计算从 1 到 n 的所有数字的数字总和

给定一个数 n，求从 1 到 n 的所有数的数字之和。
例子：

Input: n = 5
Output: Sum of digits in numbers from 1 to 5 = 15

Input: n = 12
Output: Sum of digits in numbers from 1 to 12 = 51

Input: n = 328
Output: Sum of digits in numbers from 1 to 328 = 3241

天真的解决方案：
一个简单的解决方案是遍历从 1 到 n 的每个数字 x，并通过遍历 x 的所有数字来计算 x 中的总和。下面是这个想法的实现。

C++

// A Simple C++ program to compute sum of digits in numbers from 1 to n
#include
using namespace std;
 
int sumOfDigits(int );
 
// Returns sum of all digits in numbers from 1 to n
int sumOfDigitsFrom1ToN(int n)
{
    int result = 0; // initialize result
 
    // One by one compute sum of digits in every number from
    // 1 to n
    for (int x = 1; x <= n; x++)
        result += sumOfDigits(x);
 
    return result;
}
 
// A utility function to compute sum of digits in a
// given number x
int sumOfDigits(int x)
{
    int sum = 0;
    while (x != 0)
    {
        sum += x %10;
        x   = x /10;
    }
    return sum;
}
 
// Driver Program
int main()
{
    int n = 328;
    cout << "Sum of digits in numbers from 1 to " << n << " is "
         << sumOfDigitsFrom1ToN(n);
    return 0;
}

Java

// A Simple JAVA program to compute sum of
// digits in numbers from 1 to n
import java.io.*;
 
class GFG {
     
    // Returns sum of all digits in numbers
    // from 1 to n
    static int sumOfDigitsFrom1ToN(int n)
    {
        int result = 0; // initialize result
      
        // One by one compute sum of digits
        // in every number from 1 to n
        for (int x = 1; x <= n; x++)
            result += sumOfDigits(x);
      
        return result;
    }
      
    // A utility function to compute sum
    // of digits in a given number x
    static int sumOfDigits(int x)
    {
        int sum = 0;
        while (x != 0)
        {
            sum += x % 10;
            x   = x / 10;
        }
        return sum;
    }
      
    // Driver Program
    public static void main(String args[])
    {
        int n = 328;
        System.out.println("Sum of digits in numbers"
                          +" from 1 to " + n + " is "
                          + sumOfDigitsFrom1ToN(n));
    }
}
 
/*This code is contributed by Nikita Tiwari.*/

Python3

# A Simple Python program to compute sum
# of digits in numbers from 1 to n
 
# Returns sum of all digits in numbers
# from 1 to n
def sumOfDigitsFrom1ToN(n) :
 
    result = 0   # initialize result
  
    # One by one compute sum of digits
    # in every number from 1 to n
    for x in range(1, n+1) :
        result = result + sumOfDigits(x)
  
    return result
 
# A utility function to compute sum of
# digits in a given number x
def sumOfDigits(x) :
    sum = 0
    while (x != 0) :
        sum = sum + x % 10
        x   = x // 10
     
    return sum
 
 
# Driver Program
n = 328
print("Sum of digits in numbers from 1 to", n, "is", sumOfDigitsFrom1ToN(n))
 
 
# This code is contributed by Nikita Tiwari.

C#

// A Simple C# program to compute sum of
// digits in numbers from 1 to n
 
using System;
 
public class GFG {
     
    // Returns sum of all digits in numbers
    // from 1 to n
    static int sumOfDigitsFrom1ToN(int n)
    {
         
        // initialize result
        int result = 0;
     
        // One by one compute sum of digits
        // in every number from 1 to n
        for (int x = 1; x <= n; x++)
            result += sumOfDigits(x);
     
        return result;
    }
     
    // A utility function to compute sum
    // of digits in a given number x
    static int sumOfDigits(int x)
    {
        int sum = 0;
         
        while (x != 0)
        {
            sum += x % 10;
            x = x / 10;
        }
         
        return sum;
    }
     
    // Driver Program
    public static void Main()
    {
        int n = 328;
         
        Console.WriteLine("Sum of digits"
               + " in numbers from 1 to "
                             + n + " is "
                + sumOfDigitsFrom1ToN(n));
    }
}
 
// This code is contributed by shiv_bhakt.

Javascript

    // A Simple Javascript program to compute sum of
    // digits in numbers from 1 to n
     
    // Returns sum of all digits in numbers
    // from 1 to n
    function sumOfDigitsFrom1ToN(n)
    {
          
        // initialize result
        let result = 0;
      
        // One by one compute sum of digits
        // in every number from 1 to n
        for (let x = 1; x <= n; x++)
            result += sumOfDigits(x);
      
        return result;
    }
      
    // A utility function to compute sum
    // of digits in a given number x
    function sumOfDigits(x)
    {
        let sum = 0;
          
        while (x != 0)
        {
            sum += x % 10;
            x = parseInt(x / 10, 10);
        }
          
        return sum;
    }
     
    let n = 328;
          
    document.write("Sum of digits"
                      + " in numbers from 1 to "
                      + n + " is "
                      + sumOfDigitsFrom1ToN(n));
                       
    // This code is contributed by divyeshrabadiya07.                 

输出：

Sum of digits in numbers from 1 to 328 is 3241

有效的解决方案：
以上是一个幼稚的解决方案。我们可以通过找到一种模式来更有效地做到这一点。
让我们举几个例子。

sum(9) = 1 + 2 + 3 + 4 ........... + 9
       = 9*10/2 
       = 45

sum(99)  = 45 + (10 + 45) + (20 + 45) + ..... (90 + 45)
         = 45*10 + (10 + 20 + 30 ... 90)
         = 45*10 + 10(1 + 2 + ... 9)
         = 45*10 + 45*10
         = sum(9)*10 + 45*10 

sum(999) = sum(99)*10 + 45*100

通常，我们可以使用以下公式计算 sum(10 ^d – 1)

sum(10d - 1) = sum(10d-1 - 1) * 10 + 45*(10d-1) 

在下面的实现中，上面的公式是使用动态规划实现的，因为存在重叠的子问题。
上述公式是该思想的核心步骤之一。下面是完整的算法

算法：sum(n)

1) Find number of digits minus one in n. Let this value be 'd'.  
   For 328, d is 2.

2) Compute some of digits in numbers from 1 to 10d - 1.  
   Let this sum be w. For 328, we compute sum of digits from 1 to 
   99 using above formula.

3) Find Most significant digit (msd) in n. For 328, msd is 3.

4) Overall sum is sum of following terms

    a) Sum of digits in 1 to "msd * 10d - 1".  For 328, sum of 
       digits in numbers from 1 to 299.
        For 328, we compute 3*sum(99) + (1 + 2)*100.  Note that sum of
        sum(299) is sum(99) + sum of digits from 100 to 199 + sum of digits
        from 200 to 299.  
        Sum of 100 to 199 is sum(99) + 1*100 and sum of 299 is sum(99) + 2*100.
        In general, this sum can be computed as w*msd + (msd*(msd-1)/2)*10d

    b) Sum of digits in msd * 10d to n.  For 328, sum of digits in 
       300 to 328.
        For 328, this sum is computed as 3*29 + recursive call "sum(28)"
        In general, this sum can be computed as  msd * (n % (msd*10d) + 1) 
        + sum(n % (10d))

下面是上述算法的实现。

C++

// C++ program to compute sum of digits in numbers from 1 to n
#include
using namespace std;
 
// Function to computer sum of digits in numbers from 1 to n
// Comments use example of 328 to explain the code
int sumOfDigitsFrom1ToN(int n)
{
    // base case: if n<10 return sum of
    // first n natural numbers
    if (n<10)
      return n*(n+1)/2;
 
    // d = number of digits minus one in n. For 328, d is 2
    int d = log10(n);
 
    // computing sum of digits from 1 to 10^d-1,
    // d=1 a[0]=0;
    // d=2 a[1]=sum of digit from 1 to 9 = 45
    // d=3 a[2]=sum of digit from 1 to 99 = a[1]*10 + 45*10^1 = 900
    // d=4 a[3]=sum of digit from 1 to 999 = a[2]*10 + 45*10^2 = 13500
    int *a = new int[d+1];
    a[0] = 0, a[1] = 45;
    for (int i=2; i<=d; i++)
        a[i] = a[i-1]*10 + 45*ceil(pow(10,i-1));
 
    // computing 10^d
    int p = ceil(pow(10, d));
 
    // Most significant digit (msd) of n,
    // For 328, msd is 3 which can be obtained using 328/100
    int msd = n/p;
 
    // EXPLANATION FOR FIRST and SECOND TERMS IN BELOW LINE OF CODE
    // First two terms compute sum of digits from 1 to 299
    // (sum of digits in range 1-99 stored in a[d]) +
    // (sum of digits in range 100-199, can be calculated as 1*100 + a[d]
    // (sum of digits in range 200-299, can be calculated as 2*100 + a[d]
    //  The above sum can be written as 3*a[d] + (1+2)*100
 
    // EXPLANATION FOR THIRD AND FOURTH TERMS IN BELOW LINE OF CODE
    // The last two terms compute sum of digits in number from 300 to 328
    // The third term adds 3*29 to sum as digit 3 occurs in all numbers
    //                from 300 to 328
    // The fourth term recursively calls for 28
    return msd*a[d] + (msd*(msd-1)/2)*p + 
           msd*(1+n%p) + sumOfDigitsFrom1ToN(n%p);
}
 
// Driver Program
int main()
{
    int n = 328;
    cout << "Sum of digits in numbers from 1 to " << n << " is "
         << sumOfDigitsFrom1ToN(n);
    return 0;
}

Java

// JAVA program to compute sum of digits
// in numbers from 1 to n
import java.io.*;
import java.math.*;
 
class GFG{
     
    // Function to computer sum of digits in
    // numbers from 1 to n. Comments use
    // example of 328 to explain the code
    static int sumOfDigitsFrom1ToN(int n)
    {
        // base case: if n<10 return sum of
        // first n natural numbers
        if (n < 10)
          return (n * (n + 1) / 2);
      
        // d = number of digits minus one in
        // n. For 328, d is 2
        int d = (int)(Math.log10(n));
      
        // computing sum of digits from 1 to 10^d-1,
        // d=1 a[0]=0;
        // d=2 a[1]=sum of digit from 1 to 9 = 45
        // d=3 a[2]=sum of digit from 1 to 99 =
        // a[1]*10 + 45*10^1 = 900
        // d=4 a[3]=sum of digit from 1 to 999 =
        // a[2]*10 + 45*10^2 = 13500
        int a[] = new int[d+1];
        a[0] = 0; a[1] = 45;
        for (int i = 2; i <= d; i++)
            a[i] = a[i-1] * 10 + 45 *
                 (int)(Math.ceil(Math.pow(10, i-1)));
      
        // computing 10^d
        int p = (int)(Math.ceil(Math.pow(10, d)));
      
        // Most significant digit (msd) of n,
        // For 328, msd is 3 which can be obtained
        // using 328/100
        int msd = n / p;
      
        // EXPLANATION FOR FIRST and SECOND TERMS IN
        // BELOW LINE OF CODE
        // First two terms compute sum of digits from
        // 1 to 299
        // (sum of digits in range 1-99 stored in a[d]) +
        // (sum of digits in range 100-199, can be
        // calculated as 1*100 + a[d]
        // (sum of digits in range 200-299, can be
        // calculated as 2*100 + a[d]
        //  The above sum can be written as 3*a[d] +
        // (1+2)*100
      
        // EXPLANATION FOR THIRD AND FOURTH TERMS IN
        // BELOW LINE OF CODE
        // The last two terms compute sum of digits in
        // number from 300 to 328. The third term adds
        // 3*29 to sum as digit 3 occurs in all numbers
        // from 300 to 328. The fourth term recursively
        // calls for 28
        return (msd * a[d] + (msd * (msd - 1) / 2) * p + 
              msd * (1 + n % p) + sumOfDigitsFrom1ToN(n % p));
    }
      
    // Driver Program
    public static void main(String args[])
    {
        int n = 328;
        System.out.println("Sum of digits in numbers " +
                          "from 1 to " +n + " is " +
                          sumOfDigitsFrom1ToN(n));
    }
}
 
/*This code is contributed by Nikita Tiwari.*/

蟒蛇3

# PYTHON 3 program to compute sum of digits
# in numbers from 1 to n
import math
 
# Function to computer sum of digits in
# numbers from 1 to n. Comments use example
# of 328 to explain the code
def sumOfDigitsFrom1ToN( n) :
   
    # base case: if n<10 return sum of
    # first n natural numbers
    if (n<10) :
        return (n*(n+1)/2)
  
    # d = number of digits minus one in n.
    # For 328, d is 2
    d = (int)(math.log10(n))
  
    """computing sum of digits from 1 to 10^d-1,
    d=1 a[0]=0;
    d=2 a[1]=sum of digit from 1 to 9 = 45
    d=3 a[2]=sum of digit from 1 to 99 = a[1]*10
    + 45*10^1 = 900
    d=4 a[3]=sum of digit from 1 to 999 = a[2]*10
    + 45*10^2 = 13500"""
    a = [0] * (d + 1)
    a[0] = 0
    a[1] = 45
    for i in range(2, d+1) :
        a[i] = a[i-1] * 10 + 45 * (int)(math.ceil(math.pow(10,i-1)))
  
    # computing 10^d
    p = (int)(math.ceil(math.pow(10, d)))
  
    # Most significant digit (msd) of n,
    # For 328, msd is 3 which can be obtained
    # using 328/100
    msd = n//p
  
    """EXPLANATION FOR FIRST and SECOND TERMS IN
    BELOW LINE OF CODE
    First two terms compute sum of digits from 1 to 299
    (sum of digits in range 1-99 stored in a[d]) +
    (sum of digits in range 100-199, can be calculated
    as 1*100 + a[d]. (sum of digits in range 200-299,
    can be calculated as 2*100 + a[d]
    The above sum can be written as 3*a[d] + (1+2)*100
  
    EXPLANATION FOR THIRD AND FOURTH TERMS IN BELOW
    LINE OF CODE
    The last two terms compute sum of digits in number
    from 300 to 328. The third term adds 3*29 to sum
    as digit 3 occurs in all numbers from 300 to 328.
    The fourth term recursively calls for 28"""
    return (int)(msd * a[d] + (msd*(msd-1) // 2) * p + 
           msd * (1 + n % p) + sumOfDigitsFrom1ToN(n % p))
 
# Driver Program
n = 328
print("Sum of digits in numbers from 1 to",
      n ,"is",sumOfDigitsFrom1ToN(n))
 
 
# This code is contributed by Nikita Tiwari.

C＃

// C# program to compute sum of digits
// in numbers from 1 to n
 
using System;
 
public class GFG {
     
    // Function to computer sum of digits in
    // numbers from 1 to n. Comments use
    // example of 328 to explain the code
    static int sumOfDigitsFrom1ToN(int n)
    {
         
        // base case: if n<10 return sum of
        // first n natural numbers
        if (n < 10)
            return (n * (n + 1) / 2);
     
        // d = number of digits minus one in
        // n. For 328, d is 2
        int d = (int)(Math.Log(n) / Math.Log(10));
     
        // computing sum of digits from 1 to 10^d-1,
        // d=1 a[0]=0;
        // d=2 a[1]=sum of digit from 1 to 9 = 45
        // d=3 a[2]=sum of digit from 1 to 99 =
        // a[1]*10 + 45*10^1 = 900
        // d=4 a[3]=sum of digit from 1 to 999 =
        // a[2]*10 + 45*10^2 = 13500
        int[] a = new int[d+1];
        a[0] = 0; a[1] = 45;
         
        for (int i = 2; i <= d; i++)
            a[i] = a[i-1] * 10 + 45 *
                (int)(Math.Ceiling(Math.Pow(10, i-1)));
     
        // computing 10^d
        int p = (int)(Math.Ceiling(Math.Pow(10, d)));
     
        // Most significant digit (msd) of n,
        // For 328, msd is 3 which can be obtained
        // using 328/100
        int msd = n / p;
     
        // EXPLANATION FOR FIRST and SECOND TERMS IN
        // BELOW LINE OF CODE
        // First two terms compute sum of digits from
        // 1 to 299
        // (sum of digits in range 1-99 stored in a[d]) +
        // (sum of digits in range 100-199, can be
        // calculated as 1*100 + a[d]
        // (sum of digits in range 200-299, can be
        // calculated as 2*100 + a[d]
        // The above sum can be written as 3*a[d] +
        // (1+2)*100
     
        // EXPLANATION FOR THIRD AND FOURTH TERMS IN
        // BELOW LINE OF CODE
        // The last two terms compute sum of digits in
        // number from 300 to 328. The third term adds
        // 3*29 to sum as digit 3 occurs in all numbers
        // from 300 to 328. The fourth term recursively
        // calls for 28
        return (msd * a[d] + (msd * (msd - 1) / 2) * p +
            msd * (1 + n % p) + sumOfDigitsFrom1ToN(n % p));
    }
     
    // Driver Program
    public static void Main()
    {
        int n = 328;
        Console.WriteLine("Sum of digits in numbers " +
                             "from 1 to " +n + " is " +
                               sumOfDigitsFrom1ToN(n));
    }
}
 
// This code is contributed by shiv_bhakt.

Javascript

 
    // Javascript program to compute sum of digits
    // in numbers from 1 to n
     
    // Function to computer sum of digits in
    // numbers from 1 to n. Comments use
    // example of 328 to explain the code
    function sumOfDigitsFrom1ToN(n)
    {
          
        // base case: if n<10 return sum of
        // first n natural numbers
        if (n < 10)
            return (n * (n + 1) / 2);
      
        // d = number of digits minus one in
        // n. For 328, d is 2
        let d = parseInt(Math.log(n) / Math.log(10), 10);
      
        // computing sum of digits from 1 to 10^d-1,
        // d=1 a[0]=0;
        // d=2 a[1]=sum of digit from 1 to 9 = 45
        // d=3 a[2]=sum of digit from 1 to 99 =
        // a[1]*10 + 45*10^1 = 900
        // d=4 a[3]=sum of digit from 1 to 999 =
        // a[2]*10 + 45*10^2 = 13500
        let a = new Array(d+1);
        a[0] = 0; a[1] = 45;
          
        for (let i = 2; i <= d; i++)
            a[i] = a[i-1] * 10 + 45 *
            parseInt(Math.ceil(Math.pow(10, i-1)), 10);
      
        // computing 10^d
        let p = parseInt(Math.ceil(Math.pow(10, d)), 10);
      
        // Most significant digit (msd) of n,
        // For 328, msd is 3 which can be obtained
        // using 328/100
        let msd = parseInt(n / p, 10);
      
        // EXPLANATION FOR FIRST and SECOND TERMS IN
        // BELOW LINE OF CODE
        // First two terms compute sum of digits from
        // 1 to 299
        // (sum of digits in range 1-99 stored in a[d]) +
        // (sum of digits in range 100-199, can be
        // calculated as 1*100 + a[d]
        // (sum of digits in range 200-299, can be
        // calculated as 2*100 + a[d]
        // The above sum can be written as 3*a[d] +
        // (1+2)*100
      
        // EXPLANATION FOR THIRD AND FOURTH TERMS IN
        // BELOW LINE OF CODE
        // The last two terms compute sum of digits in
        // number from 300 to 328. The third term adds
        // 3*29 to sum as digit 3 occurs in all numbers
        // from 300 to 328. The fourth term recursively
        // calls for 28
        return (msd * a[d] + (msd * (msd - 1) / 2) * p +
            msd * (1 + n % p) + sumOfDigitsFrom1ToN(n % p));
    }
     
    let n = 328;
    document.write("Sum of digits in numbers from 1 to " + n +
                     " is " + sumOfDigitsFrom1ToN(n));
                      

输出：

Sum of digits in numbers from 1 to 328 is 3241

高效算法还有一个优点，即即使给定多个输入，我们也只需要计算数组 ‘a[]’ 一次。

改进：
上述实现需要 O(d ² ) 时间，因为每次递归调用都会再次计算 dp[] 数组。第一次调用需要 O(d)，第二次调用需要 O(d-1)，第三次调用 O(d-2)，依此类推。我们不需要在每次递归调用中重新计算 dp[] 数组。下面是在 O(d) 时间内工作的修改后的实现。其中d是输入数字中的位数。

C++

// C++ program to compute sum of digits
// in numbers from 1 to n
#include
using namespace std;
 
int sumOfDigitsFrom1ToNUtil(int n, int a[])
{
    if (n < 10)
        return (n * (n + 1) / 2);
     
    int d = (int)(log10(n));
    int p = (int)(ceil(pow(10, d)));
    int msd = n / p;
     
    return (msd * a[d] + (msd * (msd - 1) / 2) * p +
            msd * (1 + n % p) +
            sumOfDigitsFrom1ToNUtil(n % p, a));
}
 
// Function to computer sum of digits in
// numbers from 1 to n
int sumOfDigitsFrom1ToN(int n)
{
    int d = (int)(log10(n));
    int a[d + 1];
    a[0] = 0; a[1] = 45;
     
    for(int i = 2; i <= d; i++)
        a[i] = a[i - 1] * 10 + 45 *
               (int)(ceil(pow(10, i - 1)));
 
    return sumOfDigitsFrom1ToNUtil(n, a);
}
  
// Driver code
int main()
{
    int n = 328;
     
    cout << "Sum of digits in numbers from 1 to "
         << n << " is "<< sumOfDigitsFrom1ToN(n);
}
 
// This code is contributed by ajaykr00kj

Java

// JAVA program to compute sum of digits
// in numbers from 1 to n
import java.io.*;
import java.math.*;
 
class GFG{
     
    // Function to computer sum of digits in
    // numbers from 1 to n
    static int sumOfDigitsFrom1ToN(int n)
    {
        int d = (int)(Math.log10(n));
        int a[] = new int[d+1];
        a[0] = 0; a[1] = 45;
        for (int i = 2; i <= d; i++)
            a[i] = a[i-1] * 10 + 45 *
                (int)(Math.ceil(Math.pow(10, i-1)));
     
        return sumOfDigitsFrom1ToNUtil(n, a);
    }
    
    static int sumOfDigitsFrom1ToNUtil(int n, int a[])
    {
        if (n < 10)
            return (n * (n + 1) / 2);
       
        int d = (int)(Math.log10(n));
        int p = (int)(Math.ceil(Math.pow(10, d)));
        int msd = n / p;
        return (msd * a[d] + (msd * (msd - 1) / 2) * p +
            msd * (1 + n % p) + sumOfDigitsFrom1ToNUtil(n % p, a));
    }
     
    // Driver Program
    public static void main(String args[])
    {
        int n = 328;
        System.out.println("Sum of digits in numbers " +
                        "from 1 to " +n + " is " +
                        sumOfDigitsFrom1ToN(n));
    }
}
 
/*This code is contributed by Narendra Jha.*/

蟒蛇3

# Python program to compute sum of digits
# in numbers from 1 to n
import math
 
# Function to computer sum of digits in
# numbers from 1 to n
def sumOfDigitsFrom1ToN(n):
     
    d = int(math.log(n, 10))
    a = [0]*(d + 1)
    a[0] = 0
    a[1] = 45
    for i in range(2, d + 1):
        a[i] = a[i - 1] * 10 + 45 * \
                int(math.ceil(pow(10, i - 1)))
         
    return sumOfDigitsFrom1ToNUtil(n, a)
 
def sumOfDigitsFrom1ToNUtil(n, a):
    if (n < 10):
        return (n * (n + 1)) // 2
     
    d = int(math.log(n,10))
    p = int(math.ceil(pow(10, d)))
    msd = n // p
    return (msd * a[d] + (msd * (msd - 1) // 2) * p +
    msd * (1 + n % p) + sumOfDigitsFrom1ToNUtil(n % p, a))
 
# Driver code
n = 328
print("Sum of digits in numbers from 1 to",n,"is",sumOfDigitsFrom1ToN(n))
 
# This code is contributed by shubhamsingh10

C＃

// C# program to compute sum of digits
// in numbers from 1 to n
using System;
 
class GFG
{
     
    // Function to computer sum of digits in
    // numbers from 1 to n
    static int sumOfDigitsFrom1ToN(int n)
    {
        int d = (int)(Math.Log10(n));
        int []a = new int[d+1];
        a[0] = 0; a[1] = 45;
        for (int i = 2; i <= d; i++)
            a[i] = a[i-1] * 10 + 45 *
                (int)(Math.Ceiling(Math.Pow(10, i-1)));
     
        return sumOfDigitsFrom1ToNUtil(n, a);
    }
     
    static int sumOfDigitsFrom1ToNUtil(int n, int []a)
    {
        if (n < 10)
            return (n * (n + 1) / 2);
         
        int d = (int)(Math.Log10(n));
        int p = (int)(Math.Ceiling(Math.Pow(10, d)));
        int msd = n / p;
        return (msd * a[d] + (msd * (msd - 1) / 2) * p +
            msd * (1 + n % p) + sumOfDigitsFrom1ToNUtil(n % p, a));
    }
     
    // Driver code
    public static void Main(String []args)
    {
        int n = 328;
        Console.WriteLine("Sum of digits in numbers " +
                        "from 1 to " +n + " is " +
                        sumOfDigitsFrom1ToN(n));
    }
}
 
// This code contributed by Rajput-Ji

Javascript

 
// JavaScript program to compute sum of digits
// in numbers from 1 to n
 
// Function to computer sum of digits in
// numbers from 1 to n
function sumOfDigitsFrom1ToNUtil( n,a)
{
    if (n < 10)
        return (n * (n + 1) / 2);
     
    var d = (parseInt)(Math.log10(n));
    var p = (Math.ceil(Math.pow(10, d)));
    var msd =(parseInt) (n / p);
     
    return (msd * a[d] + (msd * (msd - 1) / 2) * p +
            msd * (1 + n % p) +
            sumOfDigitsFrom1ToNUtil(n % p, a));
}
 
// Function to computer sum of digits in
// numbers from 1 to n
function sumOfDigitsFrom1ToN( n)
{
    var d =(parseInt)(Math.log10(n));
   var a=new Array(d + 1).fill(0);
    a[0] = 0; a[1] = 45;
     
    for(var i = 2; i <= d; i++)
        a[i] = a[i - 1] * 10 + 45 *
               (parseInt)(Math.ceil(Math.pow(10, i - 1)));
 
    return sumOfDigitsFrom1ToNUtil(n, a);
}
  
var n = 328;
     
    document.write( "Sum of digits in numbers from 1 to "
         + n + " is "+ sumOfDigitsFrom1ToN(n))
 
  

输出：

Sum of digits in numbers from 1 to 328 is 3241

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。