给定数字根范围内的数字

给定整数K和一系列连续数字[L，R] 。任务是计算给定范围内数字根为K(1≤K≤9)的数字。数字根是一个数字的总和，直到成为单个数字为止。例如，256-> 2 + 5 + 6 = 13-> 1 + 3 = 4。

例子：

Input: L = 10, R = 22, K = 3 
Output: 2 
12 and 21 are the only numbers from the range whose digit sum is 3.

Input: L = 100, R = 200, K = 5 
Output: 11 
 

方法：

• 首先要注意的是，对于任何数字，数字总和等于数字％9。如果余数为0，则数字总和为9。
• 因此，如果K = 9，则将K替换为0。
• 现在的任务是找到模数等于K的L到R范围内的数字计数。
• 从L(TotalRange / 9)开始，将整个范围分成9个最大可能的组，因为在每个范围内，模数9等于K时将恰好有一个数字。
• 循环从R到R的剩余元素数–剩余元素数，并检查是否有满足条件的元素。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
#define ll long long int
using namespace std;
 
// Function to return the count
// of required numbers
int countNumbers(int L, int R, int K)
{
    if (K == 9)
        K = 0;
 
    // Count of numbers present
    // in given range
    int totalnumbers = R - L + 1;
 
    // Number of groups of 9 elements
    // starting from L
    int factor9 = totalnumbers / 9;
 
    // Left over elements not covered
    // in factor 9
    int rem = totalnumbers % 9;
 
    // One Number in each group of 9
    int ans = factor9;
 
    // To check if any number in rem
    // satisfy the property
    for (int i = R; i > R - rem; i--) {
        int rem1 = i % 9;
        if (rem1 == K)
            ans++;
    }
 
    return ans;
}
 
// Driver code
int main()
{
    int L = 10;
    int R = 22;
    int K = 3;
    cout << countNumbers(L, R, K);
 
    return 0;
}

Java

// Java implementation of the approach
 
class GFG {
 
// Function to return the count
// of required numbers
    static int countNumbers(int L, int R, int K) {
        if (K == 9) {
            K = 0;
        }
 
        // Count of numbers present
        // in given range
        int totalnumbers = R - L + 1;
 
        // Number of groups of 9 elements
        // starting from L
        int factor9 = totalnumbers / 9;
 
        // Left over elements not covered
        // in factor 9
        int rem = totalnumbers % 9;
 
        // One Number in each group of 9
        int ans = factor9;
 
        // To check if any number in rem
        // satisfy the property
        for (int i = R; i > R - rem; i--) {
            int rem1 = i % 9;
            if (rem1 == K) {
                ans++;
            }
        }
 
        return ans;
    }
 
// Driver code
    public static void main(String[] args) {
        int L = 10;
        int R = 22;
        int K = 3;
        System.out.println(countNumbers(L, R, K));
    }
}
/* This code contributed by PrinciRaj1992 */

Python3

# Python3 implementation of the approach
 
# Function to return the count
# of required numbers
def countNumbers(L, R, K):
 
    if (K == 9):
        K = 0
 
    # Count of numbers present
    # in given range
    totalnumbers = R - L + 1
 
    # Number of groups of 9 elements
    # starting from L
    factor9 = totalnumbers // 9
 
    # Left over elements not covered
    # in factor 9
    rem = totalnumbers % 9
 
    # One Number in each group of 9
    ans = factor9
 
    # To check if any number in rem
    # satisfy the property
    for i in range(R, R - rem, -1):
        rem1 = i % 9
        if (rem1 == K):
            ans += 1
     
    return ans
 
# Driver code
L = 10
R = 22
K = 3
print(countNumbers(L, R, K))
 
# This code is contributed
# by mohit kumar

C#

// C# implementation of the approach
using System ;
 
class GFG
{
 
    // Function to return the count
    // of required numbers
    static int countNumbers(int L, int R, int K)
    {
        if (K == 9)
        {
            K = 0;
        }
 
        // Count of numbers present
        // in given range
        int totalnumbers = R - L + 1;
 
        // Number of groups of 9 elements
        // starting from L
        int factor9 = totalnumbers / 9;
 
        // Left over elements not covered
        // in factor 9
        int rem = totalnumbers % 9;
 
        // One Number in each group of 9
        int ans = factor9;
 
        // To check if any number in rem
        // satisfy the property
        for (int i = R; i > R - rem; i--)
        {
            int rem1 = i % 9;
            if (rem1 == K)
            {
                ans++;
            }
        }
 
        return ans;
    }
 
    // Driver code
    public static void Main()
    {
        int L = 10;
        int R = 22;
        int K = 3;
         
        Console.WriteLine(countNumbers(L, R, K));
    }
}
 
/* This code is contributed by Ryuga */

PHP

 $R - $rem; $i--)
    {
        $rem1 = $i % 9;
        if ($rem1 == $K)
            $ans++;
    }
 
    return $ans;
}
 
// Driver code
$L = 10;
$R = 22;
$K = 3;
echo countNumbers($L, $R, $K);
 
// This code is contributed by Ita_c
?>

Javascript

 
// Javascript implementation of the approach
 
// Function to return the count
// of required numbers
function countNumbers(L, R, K)
{
    if (K == 9)
    {
        K = 0;
    }
 
    // Count of numbers present
    // in given range
    var totalnumbers = R - L + 1;
 
    // Number of groups of 9 elements
    // starting from L
    var factor9 = totalnumbers / 9;
 
    // Left over elements not covered
    // in factor 9
    var rem = totalnumbers % 9;
 
    // One Number in each group of 9
    var ans = factor9;
 
    // To check if any number in rem
    // satisfy the property
    for(var i = R; i > R - rem; i--)
    {
        var rem1 = i % 9;
        if (rem1 == K)
        {
            ans++;
        }
    }
 
    return ans;
}
  
// Driver Code
var L = 10;
var R = 22;
var K = 3;
 
document.write(Math.round(countNumbers(L, R, K)));
 
// This code is contributed by Ankita saini
                     

输出：

2