给定范围内的数字计数，可以表示为加和后的数字乘以数字的乘方

给定一个由{L，R}形式的查询组成的数组arr [] ，每个查询的任务是对[L，R]范围内的数字进行计数，该数字可以表示为其幂加和的数字的总和数位数。

例子：

Input: arr[][] = {{8, 11}}
Output: 2
Explanation:
From the given range [1, 9], the numbers that can be expressed as the sum of its digit raised to the power of count of digits are: 

1. 8: Sum of digits = 8, Count of digit = 1. Therefore, 8¹ is equal to the given number.
2. 9: Sum of digits = 9, Count of digit = 1. Therefore, 9¹ is equal to the given number.

Therefore, the count of such numbers from the given range is 2.

Input: arr[][] = {{10, 100}, {1, 400}}
Output: 0 11 

天真的方法：最简单的方法是针对每个查询在arr [i] [0]到arr [i] [1]的范围内进行迭代，并打印这些数字的计数。

时间复杂度： O(Q *(R – L)* log ₁₀ R)，其中R和L表示最长范围的极限。
辅助空间： O(1)

高效的方法：可以通过预先计算并存储所有数字来优化上述方法，无论这些数字是否可以表示为其数字的和加到数字的幂上。最后，有效地打印每个查询的计数。
请按照以下步骤解决问题：

• 初始化一个辅助数组，例如ans [] ，以存储在ans [i]上，是否可以将i表示为其位数的和，即升为位数的幂。
• 迭代范围[ ^{1，10 6} ]并相应地更新数组ans [] 。
• 将数组ans []转换为前缀和数组。
• 遍历给定的查询数组arr [] ，对于每个查询{arr [i] [0]，arr [i] [1]} ，打印(ans [arr [i] [1]] – ans [arr [i] [1] – 1])作为数字的结果计数，可以表示为数字的总和乘以数字的幂。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
#define R 100005
int arr[R];
 
// Function to check if a number N can be
// expressed as sum of its digits raised
// to the power of the count of digits
bool canExpress(int N)
{
    int temp = N;
 
    // Stores the number of digits
    int n = 0;
 
    while (N != 0) {
        N /= 10;
        n++;
    }
 
    // Stores the resultant number
    N = temp;
 
    int sum = 0;
 
    while (N != 0) {
        sum += pow(N % 10, n);
        N /= 10;
    }
 
    // Return true if both the
    // numbers are same
    return (sum == temp);
}
 
// Function to precompute and store
// for all numbers whether they can
// be expressed
void precompute()
{
    // Mark all the index which
    // are plus perfect number
    for (int i = 1; i < R; i++) {
 
        // If true, then update the
        // value at this index
        if (canExpress(i)) {
            arr[i] = 1;
        }
    }
 
    // Compute prefix sum of the array
    for (int i = 1; i < R; i++) {
        arr[i] += arr[i - 1];
    }
}
 
// Function to count array elements that
// can be expressed as the sum of digits
// raised to the power of count of digits
void countNumbers(int queries[][2], int N)
{
    // Precompute the results
    precompute();
 
    // Traverse the queries
    for (int i = 0; i < N; i++) {
 
        int L1 = queries[i][0];
        int R1 = queries[i][1];
 
        // Print the resultant count
        cout << (arr[R1] - arr[L1 - 1])
             << ' ';
    }
}
 
// Driver Code
int main()
{
    int queries[][2] = {
        { 1, 400 },
        { 1, 9 }
    };
    int N = sizeof(queries)
            / sizeof(queries[0]);
    countNumbers(queries, N);
 
    return 0;
}

C#

// C# program for the above approach
using System;
 
class GFG{
     
static int R = 100005;
static int[] arr = new int[R];
 
// Function to check if a number N can be
// expressed as sum of its digits raised
// to the power of the count of digits
public static bool canExpress(int N)
{
    int temp = N;
 
    // Stores the number of digits
    int n = 0;
 
    while (N != 0)
    {
        N /= 10;
        n++;
    }
 
    // Stores the resultant number
    N = temp;
 
    int sum = 0;
 
    while (N != 0)
    {
        sum += ((int)Math.Pow(N % 10, n));
        N /= 10;
    }
 
    // Return true if both the
    // numbers are same
      if (sum == temp)
      return true;
   
    return false;
}
 
// Function to precompute and store
// for all numbers whether they can
// be expressed
public static void precompute()
{
 
    // Mark all the index which
    // are plus perfect number
    for(int i = 1; i < R; i++)
    {
         
        // If true, then update the
        // value at this index
        if (canExpress(i))
        {
            arr[i] = 1;
        }
    }
 
    // Compute prefix sum of the array
    for(int i = 1; i < R; i++)
    {
        arr[i] += arr[i - 1];
    }
}
 
// Function to count array elements that
// can be expressed as the sum of digits
// raised to the power of count of digits
public static void countNumbers(int[,] queries, int N)
{
     
    // Precompute the results
    precompute();
 
    // Traverse the queries
    for(int i = 0; i < N; i++)
    {
        int L1 = queries[i, 0];
        int R1 = queries[i, 1];
 
        // Print the resultant count
        Console.Write((arr[R1] - arr[L1 - 1]) + " ");
    }
}
 
// Driver Code
static public void Main()
{
    int[,] queries = { { 1, 400 }, { 1, 9 } };
    int N = queries.GetLength(0);
 
    // Function call
    countNumbers(queries, N);
}
}
 
// This code is contributed by Dharanendra L V.

Python3

# Python 3 program for the above approach
R = 100005
arr = [0 for i in range(R)]
 
# Function to check if a number N can be
# expressed as sum of its digits raised
# to the power of the count of digits
def canExpress(N):
    temp = N
 
    # Stores the number of digits
    n = 0
    while (N != 0):
        N //= 10
        n += 1
 
    # Stores the resultant number
    N = temp
    sum = 0
    while (N != 0):
        sum += pow(N % 10, n)
        N //= 10
 
    # Return true if both the
    # numbers are same
    return (sum == temp)
 
# Function to precompute and store
# for all numbers whether they can
# be expressed
def precompute():
   
    # Mark all the index which
    # are plus perfect number
    for i in range(1, R, 1):
       
        # If true, then update the
        # value at this index
        if(canExpress(i)):
            arr[i] = 1
 
    # Compute prefix sum of the array
    for i in range(1,R,1):
        arr[i] += arr[i - 1]
 
# Function to count array elements that
# can be expressed as the sum of digits
# raised to the power of count of digits
def countNumbers(queries, N):
   
    # Precompute the results
    precompute()
 
    # Traverse the queries
    for i in range(N):
        L1 = queries[i][0]
        R1 = queries[i][1]
 
        # Print the resultant count
        print((arr[R1] - arr[L1 - 1]),end = " ")
 
# Driver Code
if __name__ == '__main__':
    queries = [[1, 400],[1, 9]]
    N = len(queries)
    countNumbers(queries, N)
 
    # This code is contributed by SURENDRA_GANGWAR.

Javascript

// javascript program for the above approach
    var R = 100005;
    var arr = Array(R).fill(0);
 
    // Function to check if a number N can be
    // expressed as sum of its digits raised
    // to the power of the count of digits
    function canExpress(N) {
        var temp = N;
 
        // Stores the number of digits
        var n = 0;
 
        while (N != 0) {
            N = parseInt(N/10);
            n++;
        }
 
        // Stores the resultant number
        N = temp;
 
        var sum = 0;
 
        while (N != 0) {
            sum += Math.pow(N % 10, n);
            N = parseInt(N/10);
        }
 
        // Return true if both the
        // numbers are same
        return (sum == temp);
    }
 
    // Function to precompute and store
    // for all numbers whether they can
    // be expressed
    function precompute() {
 
        // Mark all the index which
        // are plus perfect number
        for (var i = 1; i < R; i++) {
 
            // If true, then update the
            // value at this index
            if (canExpress(i)) {
                arr[i] = 1;
            }
        }
 
        // Compute prefix sum of the array
        for (var i = 1; i < R; i++) {
            arr[i] += arr[i - 1];
        }
    }
 
    // Function to count array elements that
    // can be expressed as the sum of digits
    // raised to the power of count of digits
    function countNumbers(queries , N) {
        // Precompute the results
        precompute();
 
        // Traverse the queries
        for (var i = 0; i < N; i++) {
 
            var L1 = queries[i][0];
            var R1 = queries[i][1];
 
            // Prvar the resultant count
            document.write((arr[R1] - arr[L1 - 1]) + " ");
        }
    }
 
    // Driver Code
    var queries = [ [ 1, 400 ], [ 1, 9 ] ];
    var N = queries.length;
 
    // function call
    countNumbers(queries, N);
 
// This code is contributed by Princi Singh

输出：

12 9

时间复杂度： O(Q + 10 ⁶ )
辅助空间： O(10 ⁶ )