}
 
// Function to find the Nth number with
// digital root as X
void findAnswer(int X, int N)
{
    // Counter variable to keep the
    // count of valid numbers
    int counter = 0;
 
    for (int i = 1; counter < N; ++i) {
 
        // Find digital root
        int digitalRoot = findDigitalRoot(i);
 
        // Check if is required answer or not
        if (digitalRoot == X) {
            ++counter;
        }
 
        // Print the answer if you have found it
        // and breakout of the loop
        if (counter == N) {
            cout << i;
 
            break;
        }
    }
}
 
// Driver Code
int main()
{
    int X = 1, N = 3;
 
    findAnswer(X, N);
 
    return 0;
}

Java

// Java program to find the N-th number whose
// digital root is X
class GFG
{
 
// Function to find the digital root of
// a number
static int findDigitalRoot(int num)
{
    int sum = Integer.MAX_VALUE, tempNum = num;
 
    while (sum >= 10)
    {
        sum = 0;
 
        while (tempNum > 0)
        {
            sum += tempNum % 10;
            tempNum /= 10;
        }
 
        tempNum = sum;
    }
 
    return sum;
}
 
// Function to find the Nth number with
// digital root as X
static void findAnswer(int X, int N)
{
    // Counter variable to keep the
    // count of valid numbers
    int counter = 0;
 
    for (int i = 1; counter < N; ++i)
    {
 
        // Find digital root
        int digitalRoot = findDigitalRoot(i);
 
        // Check if is required answer or not
        if (digitalRoot == X)
        {
            ++counter;
        }
 
        // Print the answer if you have found it
        // and breakout of the loop
        if (counter == N)
        {
            System.out.print( i);
 
            break;
        }
    }
}
 
// Driver Code
public static void main(String args[])
{
    int X = 1, N = 3;
 
    findAnswer(X, N);
}
}
 
// This code is contributed by Arnab Kundu

Python3

# Python3 program to find the N-th number whose
# digital root is X
import sys
 
# Function to find the digital root of
# a number
def findDigitalRoot(num):
    sum = sys.maxsize;
    tempNum = num;
 
    while (sum >= 10):
        sum = 0;
 
        while (tempNum > 0):
            sum += tempNum % 10;
            tempNum //= 10;
 
        tempNum = sum;
 
    return sum;
 
# Function to find the Nth number with
# digital root as X
def findAnswer(X, N):
     
    # Counter variable to keep the
    # count of valid numbers
    counter = 0;
    i = 0;
    while (counter < N):
        i += 1;
         
        # Find digital root
        digitalRoot = findDigitalRoot(i);
         
        # Check if is required answer or not
        if (digitalRoot == X):
            counter += 1;
         
        # Print the answer if you have found it
        # and breakout of the loop
        if (counter == N):
            print(i);
            break;
 
# Driver Code
if __name__ == '__main__':
    X = 1;
    N = 3;
 
    findAnswer(X, N);
 
# This code is contributed by 29AjayKumar

C#

// C# program to find the N-th number whose
// digital root is X
using System;
 
class GFG
{
 
// Function to find the digital root of
// a number
static int findDigitalRoot(int num)
{
    int sum = int.MaxValue, tempNum = num;
 
    while (sum >= 10)
    {
        sum = 0;
 
        while (tempNum > 0)
        {
            sum += tempNum % 10;
            tempNum /= 10;
        }
 
        tempNum = sum;
    }
 
    return sum;
}
 
// Function to find the Nth number with
// digital root as X
static void findAnswer(int X, int N)
{
    // Counter variable to keep the
    // count of valid numbers
    int counter = 0;
 
    for (int i = 1; counter < N; ++i)
    {
 
        // Find digital root
        int digitalRoot = findDigitalRoot(i);
 
        // Check if is required answer or not
        if (digitalRoot == X)
        {
            ++counter;
        }
 
        // Print the answer if you have found it
        // and breakout of the loop
        if (counter == N)
        {
            Console.Write( i);
 
            break;
        }
    }
}
 
// Driver Code
public static void Main(String []args)
{
    int X = 1, N = 3;
 
    findAnswer(X, N);
}
}
 
// This code has been contributed by 29AjayKumar

PHP

= 10)
    {
        $sum = 0;
 
        while ($tempNum > 0)
        {
            $sum += $tempNum % 10;
            $tempNum /= 10;
        }
 
        $tempNum = $sum;
    }
 
    return $sum;
}
 
// Function to find the Nth number 
// with digital root as X
function findAnswer($X, $N)
{
     
    // Counter variable to keep the
    // count of valid numbers
    $counter = 0;
 
    for ($i = 1; $counter < $N; ++$i)
    {
 
        // Find digital root
        $digitalRoot = findDigitalRoot($i);
 
        // Check if is required answer or not
        if ($digitalRoot == $X)
        {
            ++$counter;
        }
 
        // Print the answer if you have found
        // it and breakout of the loop
        if ($counter == $N)
        {
            echo( $i);
 
            break;
        }
    }
}
 
// Driver Code
$X = 1; $N = 3;
 
findAnswer($X, $N);
 
// This code is contributed by Code_Mech.

输出：

19

高效方法：我们可以使用以下公式直接找到数字K的数字根：

digitalRoot(k) = (k - 1)mod 9 +1

由此我们可以找到数字根为K的第N个数字，

Nth number = (N - 1)*9 + K

下面是上述方法的实现：

C++

// C++ program to find the N-th number with
// digital root as X
 
#include 
using namespace std;
 
// Function to find the N-th number with
// digital root as X
int findAnswer(int X, int N)
{
    return (N - 1) * 9 + X;
}
 
// Driver Code
int main()
{
    int X = 7, N = 43;
 
    cout << findAnswer(X, N);
 
    return 0;
}

Java

// Java program to find the N-th number with
// digital root as X
 
class GfG
{
 
    // Function to find the N-th number with
    // digital root as X
    static int findAnswer(int X, int N)
    {
        return (N - 1) * 9 + X;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int X = 7, N = 43;
        System.out.println(findAnswer(X, N));
    }
}
 
// This code contributed by Rajput-Ji

Python3

# Python3 program to find the N-th 
# number with digital root as X
 
# Function to find the N-th number
# with digital root as X
def findAnswer(X, N):
 
    return (N - 1) * 9 + X;
 
# Driver Code
X = 7;
N = 43;
print(findAnswer(X, N));
 
# This code is contributed by mits

C＃

// C# program to find the N-th number
// with digital root as X
using System;
 
class GFG
{
 
// Function to find the N-th
// number with digital root as X
static int findAnswer(int X, int N)
{
    return (N - 1) * 9 + X;
}
 
// Driver Code
public static void Main()
{
    int X = 7, N = 43;
    Console.WriteLine(findAnswer(X, N));
}
}
 
// This code contributed by Ryuga

Java脚本

 
// Javascript program to find the N-th number
// with digital root as X
 
// Function to find the N-th number
// with digital root as X
function findAnswer(X, N)
{
    return (N - 1) * 9 + X;
}
 
// Driver Code
let X = 7;
let N = 43;
 
document.write(findAnswer(X, N));
 
// This code is contributed by mohan1240760
 

输出：

385