计算给定范围内数字积为K的数字

给定三个正整数L ， R和K ，任务是计算数字乘积等于K的[L，R]范围内的数字

例子：

Input: L = 1, R = 130, K = 14
Output: 3
Explanation: 
Numbers in the range [1, 100] whose sum of digits is K(= 14) are: 
27 => 2 * 7 = 14 
72 => 7 * 2 = 14 
127 => 1 * 2 * 7 = 14 
Therefore, the required output is 3.

Input: L = 20, R = 10000, K = 14
Output: 20

天真的方法：解决此问题的最简单方法是遍历[L，R]范围内的所有数字，并对每个数字检查其数字乘积是否等于K。如果发现为真，则增加计数。最后，打印获得的计数。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to find the product
// of digits of a number
int prodOfDigit(int N)
{
    // Stores product of
    // digits of N
    int res = 1;
 
    while (N) {
 
        // Update res
        res = res * (N % 10);
 
        // Update N
        N /= 10;
    }
 
    return res;
}
 
// Function to count numbers in the range
// [0, X] whose product of digit is K
int cntNumRange(int L, int R, int K)
{
    // Stores count of numbers in the range
    // [L, R] whose product of digit is K
    int cnt = 0;
 
    // Iterate over the range [L, R]
    for (int i = L; i <= R; i++) {
 
        // If product of digits of
        // i equal to K
        if (prodOfDigit(i) == K) {
 
            // Update cnt
            cnt++;
        }
    }
 
    return cnt;
}
 
// Driver Code
int main()
{
    int L = 20, R = 10000, K = 14;
    cout << cntNumRange(L, R, K);
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
// Function to find the product
// of digits of a number
static int prodOfDigit(int N)
{
   
    // Stores product of
    // digits of N
    int res = 1;
    while (N > 0)
    {
 
        // Update res
        res = res * (N % 10);
 
        // Update N
        N /= 10;
    }
 
    return res;
}
 
// Function to count numbers in the range
// [0, X] whose product of digit is K
static int cntNumRange(int L, int R, int K)
{
   
    // Stores count of numbers in the range
    // [L, R] whose product of digit is K
    int cnt = 0;
 
    // Iterate over the range [L, R]
    for (int i = L; i <= R; i++)
    {
 
        // If product of digits of
        // i equal to K
        if (prodOfDigit(i) == K)
        {
 
            // Update cnt
            cnt++;
        }
    }
    return cnt;
}
 
// Driver Code
public static void main(String[] args)
{
    int L = 20, R = 10000, K = 14;
    System.out.print(cntNumRange(L, R, K));
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program to implement
# the above approach
 
# Function to find the product
# of digits of a number
def prodOfDigit(N):
     
    # Stores product of
    # digits of N
    res = 1
 
    while (N):
         
        # Update res
        res = res * (N % 10)
 
        # Update N
        N //= 10
 
    return res
 
# Function to count numbers in the range
# [0, X] whose product of digit is K
def cntNumRange(L, R, K):
     
    # Stores count of numbers in the range
    # [L, R] whose product of digit is K
    cnt = 0
 
    # Iterate over the range [L, R]
    for i in range(L, R + 1):
         
        # If product of digits of
        # i equal to K
        if (prodOfDigit(i) == K):
 
            # Update cnt
            cnt += 1
 
    return cnt
 
# Driver Code
if __name__ == '__main__':
     
    L, R, K = 20, 10000, 14
     
    print(cntNumRange(L, R, K))
 
# This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach
using System;
  
class GFG{
      
// Function to find the product
// of digits of a number
static int prodOfDigit(int N)
{
     
    // Stores product of
    // digits of N
    int res = 1;
     
    while (N > 0)
    {
         
        // Update res
        res = res * (N % 10);
  
        // Update N
        N /= 10;
    }
    return res;
}
  
// Function to count numbers in the range
// [0, X] whose product of digit is K
static int cntNumRange(int L, int R, int K)
{
    
    // Stores count of numbers in the range
    // [L, R] whose product of digit is K
    int cnt = 0;
  
    // Iterate over the range [L, R]
    for(int i = L; i <= R; i++)
    {
  
        // If product of digits of
        // i equal to K
        if (prodOfDigit(i) == K)
        {
  
            // Update cnt
            cnt++;
        }
    }
    return cnt;
}
  
// Driver Code
static void Main()
{
    int L = 20, R = 10000, K = 14;
     
    Console.WriteLine(cntNumRange(L, R, K));
}
}
 
// This code is contributed by code_hunt

Javascript

// Javascript program to implement
// the above approach
 
// Function to find the product
// of digits of a number
function prodOfDigit(N)
{
    // Stores product of
    // digits of N
    let res = 1;
 
    while (N) {
 
        // Update res
        res = res * (N % 10);
 
        // Update N
        N = Math.floor(N/10);
    }
 
    return res;
}
 
// Function to count numbers in the range
// [0, X] whose product of digit is K
function cntNumRange(L, R, K)
{
 
    // Stores count of numbers in the range
    // [L, R] whose product of digit is K
    let cnt = 0;
 
    // Iterate over the range [L, R]
    for (let i = L; i <= R; i++) {
 
        // If product of digits of
        // i equal to K
        if (prodOfDigit(i) == K)
        {
 
            // Update cnt
            cnt++;
        }
    }
 
    return cnt;
}
 
// Driver Code
    let L = 20, R = 10000, K = 14;
    document.write(cntNumRange(L, R, K));
 
// This code is contributed by manoj.

---

输出：

20

时间复杂度： O(R – L +1)* log ₁₀ (R)
辅助空间： O(1)

高效的方法：为了优化上述方法，我们的想法是使用Digit DP。以下是Digit DP的动态编程状态：

Dynamic programming states are: 
prod: Represents sum of digits. 
tight: Check if sum of digits exceed K or not. 
end: Stores the maximum possible value of i^th digit of a number. 
st: Check if a number contains leading 0 or not. 
 

以下是动态编程状态的递归关系：

• 如果i == 0且st == 0：

$cntNum(i, prod, st, tight)$
$= \sum^{9}_{j=0} cntNum(i + 1, prod, false, tight & (j == end)$
 

cntNum(N, K, st, tight): Returns the count of numbers in the range [0, X] whose product of digits is K.

 

• 除此以外，

$cntNum(i, prod, st, tight)$
$= \sum^{9}_{j=0} cntNum(i + 1, (prod * j), true, tight & (j == end)$
 

cntNum(N, K, st, tight): Returns the count of numbers in the range [0, X] whose product of digits is K.

 

请按照以下步骤解决问题：

1. 初始化3D数组dp [N] [K] [tight]，以计算和存储上述递归关系的所有子问题的值。
2. 最后，返回dp [N] [sum] [tight]的值。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
#define M 100
 
// Function to count numbers in the range
// [0, X] whose product of digit is K
int cntNum(string X, int i, int prod, int K,
           int st, int tight, int dp[M][M][2][2])
{
    // If count of digits in a number
    // greater than count of digits in X
    if (i >= X.length() || prod > K) {
 
        // If product of digits of a
        // number equal to K
        return prod == K;
    }
 
    // If overlapping subproblems
    // already occurred
    if (dp[prod][i][tight][st] != -1) {
        return dp[prod][i][tight][st];
    }
 
    // Stores count of numbers whose
    // product of digits is K
    int res = 0;
 
    // Check if the numbers
    // exceeds K or not
    int end = tight ? X[i] - '0' : 9;
 
    // Iterate over all possible
    // value of i-th digits
    for (int j = 0; j <= end; j++) {
 
        // if number contains leading 0
        if (j == 0 && !st) {
 
            // Update res
            res += cntNum(X, i + 1, prod, K,
                          false, (tight & (j == end)), dp);
        }
 
        else {
 
            // Update res
            res += cntNum(X, i + 1, prod * j, K,
                          true, (tight & (j == end)), dp);
        }
 
        // Update res
    }
 
    // Return res
    return dp[prod][i][tight][st] = res;
}
 
// Utility function to count the numbers in
// the range [L, R] whose prod of digits is K
int UtilCntNumRange(int L, int R, int K)
{
    // Stores numbers in the form
    // of string
    string str = to_string(R);
 
    // Stores overlapping subproblems
    int dp[M][M][2][2];
 
    // Initialize dp[][][] to -1
    memset(dp, -1, sizeof(dp));
 
    // Stores count of numbers in
    // the range [0, R] whose
    // product of  digits is k
    int cntR = cntNum(str, 0, 1, K,
                      false, true, dp);
 
    // Update str
    str = to_string(L - 1);
 
    // Initialize dp[][][] to -1
    memset(dp, -1, sizeof(dp));
 
    // Stores count of numbers in
    // the range [0, L - 1] whose
    // product of  digits is k
    int cntL = cntNum(str, 0, 1, K,
                      false, true, dp);
 
    return (cntR - cntL);
}
 
// Driver Code
int main()
{
    int L = 20, R = 10000, K = 14;
    cout << UtilCntNumRange(L, R, K);
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
     
static final int M = 100;
 
// Function to count numbers in the range
// [0, X] whose product of digit is K
static int cntNum(String X, int i, int prod, int K,
                    int st, int tight, int [][][][]dp)
{
     
    // If count of digits in a number
    // greater than count of digits in X
    if (i >= X.length() || prod > K)
    {
         
        // If product of digits of a
        // number equal to K
        return prod == K ? 1 : 0;
    }
 
    // If overlapping subproblems
    // already occurred
    if (dp[prod][i][tight][st] != -1)
    {
        return dp[prod][i][tight][st];
    }
 
    // Stores count of numbers whose
    // product of digits is K
    int res = 0;
 
    // Check if the numbers
    // exceeds K or not
    int end = tight > 0 ? X.charAt(i) - '0' : 9;
 
    // Iterate over all possible
    // value of i-th digits
    for(int j = 0; j <= end; j++)
    {
         
        // If number contains leading 0
        if (j == 0 && st == 0)
        {
             
            // Update res
            res += cntNum(X, i + 1, prod, K, 0,
                  (tight & ((j == end) ? 1 : 0)), dp);
        }
        else
        {
             
            // Update res
            res += cntNum(X, i + 1, prod * j, K, 1,
                  (tight & ((j == end) ? 1 : 0)), dp);
        }
    }
 
    // Return res
    return dp[prod][i][tight][st] = res;
}
 
// Utility function to count the numbers in
// the range [L, R] whose prod of digits is K
static int UtilCntNumRange(int L, int R, int K)
{
     
    // Stores numbers in the form
    // of String
    String str = String.valueOf(R);
 
    // Stores overlapping subproblems
    int [][][][]dp = new int[M][M][2][2];
 
    // Initialize dp[][][] to -1
    for(int i = 0; i < M; i++)
    {
        for(int j = 0; j < M; j++)
        {
            for(int k = 0; k < 2; k++)
                for(int l = 0; l < 2; l++)
                    dp[i][j][k][l] = -1;
        }
    }
 
    // Stores count of numbers in
    // the range [0, R] whose
    // product of  digits is k
    int cntR = cntNum(str, 0, 1, K,
                      0, 1, dp);
 
    // Update str
    str = String.valueOf(L - 1);
 
    // Initialize dp[][][] to -1
    for(int i = 0;i

Python3

# Python 3 program to implement
# the above approach
M = 100
 
# Function to count numbers in the range
# [0, X] whose product of digit is K
def cntNum(X, i, prod, K, st, tight, dp):
    end = 0
     
    # If count of digits in a number
    # greater than count of digits in X
    if (i >= len(X) or prod > K):
       
        # If product of digits of a
        # number equal to K
        if(prod == K):
          return 1
        else:
          return 0
 
    # If overlapping subproblems
    # already occurred
    if (dp[prod][i][tight][st] != -1):
        return dp[prod][i][tight][st]
 
    # Stores count of numbers whose
    # product of digits is K
    res = 0
 
    # Check if the numbers
    # exceeds K or not
    if(tight != 0):
        end = ord(X[i]) - ord('0')
 
    # Iterate over all possible
    # value of i-th digits
    for j in range(end + 1):
       
        # if number contains leading 0
        if (j == 0 and st == 0):
            # Update res
            res += cntNum(X, i + 1, prod, K, False, (tight & (j == end)), dp)
 
        else:
            # Update res
            res += cntNum(X, i + 1, prod * j, K, True, (tight & (j == end)), dp)
        # Update res
 
    # Return res
    dp[prod][i][tight][st] = res
    return res
 
# Utility function to count the numbers in
# the range [L, R] whose prod of digits is K
def UtilCntNumRange(L, R, K):
    global M
     
    # Stores numbers in the form
    # of string
    str1 = str(R)
 
    # Stores overlapping subproblems
    dp = [[[[-1 for i in range(2)] for j in range(2)] for k in range(M)] for l in range(M)]
 
    # Stores count of numbers in
    # the range [0, R] whose
    # product of  digits is k
    cntR = cntNum(str1, 0, 1, K, False, True, dp)
 
    # Update str
    str1 = str(L - 1)
    dp = [[[[-1 for i in range(2)] for j in range(2)] for k in range(M)] for l in range(M)]
 
    # Stores count of numbers in
    cntR = 20
     
    # the range [0, L - 1] whose
    # product of  digits is k
    cntL = cntNum(str1, 0, 1, K, False, True, dp)
    return (cntR - cntL)
 
# Driver Code
if __name__ == '__main__':
    L = 20
    R = 10000
    K = 14
    print(UtilCntNumRange(L, R, K))
 
    # This code is contributed by SURENDRA_GANGWAR.

C＃

// C# program to implement
// the above approach
using System;
class GFG
{
 
  static readonly int M = 100;
 
  // Function to count numbers in the range
  // [0, X] whose product of digit is K
  static int cntNum(String X, int i, int prod, int K,
                    int st, int tight, int [,,,]dp)
  {
 
    // If count of digits in a number
    // greater than count of digits in X
    if (i >= X.Length || prod > K)
    {
 
      // If product of digits of a
      // number equal to K
      return prod == K ? 1 : 0;
    }
 
    // If overlapping subproblems
    // already occurred
    if (dp[prod, i, tight, st] != -1)
    {
      return dp[prod, i, tight, st];
    }
 
    // Stores count of numbers whose
    // product of digits is K
    int res = 0;
 
    // Check if the numbers
    // exceeds K or not
    int end = tight > 0 ? X[i] - '0' : 9;
 
    // Iterate over all possible
    // value of i-th digits
    for(int j = 0; j <= end; j++)
    {
 
      // If number contains leading 0
      if (j == 0 && st == 0)
      {
 
        // Update res
        res += cntNum(X, i + 1, prod, K, 0,
                      (tight & ((j == end) ? 1 : 0)), dp);
      }
      else
      {
 
        // Update res
        res += cntNum(X, i + 1, prod * j, K, 1,
                      (tight & ((j == end) ? 1 : 0)), dp);
      }
    }
 
    // Return res
    return dp[prod, i, tight, st] = res;
  }
 
  // Utility function to count the numbers in
  // the range [L, R] whose prod of digits is K
  static int UtilCntNumRange(int L, int R, int K)
  {
 
    // Stores numbers in the form
    // of String
    String str = String.Join("", R);
 
    // Stores overlapping subproblems
    int [,,,]dp = new int[M, M, 2, 2];
 
    // Initialize [,]dp[] to -1
    for(int i = 0; i < M; i++)
    {
      for(int j = 0; j < M; j++)
      {
        for(int k = 0; k < 2; k++)
          for(int l = 0; l < 2; l++)
            dp[i, j, k, l] = -1;
      }
    }
 
    // Stores count of numbers in
    // the range [0, R] whose
    // product of  digits is k
    int cntR = cntNum(str, 0, 1, K,
                      0, 1, dp);
 
    // Update str
    str = String.Join("", L - 1);
 
    // Initialize [,]dp[] to -1
    for(int i = 0; i < M; i++)
    {
      for(int j = 0; j < M; j++)
      {
        for(int k = 0; k < 2; k++)
          for(int l = 0; l < 2; l++)
            dp[i, j, k, l] = -1;
      }
    }
 
    // Stores count of numbers in
    // the range [0, L - 1] whose
    // product of  digits is k
    int cntL = cntNum(str, 0, 1, K,
                      0, 1, dp);
 
    return (cntR - cntL);
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    int L = 20, R = 10000, K = 14;
 
    Console.Write(UtilCntNumRange(L, R, K));
  }
}
 
 
 
// This code is contributed by 29AjayKumar

输出：

20

时间复杂度： O(K * log ₁₀ (R)* 10 * 4)
辅助空间： O(K * log ₁₀ (R)* 4)