[L，R]范围内所有i的i * countDigits(i)^ countDigits(i)之和

给定两个表示范围[L，R]的数字L，R ，任务是找到所有i∈[L，R]的总和i * countDigits(i) ^{countDigits(i)}其中countDigits(i)是计数i中的位数。
例子：

Input: L = 8, R = 11 
Output: 101 
Explanation: 
For the given numbers, we need to find the values for all the numbers in the range [L, R]. 
=> 8 * 1¹ + 9 * 1¹ + 10 * 2² + 11 * 2² 
=> 8 + 9 + 40 + 44 
=> 101
Input: L = 98, R = 102 
Output: 8969 
Explanation: 
For the given numbers, we need to find the values for all the numbers in the range [L, R]. 
=> 98 * 2² + 99 * 2² + 100 * 3³ + 101 * 3³ + 102 * 3³ 
=> 8969 
 

方法：想法是根据数字位数将范围分为几部分。也就是说，每个段都包含具有相同位数的数字：

[1 - 9], [10 - 99], [100 - 999], [1000 - 9999], [10000 - 99999] ...

从上面的片段中可以明显看出，每个片段的[L，R]都具有相同的长度。因此，该细分的要求总和为：

countDigits(L)countDigits(L) * (L + R) * (R - L + 1) / 2

证明：

• 令[L，R] = [10，14] ，其中L和R的长度相同，即2。
• 因此，段[L，R]的总和为：

10 * 22 + 11 * 22 + 12 * 22 + 13 * 22 + 14 * 22

• 在采取2 ²共同：

22 * (10 + 11 + 12 + 13 + 14) 
=> totalDigitstotalDigits * (Sum of AP)

• AP的总和=(项数/ 2)*(第一项+最后一项)，即(R–L + 1)*(L + R)/ 2 。
• 因此，所需的总和为：

countDigits(L)countDigits(L) * (L + R) * (R - L + 1) / 2

下面是上述方法的实现：

C++

// C++ program to find the required sum
 
#include 
using namespace std;
#define MOD 1000000007
 
// Function to return the required sum
int rangeSum(int l, int r)
{
    int a = 1, b = 9, res = 0;
 
    // Iterating for all the number
    // of digits from 1 to 10
    for (int i = 1; i <= 10; i++) {
        int L = max(l, a);
        int R = min(r, b);
 
        // If the range is valid
        if (L <= R) {
 
            // Sum of AP
            int sum = (L + R) * (R - L + 1) / 2;
            res += pow(i, i) * (sum % MOD);
            res %= MOD;
        }
 
        // Computing the next minimum and maximum
        // numbers by for the (i+1)-th digit
        a = a * 10;
        b = b * 10 + 9;
    }
    return res;
}
 
// Driver code
int main()
{
    int l = 98, r = 102;
    cout << rangeSum(l, r);
    return 0;
}

Java

// Java program to find the required sum
import java.util.*;
 
class GFG{
static final int MOD = 1000000007;
  
// Function to return the required sum
static int rangeSum(int l, int r)
{
    int a = 1, b = 9, res = 0;
  
    // Iterating for all the number
    // of digits from 1 to 10
    for (int i = 1; i <= 10; i++) {
        int L = Math.max(l, a);
        int R = Math.min(r, b);
  
        // If the range is valid
        if (L <= R) {
  
            // Sum of AP
            int sum = (L + R) * (R - L + 1) / 2;
            res += Math.pow(i, i) * (sum % MOD);
            res %= MOD;
        }
  
        // Computing the next minimum and maximum
        // numbers by for the (i+1)-th digit
        a = a * 10;
        b = b * 10 + 9;
    }
    return res;
}
  
// Driver code
public static void main(String[] args)
{
    int l = 98, r = 102;
    System.out.print(rangeSum(l, r));
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python 3 program to find the required sum
  
MOD = 1000000007
  
# Function to return the required sum
def rangeSum(l, r):
 
    a = 1
    b = 9
    res = 0
  
    # Iterating for all the number
    # of digits from 1 to 10
    for i in range(1, 11):
        L = max(l, a)
        R = min(r, b)
  
        # If the range is valid
        if (L <= R):
  
            # Sum of AP
            sum = (L + R) * (R - L + 1) // 2
            res += pow(i, i) * (sum % MOD)
            res %= MOD
  
        # Computing the next minimum and maximum
        # numbers by for the (i+1)-th digit
        a = a * 10
        b = b * 10 + 9
    return res
  
# Driver code
if __name__ == "__main__":
     
    l = 98
    r = 102
    print(rangeSum(l, r))
     
# This code is contributed by chitranayal

C#

// C# program to find the required sum
using System;
 
class GFG{
static readonly int MOD = 1000000007;
   
// Function to return the required sum
static int rangeSum(int l, int r)
{
    int a = 1, b = 9, res = 0;
   
    // Iterating for all the number
    // of digits from 1 to 10
    for (int i = 1; i <= 10; i++) {
        int L = Math.Max(l, a);
        int R = Math.Min(r, b);
   
        // If the range is valid
        if (L <= R) {
   
            // Sum of AP
            int sum = (L + R) * (R - L + 1) / 2;
            res += (int)Math.Pow(i, i) * (sum % MOD);
            res %= MOD;
        }
   
        // Computing the next minimum and maximum
        // numbers by for the (i+1)-th digit
        a = a * 10;
        b = b * 10 + 9;
    }
    return res;
}
   
// Driver code
public static void Main(String[] args)
{
    int l = 98, r = 102;
    Console.Write(rangeSum(l, r));
}
}
  
// This code is contributed by Rajput-Ji

Javascript

 
// Javascript program to find the required sum
 
MOD=1000000007
 
// Function to return the required sum
function rangeSum(l, r)
{
    var a = 1, b = 9, res = 0;
 
    // Iterating for all the number
    // of digits from 1 to 10
    for (var i = 1; i <= 10; i++) {
        var L = Math.max(l, a);
        var R = Math.min(r, b);
 
        // If the range is valid
        if (L <= R) {
 
            // Sum of AP
            var sum = (L + R) * (R - L + 1) / 2;
            res += Math.pow(i, i) * (sum % MOD);
            res %= MOD;
        }
 
        // Computing the next minimum and maximum
        // numbers by for the (i+1)-th digit
        a = a * 10;
        b = b * 10 + 9;
    }
    return res;
}
 
// Driver code
var l = 98, r = 102;
document.write( rangeSum(l, r));
 
// This code is contributed by noob2000.

输出：

8969