n 位非减数的总数

如果每个数字(第一个除外)都大于或等于前一个数字，则该数字是非递减的。例如，223、4455567、899 是非递减的数字。
因此，给定位数 n，您需要找到具有 n 位数的总非减少数字的计数。
例子：

Input:  n = 1
Output: count  = 10

Input:  n = 2
Output: count  = 55

Input:  n = 3
Output: count  = 220

我们强烈建议您将浏览器最小化，然后自己先尝试一下。
看待问题的一种方法是，数字计数等于以 9 结尾的 n 位数字加上以数字 8 结尾的计数加上 7 的计数，依此类推。如何获得以特定数字结尾的计数?我们可以重复 n-1 个长度和小于或等于最后一位的数字。所以下面是递归公式。

Count of n digit numbers = (Count of (n-1) digit numbers Ending with digit 9) +
                           (Count of (n-1) digit numbers Ending with digit 8) +
                           .............................................+ 
                           .............................................+
                           (Count of (n-1) digit numbers Ending with digit 0) 

让以数字 ‘d’ 和长度 n 结尾的计数为 count(n, d)

count(n, d) = ∑(count(n-1, i)) where i varies from 0 to d

Total count = ∑count(n-1, d) where d varies from 0 to n-1

上述递归解决方案将有许多重叠的子问题。因此，我们可以使用动态规划以自底向上的方式构建表格。
以下是上述想法的实现：

C++

// C++ program to count non-decreasing number with n digits
#include
using namespace std;
 
long long int countNonDecreasing(int n)
{
    // dp[i][j] contains total count of non decreasing
    // numbers ending with digit i and of length j
    long long int dp[10][n+1];
    memset(dp, 0, sizeof dp);
 
    // Fill table for non decreasing numbers of length 1
    // Base cases 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
    for (int i = 0; i < 10; i++)
        dp[i][1] = 1;
 
    // Fill the table in bottom-up manner
    for (int digit = 0; digit <= 9; digit++)
    {
        // Compute total numbers of non decreasing
        // numbers of length 'len'
        for (int len = 2; len <= n; len++)
        {
            // sum of all numbers of length of len-1
            // in which last digit x is <= 'digit'
            for (int x = 0; x <= digit; x++)
                dp[digit][len] += dp[x][len-1];
        }
    }
 
    long long int count = 0;
 
    // There total nondecreasing numbers of length n
    // wiint be dp[0][n] +  dp[1][n] ..+ dp[9][n]
    for (int i = 0; i < 10; i++)
        count += dp[i][n];
 
    return count;
}
 
// Driver program
int main()
{
    int n = 3;
    cout << countNonDecreasing(n);
    return 0;
}

Java

class NDN
{
    static int countNonDecreasing(int n)
    {
        // dp[i][j] contains total count of non decreasing
        // numbers ending with digit i and of length j
        int dp[][] = new int[10][n+1];
      
        // Fill table for non decreasing numbers of length 1
        // Base cases 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
        for (int i = 0; i < 10; i++)
            dp[i][1] = 1;
      
        // Fill the table in bottom-up manner
        for (int digit = 0; digit <= 9; digit++)
        {
            // Compute total numbers of non decreasing
            // numbers of length 'len'
            for (int len = 2; len <= n; len++)
            {
                // sum of all numbers of length of len-1
                // in which last digit x is <= 'digit'
                for (int x = 0; x <= digit; x++)
                    dp[digit][len] += dp[x][len-1];
            }
        }
      
        int count = 0;
      
        // There total nondecreasing numbers of length n
        // wiint be dp[0][n] +  dp[1][n] ..+ dp[9][n]
        for (int i = 0; i < 10; i++)
            count += dp[i][n];
      
        return count;
    }
    public static void main(String args[])
    {
       int n = 3;
       System.out.println(countNonDecreasing(n));
    }
}/* This code is contributed by Rajat Mishra */

Python3

# Python3 program to count
# non-decreasing number with n digits
def countNonDecreasing(n):
     
    # dp[i][j] contains total count
    # of non decreasing numbers ending
    # with digit i and of length j
    dp = [[0 for i in range(n + 1)]
             for i in range(10)]
              
    # Fill table for non decreasing
    # numbers of length 1.
    # Base cases 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
 
    for i in range(10):
        dp[i][1] = 1
 
    # Fill the table in bottom-up manner
    for digit in range(10):
         
        # Compute total numbers of non
        # decreasing numbers of length 'len'
        for len in range(2, n + 1):
             
            # sum of all numbers of length
            # of len-1 in which last
            # digit x is <= 'digit'
            for x in range(digit + 1):
                dp[digit][len] += dp[x][len - 1]
    count = 0
     
    # There total nondecreasing numbers
    # of length n won't be dp[0][n] +
    # dp[1][n] ..+ dp[9][n]
    for i in range(10):
        count += dp[i][n]
    return count
     
# Driver Code
n = 3
print(countNonDecreasing(n))
 
# This code is contributed
# by sahilshelangia

C#

// C# program to print sum
// triangle for a given array
using System;
 
class GFG {
     
    static int countNonDecreasing(int n)
    {
        // dp[i][j] contains total count
        // of non decreasing numbers ending
        // with digit i and of length j
        int [,]dp = new int[10,n + 1];
     
        // Fill table for non decreasing
        // numbers of length 1 Base cases
        // 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
        for (int i = 0; i < 10; i++)
            dp[i, 1] = 1;
     
        // Fill the table in bottom-up manner
        for (int digit = 0; digit <= 9; digit++)
        {
             
            // Compute total numbers of non decreasing
            // numbers of length 'len'
            for (int len = 2; len <= n; len++)
            {
                 
                // sum of all numbers of length of len-1
                // in which last digit x is <= 'digit'
                for (int x = 0; x <= digit; x++)
                    dp[digit, len] += dp[x, len - 1];
            }
        }
     
        int count = 0;
     
        // There total nondecreasing numbers
        // of length n wiint be dp[0][n]
        // + dp[1][n] ..+ dp[9][n]
        for (int i = 0; i < 10; i++)
            count += dp[i, n];
     
        return count;
    }
     
    // Driver code
    public static void Main()
    {
        int n = 3;
        Console.WriteLine(countNonDecreasing(n));
    }
}
 
// This code is contributed by Sam007.

Javascript

     
    function countNonDecreasing(n)
    {
        // dp[i][j] contains total count of non decreasing
        // numbers ending with digit i and of length j
        let dp=new Array(10);
        for(let i=0;i<10;i++)
        {
            dp[i]=new Array(n+1);
        }   
         
         
        for(let i=0;i<10;i++)
        {
            for(let j=0;j<n+1;j++)
            {
                dp[i][j]=0;
            }
        }
         
        // Fill table for non decreasing numbers of length 1
        // Base cases 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
        for (let i = 0; i < 10; i++)
            dp[i][1] = 1;
        
        // Fill the table in bottom-up manner
        for (let digit = 0; digit <= 9; digit++)
        {
            // Compute total numbers of non decreasing
            // numbers of length 'len'
            for (let len = 2; len <= n; len++)
            {
                // sum of all numbers of length of len-1
                // in which last digit x is <= 'digit'
                for (let x = 0; x <= digit; x++)
                    dp[digit][len] += dp[x][len-1];
            }
        }
        
        let count = 0;
        
        // There total nondecreasing numbers of length n
        // wiint be dp[0][n] +  dp[1][n] ..+ dp[9][n]
        for (let i = 0; i < 10; i++)
            count += dp[i][n];
        
        return count;
    }
     
    let n = 3;
    document.write(countNonDecreasing(n));
     
    // This code is contributed by avanitrachhadiya2155
 

输出：

220

感谢 Gaurav Ahirwar 提出上述方法。
另一种方法是基于下面的直接公式

Count of non-decreasing numbers with n digits = 
                                N*(N+1)/2*(N+2)/3* ....*(N+n-1)/n
Where N = 10

以下是使用上述公式计算计数的程序。

C++

// C++ program to count non-decreasing numner with n digits
#include
using namespace std;
 
long long int countNonDecreasing(int n)
{
    int N = 10;
 
    // Compute value of N*(N+1)/2*(N+2)/3* ....*(N+n-1)/n
    long long count = 1;
    for (int i=1; i<=n; i++)
    {
        count *= (N+i-1);
        count /= i;
    }
 
    return count;
}
 
// Driver program
int main()
{
    int n = 3;
    cout << countNonDecreasing(n);
    return 0;
}

Java

// java program to count non-decreasing
// numner with n digits
public class GFG {
     
    static long countNonDecreasing(int n)
    {
        int N = 10;
      
        // Compute value of N * (N+1)/2 *
        // (N+2)/3 * ....* (N+n-1)/n
        long count = 1;
          
        for (int i = 1; i <= n; i++)
        {
            count *= (N + i - 1);
            count /= i;
        }
      
        return count;
    }
 
    // Driver code
    public static void main(String args[]) {
         
        int n = 3;
        System.out.print(countNonDecreasing(n));
    }  
}
 
// This code is contributed by Sam007.

蟒蛇3

# python program to count non-decreasing
# numner with n digits
 
def countNonDecreasing(n):
    N = 10
 
    # Compute value of N*(N+1)/2*(N+2)/3
    # * ....*(N+n-1)/n
    count = 1
    for i in range(1, n+1):
        count = int(count * (N+i-1))
        count = int(count / i )
         
    return count
 
# Driver program
n = 3;
print(countNonDecreasing(n))
     
# This code is contributed by Sam007

C＃

// C# program to count non-decreasing
// numner with n digits
using System;
 
class GFG {
     
    static long countNonDecreasing(int n)
    {
        int N = 10;
     
        // Compute value of N * (N+1)/2 *
        // (N+2)/3 * ....* (N+n-1)/n
        long count = 1;
         
        for (int i = 1; i <= n; i++)
        {
            count *= (N + i - 1);
            count /= i;
        }
     
        return count;
    }
 
     
    public static void Main()
    {
        int n = 3;
         
        Console.WriteLine(countNonDecreasing(n));
    }
}
 
// This code is contributed by Sam007.

Javascript

 
// javascript program to count non-decreasing
// numner with n digits
     
    function countNonDecreasing(n)
    {
        let N = 10;
       
        // Compute value of N * (N+1)/2 *
        // (N+2)/3 * ....* (N+n-1)/n
        let count = 1;
           
        for (let i = 1; i <= n; i++)
        {
            count *= (N + i - 1);
            count = Math.floor(count/ i);
        }
       
        return count;
    }
     
    // Driver code
    let n = 3;
    document.write(countNonDecreasing(n));
     
    //  This code is contributed by rag2127.

输出：

220

感谢 Abhishek Somani 提出这种方法。
这个公式是如何工作的?

N * (N+1)/2 * (N+2)/3 * .... * (N+n-1)/n
Where N = 10 

让我们尝试不同的 n 值。

For n = 1, the value is N from formula.
Which is true as for n = 1, we have all single digit
numbers, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

For n = 2, the value is N(N+1)/2 from formula
We can have N numbers beginning with 0, (N-1) numbers 
beginning with 1, and so on.
So sum is N + (N-1) + .... + 1 = N(N+1)/2

For n = 3, the value is N(N+1)/2(N+2)/3 from formula
We can have N(N+1)/2 numbers beginning with 0, (N-1)N/2 
numbers beginning with 1 (Note that when we begin with 1, 
we have N-1 digits left to consider for remaining places),
(N-2)(N-1)/2 beginning with 2, and so on.
Count = N(N+1)/2 + (N-1)N/2 + (N-2)(N-1)/2 + 
                               (N-3)(N-2)/2 .... 3 + 1 
     [Combining first 2 terms, next 2 terms and so on]
     = 1/2[N2 + (N-2)2 + .... 4]
     = N*(N+1)*(N+2)/6  [Refer this , putting n=N/2 in the 
                         even sum formula]

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