给定一个二维数组arr [] [] ，每行以查询{L，R}的形式，任务是对范围[L，R]中的数字进行计数，以使该数字可被其所有非整数整除。 -零位数。
例子：
Input: arr[][] ={ {1, 5}, {12, 14} }
Output: 5 1
Explanation:
Query1: All the numbers in the range [1, 5] are divisible by their digits.
Query2: Numbers in the range [12, 14] which are divisible by all of its digits is 12 only.
Input: arr[][] = { {1, 20} }
Output:14
Explanation:
Numbers in the range [1, 20] which are divisible by all of its non-zero digits are: { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 15, 20}
天真的方法：解决此问题的最简单方法是遍历数组arr [] [] ，对于数组的第i行，在范围[L，R]上进行迭代。对于范围中的每个元素，请检查数字是否可以被其所有非零数字整除。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(N *(R – L))，其中N是行数
辅助空间： O(1)
高效的方法：可以通过找到arr [i] [1]的最大可能值并使用Prefix Sum技术预先计算可被其非零数字整除的数字计数，从而优化上述方法。请按照以下步骤解决问题：
- 初始化一个变量，例如Max ，以存储arr [i] [1]的最大可能值。
- 初始化一个数组，例如prefCntDiv [] ，其中prefCntDiv [i]存储范围为[1，i]的数字计数，该范围可被其非零数字整除。
- 在[1，Max]范围内迭代。对于第i次迭代，请检查i是否可被其所有非零数字整除。如果发现是真的，则更新prefCntDiv [i] = prefCntDiv [i – 1] + 1 。
- 否则，更新prefCntDiv [i] = prefCntDiv [i – 1] 。
- 遍历数组arr [] 。对于数组的第i行，打印prefCntDiv [arr [i] [1]] – prefCntDiv [arr [i] [0] – 1]。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
#define Max 1000005
// Function to check if a number is divisible
// by all of its non-zero digits or not
bool CheckDivByAllDigits(int number)
{
// Stores the number
int n = number;
// Iterate over the digits
// of the numbers
while (n > 0) {
// If digit of number
// is non-zero
if (n % 10)
// If number is not divisible
// by its current digit
if (number % (n % 10)) {
return false;
}
// Update n
n /= 10;
}
return true;
}
// Function to count of numbers which are
// divisible by all of its non-zero
// digits in the range [1, i]
void cntNumInRang(int arr[][2], int N)
{
// Stores count of numbers which are
// divisible by all of its non-zero
// digits in the range [1, i]
int prefCntDiv[Max] = { 0 };
// Iterate over the range [1, Max]
for (int i = 1; i <= Max; i++) {
// Update
prefCntDiv[i] = prefCntDiv[i - 1]
+ (CheckDivByAllDigits(i));
}
// Traverse the array, arr[]
for (int i = 0; i < N; i++)
cout << (prefCntDiv[arr[i][1]]
- prefCntDiv[arr[i][0] - 1])
<< " ";
}
// Driver Code
int main()
{
int arr[][2] = { { 1, 5 }, { 12, 14 } };
int N = sizeof(arr) / sizeof(arr[0]);
cntNumInRang(arr, N);
return 0;
} Java
// Java program to implement
// the above approach
import java.io.*;
class GFG
{
public static int Max = 1000005;
// Function to check if a number is divisible
// by all of its non-zero digits or not
static boolean CheckDivByAllDigits(int number)
{
// Stores the number
int n = number;
// Iterate over the digits
// of the numbers
while (n > 0)
{
// If digit of number
// is non-zero
if (n % 10 != 0)
// If number is not divisible
// by its current digit
if (number % (n % 10) != 0)
{
return false;
}
// Update n
n /= 10;
}
return true;
}
// Function to count of numbers which are
// divisible by all of its non-zero
// digits in the range [1, i]
static void cntNumInRang(int arr[][], int N)
{
// Stores count of numbers which are
// divisible by all of its non-zero
// digits in the range [1, i]
int prefCntDiv[] = new int[Max + 1];
// Iterate over the range [1, Max]
for (int i = 1; i <= Max; i++)
{
int ans = 0;
if (CheckDivByAllDigits(i))
ans = 1;
// Update
prefCntDiv[i] = prefCntDiv[i - 1] + ans;
}
// Traverse the array, arr[]
for (int i = 0; i < N; i++)
System.out.print((prefCntDiv[arr[i][1]]
- prefCntDiv[arr[i][0] - 1])
+ " ");
}
// Driver Code
public static void main(String[] args)
{
int arr[][] = { { 1, 5 }, { 12, 14 } };
int N = arr.length;
cntNumInRang(arr, N);
}
}
// This code is contributed by Dharanendra L VPython3
# Python3 program to implement
# the above approach
# Function to check if a number is divisible
# by all of its non-zero digits or not
def CheckDivByAllDigits(number):
# Stores the number
n = number
# Iterate over the digits
# of the numbers
while (n > 0):
# If digit of number
# is non-zero
if (n % 10):
# If number is not divisible
# by its current digit
if (number % (n % 10)):
return False
# Update n
n //= 10
return True
# Function to count of numbers which are
# divisible by all of its non-zero
# digits in the range [1, i]
def cntNumInRang(arr, N):
global Max
# Stores count of numbers which are
# divisible by all of its non-zero
# digits in the range [1, i]
prefCntDiv = [0]*Max
# Iterate over the range [1, Max]
for i in range(1, Max):
# Update
prefCntDiv[i] = prefCntDiv[i - 1] + (CheckDivByAllDigits(i))
# Traverse the array, arr[]
for i in range(N):
print(prefCntDiv[arr[i][1]]- prefCntDiv[arr[i][0] - 1], end = " ")
# Driver Code
if __name__ == '__main__':
arr =[ [ 1, 5 ], [12, 14]]
Max = 1000005
N = len(arr)
cntNumInRang(arr, N)
# This code is contributed by mohit kumar 29.C#
// C# program to implement
// the above approach
using System;
class GFG
{
public static int Max = 1000005;
// Function to check if a number is divisible
// by all of its non-zero digits or not
static bool CheckDivByAllDigits(int number)
{
// Stores the number
int n = number;
// Iterate over the digits
// of the numbers
while (n > 0) {
// If digit of number
// is non-zero
if (n % 10 != 0)
// If number is not divisible
// by its current digit
if (number % (n % 10) != 0)
{
return false;
}
// Update n
n /= 10;
}
return true;
}
// Function to count of numbers which are
// divisible by all of its non-zero
// digits in the range [1, i]
static void cntNumInRang(int[, ] arr, int N)
{
// Stores count of numbers which are
// divisible by all of its non-zero
// digits in the range [1, i]
int[] prefCntDiv = new int[Max + 1];
// Iterate over the range [1, Max]
for (int i = 1; i <= Max; i++) {
int ans = 0;
if (CheckDivByAllDigits(i))
ans = 1;
// Update
prefCntDiv[i] = prefCntDiv[i - 1] + ans;
}
// Traverse the array, arr[]
for (int i = 0; i < N; i++)
Console.Write((prefCntDiv[arr[i, 1]]
- prefCntDiv[arr[i, 0] - 1])
+ " ");
}
// Driver Code
public static void Main(string[] args)
{
int[, ] arr = { { 1, 5 }, { 12, 14 } };
int N = arr.GetLength(0);
cntNumInRang(arr, N);
}
}
// This code is contributed by chitranayal.输出：
5 1时间复杂度： O(N + Max)，其中Max是arr [i] [1]的最大值
辅助空间： O(N)