数字总和为Y | [L，R]范围内的数字计数套装2

给定三个正整数L ， R和Y ，任务是计算数字总和等于Y的[L，R]范围内的数字

例子：

Input: L = 500, R = 1000, Y = 6
Output: 3
Explanation: 
Numbers in the range [500, 600] whose sum of digits is Y(= 6) are: 
501 = 5 + 0 + 1 = 6 
510 = 5 + 1 + 0 = 6 
600 = 6 + 0 + 0 = 6 
Therefore, the required output is 3.

Input: L = 20, R = 10000, Y = 14
Output: 540

天真的方法：请参阅上一篇文章，通过遍历[L，R]范围内的所有数字来解决此问题，并对每个数字检查其数字总和是否等于Y。如果发现为真，则增加计数。最后，打印获得的计数。

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，我们的想法是通过以下递归关系使用Digit DP：

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

where, sum: Represents sum of digits. 
tight: Check if sum of digits exceed Y or not. 
end: Stores the maximum possible value of i^th digit of a number. 
cntNum(N, Y, tight): Returns the count of numbers in the range [0, X] whose sum of digits is Y.

请按照以下步骤解决问题：

1. 初始化3D数组dp [N] [Y] [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++

// CPP program for the above approach
#include 
using namespace std;
 
#define M 1000
 
// Function to find the sum of digits
// of numbers in the range [0, X]
int cntNum(string X, int i, int sum,
           int tight, int dp[M][M][2])
{
    // Check if count of digits in a number
    // greater than count of digits in X
    if (i >= X.length() || sum < 0) {
 
        // If sum of digits of a
        // number is equal to Y
        if (sum == 0) {
            return 1;
        }
 
        return 0;
    }
 
    // Check if current subproblem has
    // already been computed
    if (dp[sum][i][tight] != -1) {
        return dp[sum][i][tight];
    }
 
    // Stores count of numbers whose
    // sum of digits is Y
    int res = 0;
 
    // Check if the number
    // exceeds Y or not
    int end = tight ? X[i] - '0' : 9;
 
    // Iterate over all possible
    // values of i-th digits
    for (int j = 0; j <= end; j++) {
 
        // Update res
        res += cntNum(X, i + 1, sum - j,
                      (tight & (j == end)), dp);
    }
 
    // Return res
    return dp[sum][i][tight]=res;
}
 
// Utility function to count the numbers in
// the range [L, R] whose sum of digits is Y
int UtilCntNumRange(int L, int R, int Y)
{
    // Base Case
    if (R == 0 && Y == 0) {
 
        return 1;
    }
 
    // Stores numbers in the form
    // of its equivalent string
    string str = to_string(R);
 
    // Stores overlapping subproblems
    int dp[M][M][2];
 
    // Initialize dp[][][]
    memset(dp, -1, sizeof(dp));
 
    // Stores count of numbers
    // in the range [0, R]
    int cntR = cntNum(str, 0, Y,
                      true, dp);
 
    // Update str
    str = to_string(L - 1);
 
    // Initialize dp[][][]
    memset(dp, -1, sizeof(dp));
 
    // Stores count of numbers in
    // the range [0, L - 1]
    int cntL = cntNum(str, 0, Y,
                      true, dp);
 
    return (cntR - cntL);
}
 
// Driver Code
int main()
{
    int L = 20, R = 10000, Y = 14;
    cout << UtilCntNumRange(L, R, Y);
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
static final int M = 1000;
 
// Function to find the sum of digits
// of numbers in the range [0, X]
static int cntNum(String X, int i, int sum,
           int tight, int dp[][][])
{
    // Check if count of digits in a number
    // greater than count of digits in X
    if (i >= X.length() || sum < 0) {
 
        // Check Iif sum of digits of a
        // number is equal to Y
        if (sum == 0) {
            return 1;
        }
 
        return 0;
    }
 
    // Check if current subproblem has
    // already been computed
    if (dp[sum][i][tight] != -1) {
        return dp[sum][i][tight];
    }
 
    // Stores count of numbers whose
    // sum of digits is Y
    int res = 0;
 
    // Check if the number
    // exceeds Y or not
    int end = tight != 0 ? X.charAt(i) - '0' : 9;
 
    // Iterate over all possible
    // values of i-th digits
    for (int j = 0; j <= end; j++) {
 
        // Update res
        res += cntNum(X, i + 1, sum - j,
                      (tight > 0 & (j == end)) ==
                               true ? 1 : 0, dp);
    }
 
    // Return res
    return dp[sum][i][tight]=res;
}
 
// Utility function to count the numbers in
// the range [L, R] whose sum of digits is Y
static int UtilCntNumRange(int L, int R, int Y)
{
    // Base Case
    if (R == 0 && Y == 0) {
 
        return 1;
    }
 
    // Stores numbers in the form
    // of its equivalent String
    String str = String.valueOf(R);
 
    // Stores overlapping subproblems
    int [][][]dp = new int[M][M][2];
 
    // Initialize dp[][][]
    for(int i = 0; i < M; i++)
    {
        for (int j = 0; j < M; j++) {
            for (int k = 0; k < 2; k++)
                dp[i][j][k] = -1;
        }
    }
 
    // Stores count of numbers
    // in the range [0, R]
    int cntR = cntNum(str, 0, Y,
                      1, dp);
 
    // Update str
    str = String.valueOf(L - 1);
 
    // Initialize dp[][][]
    for(int i = 0; i < M; i++)
    {
        for (int j = 0; j < M; j++) {
            for (int k = 0; k < 2; k++)
                dp[i][j][k] = -1;
        }
    }
 
    // Stores count of numbers in
    // the range [0, L - 1]
    int cntL = cntNum(str, 0, Y,
                      1, dp);
 
    return (cntR - cntL);
}
 
// Driver Code
public static void main(String[] args)
{
    int L = 20, R = 10000, Y = 14;
    System.out.print(UtilCntNumRange(L, R, Y));
}
}
 
// This code is contributed by shikhasingrajput

Python3

# Python program for the above approach
M = 1000
 
# Function to find the sum of digits
# of numbers in the range [0, X]
def cntNum(X, i, sum, tight, dp):
   
    # Check if count of digits in a number
    # greater than count of digits in X
    if (i >= len(X) or sum < 0):
 
        # Check if sum of digits of a
        # number is equal to Y
        if (sum == 0):
            return 1
 
        return 0
 
    # Check if current subproblem has
    # already been comrputed
    if (dp[sum][i][tight] != -1):
        return dp[sum][i][tight]
 
    # Stores count of numbers whose
    # sum of digits is Y
    res, end = 0, 9
 
    # Check if the number
    # exceeds Y or not
    if tight:
        end = ord(X[i]) - ord('0')
    # end = tight ? X[i] - '0' : 9;
 
    # Iterate over all possible
    # values of i-th digits
    for j in range(end + 1):
 
        # Update res
        res += cntNum(X, i + 1, sum - j,
                      (tight & (j == end)), dp)
 
    # Return res
    dp[sum][i][tight] = res
    return res
 
# Utility function to count the numbers in
# the range [L, R] whose sum of digits is Y
def UtilCntNumRange(L, R, Y):
   
    # Base Case
    if (R == 0 and Y == 0):
 
        return 1
 
    # Stores numbers in the form
    # of its equivalent
    strr = str(R)
 
    # Stores overlapping subproblems
    dp = [[[-1 for i in range(2)] for i in range(M)]
                                  for i in range(M)]
 
    # Initialize dp[][][]
    # memset(dp, -1, sizeof(dp))
 
    # Stores count of numbers
    # in the range [0, R]
    cntR = cntNum(strr, 0, Y, True, dp)
 
    # Update str
    strr = str(L - 1)
 
    # Initialize dp[][][]
    # memset(dp, -1, sizeof(dp))
 
    # Stores count of numbers in
    # the range [0, L - 1]
    cntL = cntNum(strr, 0, Y, True, dp)
 
    return (cntR - cntL)
 
# Driver Code
if __name__ == '__main__':
    L, R, Y = 20, 10000, 14
    print(UtilCntNumRange(L, R, Y))
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
 
class GFG{
 
static readonly int M = 1000;
 
// Function to find the sum of digits
// of numbers in the range [0, X]
static int cntNum(String X, int i, int sum,
                 int tight, int [,,]dp)
{
     
    // Check if count of digits in a number
    // greater than count of digits in X
    if (i >= X.Length || sum < 0)
    {
         
        // Check if sum of digits of a
        // number is equal to Y
        if (sum == 0)
        {
            return 1;
        }
        return 0;
    }
 
    // Check if current subproblem has
    // already been computed
    if (dp[sum, i, tight] != -1)
    {
        return dp[sum, i, tight];
    }
 
    // Stores count of numbers whose
    // sum of digits is Y
    int res = 0;
 
    // Check if the number
    // exceeds Y or not
    int end = tight != 0 ? X[i] - '0' : 9;
 
    // Iterate over all possible
    // values of i-th digits
    for(int j = 0; j <= end; j++)
    {
         
        // Update res
        res += cntNum(X, i + 1, sum - j,
                    (tight > 0 & (j == end)) ==
                      true ? 1 : 0, dp);
    }
 
    // Return res
    return dp[sum][i][tight] = res;
}
 
// Utility function to count the numbers in
// the range [L, R] whose sum of digits is Y
static int UtilCntNumRange(int L, int R, int Y)
{
     
    // Base Case
    if (R == 0 && Y == 0)
    {
        return 1;
    }
 
    // Stores numbers in the form
    // of its equivalent String
    String str = String.Join("", R);
     
    // Stores overlapping subproblems
    int [,,]dp = new int[M, M, 2];
 
    // Initialize [,]dp[]
    for(int i = 0; i < M; i++)
    {
        for(int j = 0; j < M; j++)
        {
            for(int k = 0; k < 2; k++)
                dp[i, j, k] = -1;
        }
    }
 
    // Stores count of numbers
    // in the range [0, R]
    int cntR = cntNum(str, 0, Y,
                      1, dp);
 
    // Update str
    str = String.Join("",L - 1);
 
    // Initialize [,]dp[]
    for(int i = 0; i < M; i++)
    {
        for(int j = 0; j < M; j++)
        {
            for(int k = 0; k < 2; k++)
                dp[i, j, k] = -1;
        }
    }
 
    // Stores count of numbers in
    // the range [0, L - 1]
    int cntL = cntNum(str, 0, Y,
                      1, dp);
 
    return (cntR - cntL);
}
 
// Driver Code
public static void Main(String[] args)
{
    int L = 20, R = 10000, Y = 14;
     
    Console.Write(UtilCntNumRange(L, R, Y));
}
}
 
// This code is contributed by 29AjayKumar

输出

540

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