数字总和等于给定总和的 n 位数字的计数

给定两个整数 ‘n’ 和 ‘sum’，找出所有 n 位数字的计数，数字之和为 ‘sum’。前导 0 不计为数字。
1 <= n <= 100 和
1 <= 总和 <= 500

例子：

Input:  n = 2, sum = 2
Output: 2
Explanation: Numbers are 11 and 20

Input:  n = 2, sum = 5
Output: 5
Explanation: Numbers are 14, 23, 32, 41 and 50

Input:  n = 3, sum = 6
Output: 21

这个想法很简单，我们从给定的总和中减去从 0 到 9 的所有值，然后重复求和减去那个数字。下面是递归公式。

countRec(n, sum) = ∑countRec(n-1, sum-x)
                            where 0 =< x = 0

    One important observation is, leading 0's must be
    handled explicitly as they are not counted as digits.
    So our final count can be written as below.
    finalCount(n, sum) = ∑countRec(n-1, sum-x)
                           where 1 =< x = 0

以下是基于上述递归公式的简单递归解决方案。

C++

// A C++ program using recursive to count numbers
// with sum of digits as given 'sum'
#include
using namespace std;
 
// Recursive function to count 'n' digit numbers
// with sum of digits as 'sum'. This function
// considers leading 0's also as digits, that is
// why not directly called
unsigned long long int countRec(int n, int sum)
{
    // Base case
    if (n == 0)
    return sum == 0;
 
    if (sum == 0)
    return 1;
 
    // Initialize answer
    unsigned long long int ans = 0;
 
    // Traverse through every digit and count
    // numbers beginning with it using recursion
    for (int i=0; i<=9; i++)
    if (sum-i >= 0)
        ans += countRec(n-1, sum-i);
 
    return ans;
}
 
// This is mainly a wrapper over countRec. It
// explicitly handles leading digit and calls
// countRec() for remaining digits.
unsigned long long int finalCount(int n, int sum)
{
    // Initialize final answer
    unsigned long long int ans = 0;
 
    // Traverse through every digit from 1 to
    // 9 and count numbers beginning with it
    for (int i = 1; i <= 9; i++)
    if (sum-i >= 0)
        ans += countRec(n-1, sum-i);
 
    return ans;
}
 
// Driver program
int main()
{
    int n = 2, sum = 5;
    cout << finalCount(n, sum);
    return 0;
}

Java

// A Java program using recursive to count numbers
// with sum of digits as given 'sum'
class sum_dig
{
    // Recursive function to count 'n' digit numbers
    // with sum of digits as 'sum'. This function
    // considers leading 0's also as digits, that is
    // why not directly called
    static int countRec(int n, int sum)
    {
        // Base case
        if (n == 0)
        return sum == 0 ?1:0;
 
            if (sum == 0)
            return 1;
     
        // Initialize answer
        int ans = 0;
     
        // Traverse through every digit and count
        // numbers beginning with it using recursion
        for (int i=0; i<=9; i++)
        if (sum-i >= 0)
            ans += countRec(n-1, sum-i);
     
        return ans;
    }
     
    // This is mainly a wrapper over countRec. It
    // explicitly handles leading digit and calls
    // countRec() for remaining digits.
    static int finalCount(int n, int sum)
    {
        // Initialize final answer
        int ans = 0;
     
        // Traverse through every digit from 1 to
        // 9 and count numbers beginning with it
        for (int i = 1; i <= 9; i++)
        if (sum-i >= 0)
            ans += countRec(n-1, sum-i);
     
        return ans;
    }
 
    /* Driver program to test above function */
    public static void main (String args[])
    {
        int n = 2, sum = 5;
        System.out.println(finalCount(n, sum));
    }
}/* This code is contributed by Rajat Mishra */

Python3

# A python 3 program using recursive to count numbers
# with sum of digits as given 'sum'
 
# Recursive function to count 'n' digit
# numbers with sum of digits as 'sum'
# This function considers leading 0's
# also as digits, that is why not
# directly called
def countRec(n, sum) :
     
    # Base case
    if (n == 0) :
        return (sum == 0)
 
    if (sum == 0) :
        return 1
 
    # Initialize answer
    ans = 0
 
    # Traverse through every digit and
    # count numbers beginning with it
    # using recursion
    for i in range(0, 10) :
        if (sum-i >= 0) :
            ans = ans + countRec(n-1, sum-i)
 
    return ans
     
     
# This is mainly a wrapper over countRec. It
# explicitly handles leading digit and calls
# countRec() for remaining digits.
def finalCount(n, sum) :
     
    # Initialize final answer
    ans = 0
 
    # Traverse through every digit from 1 to
    # 9 and count numbers beginning with it
    for i in range(1, 10) :
        if (sum-i >= 0) :
            ans = ans + countRec(n-1, sum-i)
 
    return ans
 
 
# Driver program
n = 2
sum = 5
print(finalCount(n, sum))
 
 
# This code is contributed by Nikita tiwari.

C#

// A C# program using recursive to count numbers
// with sum of digits as given 'sum'
using System;
class GFG {
     
    // Recursive function to
    // count 'n' digit numbers
    // with sum of digits as
    // 'sum'. This function
    // considers leading 0's
    // also as digits, that is
    // why not directly called
    static int countRec(int n, int sum)
    {
         
        // Base case
        if (n == 0)
        return sum == 0 ? 1 : 0;
 
            if (sum == 0)
            return 1;
     
        // Initialize answer
        int ans = 0;
     
        // Traverse through every
        // digit and count numbers
        // beginning with it using
        // recursion
        for (int i = 0; i <= 9; i++)
        if (sum - i >= 0)
            ans += countRec(n - 1, sum - i);
     
        return ans;
    }
     
    // This is mainly a
    // wrapper over countRec. It
    // explicitly handles leading
    // digit and calls countRec()
    // for remaining digits.
    static int finalCount(int n, int sum)
    {
         
        // Initialize final answer
        int ans = 0;
     
        // Traverse through every
        // digit from 1 to 9 and
        // count numbers beginning
        // with it
        for (int i = 1; i <= 9; i++)
        if (sum - i >= 0)
            ans += countRec(n - 1, sum - i);
     
        return ans;
    }
 
    // Driver Code
    public static void Main ()
    {
        int n = 2, sum = 5;
        Console.Write(finalCount(n, sum));
    }
}
 
// This code is contributed by nitin mittal.

PHP

= 0)
        $ans += countRec($n-1, $sum-$i);
 
    return $ans;
}
 
// This is mainly a wrapper
// over countRec. It
// explicitly handles leading
// digit and calls
// countRec() for remaining digits.
function finalCount($n, $sum)
{
     
    // Initialize final answer
    $ans = 0;
 
    // Traverse through every
    // digit from 1 to
    // 9 and count numbers
    // beginning with it
    for ($i = 1; $i <= 9; $i++)
    if ($sum - $i >= 0)
        $ans += countRec($n - 1, $sum - $i);
 
    return $ans;
}
 
    // Driver Code
    $n = 2;
    $sum = 5;
    echo finalCount($n, $sum);
     
// This code is contributed by ajit
?>

Javascript

// A JavaScript program using
// recursive to count numbers 
// with sum of digits as given 'sum'
 
    // Recursive function to
    // count 'n' digit numbers
    // with sum of digits as 'sum'.
    //This function
    // considers leading 0's also as digits,
    //that is why not directly called
    function countRec(n, sum) {
        // Base case
        if (n == 0)
            return sum == 0;
   
        if (sum == 0)
            return 1;
   
        // Initialize answer
        let ans = 0;
   
    // Traverse through every
    // digit and count
    // numbers beginning with
    // it using recursion
        for (let i = 0; i <= 9; i++) {
            if (sum - i >= 0)
                ans += countRec(n - 1, sum - i);
        }
           return ans;
    }
     
    // This is mainly a wrapper over countRec.
    // It explicitly handles leading digit 
    // and calls countRec() for remaining digits.
    function finalCount(n, sum) {
        // Initialize final answer
        let ans = 0;
   
        // Traverse through every digit from 1 to
        // 9 and count numbers beginning with it
        for (let i = 1; i <= 9; i++) {
            if (sum - i >= 0)
                ans += countRec(n - 1, sum - i);
        }
             
         return ans;
    }
 
    // Driver program
    let n = 2, sum = 5;
    document.write(finalCount(n, sum));
 
//This code is contributed by Surbhi Tyagi

输出：

5

上述解决方案的时间复杂度是指数级的。如果我们绘制完整的递归树，我们可以观察到许多子问题一次又一次地被解决。例如，如果我们从 n = 3 和 sum = 10 开始，通过考虑数字序列 1,1 或 2, 0，我们可以达到 n = 1，sum = 8。
由于再次调用相同的子问题，因此该问题具有重叠子问题的属性。所以最小平方和问题具有动态规划问题的两个属性(见this和this)。

下面是基于 Memoization 的实现。

C++

// A C++ memoization based recursive program to count
// numbers with sum of n as given 'sum'
#include
using namespace std;
 
// A lookup table used for memoization
unsigned long long int lookup[101][501];
 
// Memoization based implementation of recursive
// function
unsigned long long int countRec(int n, int sum)
{
    // Base case
    if (n == 0)
    return sum == 0;
 
    // If this subproblem is already evaluated,
    // return the evaluated value
    if (lookup[n][sum] != -1)
    return lookup[n][sum];
 
    // Initialize answer
    unsigned long long int ans = 0;
 
    // Traverse through every digit and
    // recursively count numbers beginning
    // with it
    for (int i=0; i<10; i++)
    if (sum-i >= 0)
        ans += countRec(n-1, sum-i);
 
    return lookup[n][sum] = ans;
}
 
// This is mainly a wrapper over countRec. It
// explicitly handles leading digit and calls
// countRec() for remaining n.
unsigned long long int finalCount(int n, int sum)
{
    // Initialize all entries of lookup table
    memset(lookup, -1, sizeof lookup);
 
    // Initialize final answer
    unsigned long long int ans = 0;
 
    // Traverse through every digit from 1 to
    // 9 and count numbers beginning with it
    for (int i = 1; i <= 9; i++)
    if (sum-i >= 0)
        ans += countRec(n-1, sum-i);
    return ans;
}
 
// Driver program
int main()
{
    int n = 3, sum = 5;
    cout << finalCount(n, sum);
    return 0;
}

Java

// A Java memoization based recursive program to count
// numbers with sum of n as given 'sum'
class sum_dig
{
    // A lookup table used for memoization
    static int lookup[][] = new int[101][501];
     
    // Memoization based implementation of recursive
    // function
    static int countRec(int n, int sum)
    {
        // Base case
        if (n == 0)
        return sum == 0 ? 1 : 0;
     
        // If this subproblem is already evaluated,
        // return the evaluated value
        if (lookup[n][sum] != -1)
        return lookup[n][sum];
     
        // Initialize answer
        int ans = 0;
     
        // Traverse through every digit and
        // recursively count numbers beginning
        // with it
        for (int i=0; i<10; i++)
        if (sum-i >= 0)
            ans += countRec(n-1, sum-i);
     
        return lookup[n][sum] = ans;
    }
     
    // This is mainly a wrapper over countRec. It
    // explicitly handles leading digit and calls
    // countRec() for remaining n.
    static int finalCount(int n, int sum)
    {
        // Initialize all entries of lookup table
        for(int i = 0; i <= 100; ++i){
            for(int j = 0; j <= 500; ++j){
                lookup[i][j] = -1;
            }
        }
     
        // Initialize final answer
        int ans = 0;
     
        // Traverse through every digit from 1 to
        // 9 and count numbers beginning with it
        for (int i = 1; i <= 9; i++)
        if (sum-i >= 0)
            ans += countRec(n-1, sum-i);
        return ans;
    }
 
    /* Driver program to test above function */
    public static void main (String args[])
    {
        int n = 3, sum = 5;
        System.out.println(finalCount(n, sum));
    }
}/* This code is contributed by Rajat Mishra */

蟒蛇3

# A Python3 memoization based recursive
# program to count numbers with Sum of n
# as given 'Sum'
 
# A lookup table used for memoization
lookup = [[-1 for i in range(501)]
              for i in range(101)]
 
# Memoization based implementation
# of recursive function
def countRec(n, Sum):
 
    # Base case
    if (n == 0):
        return Sum == 0
 
    # If this subproblem is already evaluated,
    # return the evaluated value
    if (lookup[n][Sum] != -1):
        return lookup[n][Sum]
 
    # Initialize answer
    ans = 0
 
    # Traverse through every digit and
    # recursively count numbers beginning
    # with it
    for i in range(10):
        if (Sum-i >= 0):
            ans += countRec(n - 1, Sum-i)
    lookup[n][Sum] = ans    
 
    return lookup[n][Sum]
 
# This is mainly a wrapper over countRec. It
# explicitly handles leading digit and calls
# countRec() for remaining n.
def finalCount(n, Sum):
 
    # Initialize final answer
    ans = 0
 
    # Traverse through every digit from 1 to
    # 9 and count numbers beginning with it
    for i in range(1, 10):
        if (Sum - i >= 0):
            ans += countRec(n - 1, Sum - i)
    return ans
 
# Driver Code
n, Sum = 3, 5
print(finalCount(n, Sum))
 
# This code is contributed by mohit kumar 29

C＃

// A C# memoization based recursive program to count
// numbers with sum of n as given 'sum'
 
using System;
class sum_dig
{
    // A lookup table used for memoization
    static int [,]lookup = new int[101,501];
      
    // Memoization based implementation of recursive
    // function
    static int countRec(int n, int sum)
    {
        // Base case
        if (n == 0)
        return sum == 0 ? 1 : 0;
      
        // If this subproblem is already evaluated,
        // return the evaluated value
        if (lookup[n,sum] != -1)
        return lookup[n,sum];
      
        // Initialize answer
        int ans = 0;
      
        // Traverse through every digit and
        // recursively count numbers beginning
        // with it
        for (int i=0; i<10; i++)
        if (sum-i >= 0)
            ans += countRec(n-1, sum-i);
      
        return lookup[n,sum] = ans;
    }
      
    // This is mainly a wrapper over countRec. It
    // explicitly handles leading digit and calls
    // countRec() for remaining n.
    static int finalCount(int n, int sum)
    {
        // Initialize all entries of lookup table
        for(int i = 0; i <= 100; ++i){
            for(int j = 0; j <= 500; ++j){
                lookup[i,j] = -1;
            }
        }
      
        // Initialize final answer
        int ans = 0;
      
        // Traverse through every digit from 1 to
        // 9 and count numbers beginning with it
        for (int i = 1; i <= 9; i++)
        if (sum-i >= 0)
            ans += countRec(n-1, sum-i);
        return ans;
    }
  
    /* Driver program to test above function */
    public static void Main ()
    {
        int n = 3, sum = 5;
        Console.Write(finalCount(n, sum));
    }
}

PHP

= 0)
        $ans += countRec($n - 1, $sum - $i);
 
    return $lookup[$n][$sum] = $ans;
}
 
// This is mainly a wrapper over countRec. It
// explicitly handles leading digit and calls
// countRec() for remaining n.
function finalCount($n, $sum)
{
    // Initialize all entries of lookup table
 
    // Initialize final answer
    $ans = 0;
 
    // Traverse through every digit from 1 to
    // 9 and count numbers beginning with it
    for ($i = 1; $i <= 9; $i++)
    if ($sum-$i >= 0)
        $ans += countRec($n - 1, $sum - $i);
    return $ans;
}
 
// Driver Code
$n = 3;
$sum = 5;
echo finalCount($n, $sum);
 
// This code is contributed by mits
?>

Javascript

 
// A Javascript memoization based
// recursive program to count numbers
// with sum of n as given 'sum'
     
// A lookup table used for memoization
let lookup = new Array(101);
 
// Memoization based implementation
// of recursive function
function countRec(n, sum)
{
     
    // Base case
    if (n == 0)
        return sum == 0 ? 1 : 0;
  
    // If this subproblem is already evaluated,
    // return the evaluated value
    if (lookup[n][sum] != -1)
        return lookup[n][sum];
  
    // Initialize answer
    let ans = 0;
  
    // Traverse through every digit and
    // recursively count numbers beginning
    // with it
    for(let i = 0; i < 10; i++)
        if (sum - i >= 0)
            ans += countRec(n - 1, sum - i);
  
    return lookup[n][sum] = ans;
}
 
// This is mainly a wrapper over countRec. It
// explicitly handles leading digit and calls
// countRec() for remaining n.
function finalCount(n, sum)
{
     
    // Initialize all entries of lookup table
    for(let i = 0; i < 101; i++)
    {  
        lookup[i] = new Array(501);
        for(let j = 0; j < 501; j++)
        {
            lookup[i][j] = -1;
        }
    }
  
    // Initialize final answer
    let ans = 0;
  
    // Traverse through every digit from 1 to
    // 9 and count numbers beginning with it
    for(let i = 1; i <= 9; i++)
        if (sum - i >= 0)
            ans += countRec(n - 1, sum - i);
             
    return ans;
}
 
// Driver code
let n = 3, sum = 5;
 
document.write(finalCount(n, sum));
 
// This code is contributed by avanitrachhadiya2155
 

输出：

15

感谢 Gaurav Ahirwar 提出上述解决方案。

另一种方法
通过迭代所有 n 位数字并检查当前 n 位数字的总和是否等于给定的总和，我们可以轻松地计算出数字总和等于给定总和的 n 位数字，如果是，那么我们将开始将数字递增 9 直到达到数字其数字总和大于给定总和，然后我们将再次递增 1，直到找到具有给定总和的另一个数字。

C++

// C++ program to Count of n digit numbers
// whose sum of digits equals to given sum
#include 
#include 
using namespace std;
 void findCount(int n, int sum) {
         
        //in case n = 2 start is 10 and end is (100-1) = 99
        int start = pow(10, n-1);
        int end = pow(10, n)-1;
     
        int count = 0;
        int i = start;
         
                while(i <= end) {
             
            int cur = 0;
            int temp = i;
             
            while( temp != 0) {
                cur += temp % 10;
                temp = temp / 10;
            }
             
            if(cur == sum) {            
                count++;            
                i += 9;        
            }else
                i++;
             
        }    
            cout << count;
 
        /* This code is contributed by Anshuman */
    }
int main() {
        int n = 3;
        int sum = 5;    
        findCount(n,sum);
         
     
    return 0;
}

Java

// Java program to Count of n digit numbers
// whose sum of digits equals to given sum
 
public class GFG {
 
    public static void main(String[] args) {
         
        int n = 3;
        int sum = 5;    
        findCount(n,sum);
         
    }
     
    private static void findCount(int n, int sum) {
         
        //in case n = 2 start is 10 and end is (100-1) = 99
        int start = (int) Math.pow(10, n-1);
        int end = (int) Math.pow(10, n)-1;
     
        int count = 0;
        int i = start;
         
                while(i < end) {
             
            int cur = 0;
            int temp = i;
             
            while( temp != 0) {
                cur += temp % 10;
                temp = temp / 10;
            }
             
            if(cur == sum) {            
                count++;            
                i += 9;        
            }else
                i++;
             
        }    
        System.out.println(count);
 
        /* This code is contributed by Anshuman */
    }
}

蟒蛇3

# Python3 program to Count of n digit numbers
# whose sum of digits equals to given sum
import math
 
def findCount(n, sum):
     
    # in case n = 2 start is 10 and
    # end is (100-1) = 99
    start = math.pow(10, n - 1);
    end = math.pow(10, n) - 1;
 
    count = 0;
    i = start;
     
    while(i <= end):
         
        cur = 0;
        temp = i;
         
        while(temp != 0):
            cur += temp % 10;
            temp = temp // 10;
         
        if(cur == sum):    
            count = count + 1;        
            i += 9;    
        else:
            i = i + 1;
         
    print(count);
 
# Driver Code
n = 3;
sum = 5;    
findCount(n, sum);
 
# This code is contributed
# by Akanksha Rai

C＃

// C# program to Count of n digit numbers
// whose sum of digits equals to given sum
using System;
 
class GFG
{
private static void findCount(int n,
                              int sum)
{
     
    // in case n = 2 start is 10 and
    // end is (100-1) = 99
    int start = (int) Math.Pow(10, n - 1);
    int end = (int) Math.Pow(10, n) - 1;
 
    int count = 0;
    int i = start;
     
    while(i < end)
    {
         
        int cur = 0;
        int temp = i;
         
        while( temp != 0)
        {
            cur += temp % 10;
            temp = temp / 10;
        }
         
        if(cur == sum)
        {        
            count++;            
            i += 9;    
        }
        else
            i++;
         
    }
    Console.WriteLine(count);
}
 
// Driver Code
public static void Main()
{
    int n = 3;
    int sum = 5;    
    findCount(n,sum);
}
}
 
// This code is contributed
// by Akanksha Rai

Javascript

 
// Javascript program to Count of n digit numbers
// whose sum of digits equals to given sum
 function findCount(n, sum) {
         
        // in case n = 2 start is 10 and end is (100-1) = 99
        let start = Math.pow(10, n-1);
        let end = Math.pow(10, n)-1;
     
        let count = 0;
        let i = start;
         
        while(i <= end)
        {
             
            let cur = 0;
            let temp = i;
             
            while( temp != 0)
            {
                cur += temp % 10;
                temp = parseInt(temp / 10);
            }
             
            if(cur == sum)
            {            
                count++;            
                i += 9;        
            }else
                i++;
             
        }    
            document.write(count);
    }
        let n = 3;
        let sum = 5;    
        findCount(n,sum);
 
// This code is contributed by souravmahato348.

输出：

15

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