n个数字总和等于给定总和的n个数字的计数

给定两个整数’n’和’sum’，找到所有n位数字的计数，且位数之和为’sum’。前导0不算作数字。
1 <= n <= 100且
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0到9之间的所有值，然后递减总和减去该数字。以下是递归公式。

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的。如果绘制完整的递归树，则可以观察到许多子问题一次又一次地得到解决。例如，如果我们以n = 3且sum = 10开头，则通过考虑数字序列1,1或2，0可以达到n = 1，sum = 8。
由于再次调用了相同的问题，因此此问题具有“重叠子问题”属性。因此，最小平方和问题具有动态规划问题的两个属性(请参阅此内容)。

以下是基于记忆的实现。

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 */

Python3

# 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));
    }
}

的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
?>

Java脚本

 
// 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
我们可以通过对所有n位数字进行迭代并检查当前n位数字的总和是否等于给定总和来轻松地计算其数字总和等于给定总和的n位数字，如果是，则我们将数字开始递增9，直到达到数字其数字的总和大于给定的总和，那么我们将再次加1，直到找到另一个具有给定的总和的数字。

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 */
    }
}

Python3

# 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

Java脚本

 
// 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