给定数字n和数字d，计算所有出现的d，范围从0到n。
例子：
Input : n = 25
d = 2
Output : 9
The occurrences are 2, 12, 20, 21
22 (Two occurrences), 23, 24, 25
Input : n = 25
d = 3
Output :3
The occurrences are 3, 13, 23
Input : n = 32
d = 3
Output : 6
The occurrences are 3, 13, 23, 30, 31, 32
d的首次出现不能在数字d之前。因此，我们从d开始进行迭代，并一次又一次地进行以下检查。除了在步骤2.a和2.b中提到的情况外，我们大多数时候(在步骤2中)将数字跳了10次。
步骤1 ：检查数字的最后一位是否等于d，如果最后一位则增加计数。
第二步：
a)如果数字可以被10整除，则同时增加计数和数字(例如，如果数字30可以被10整除并且d = 3，那么我们必须对31-39之间的所有数字进行计数，这就是为什么我们将count加1并将number加1)
b)否则，如果数字的第一个数字比d小一个，则意味着我们来到了必须将数字增加10并从中减去d的行。例如，如果我们在d = 3时达到23，则将数字增加23 + 10-3 = 30)
c)否则仅将数字增加10。 (例如：d = 3，itr = 3，然后增加10 ie13，23)
步骤3：-返回计数。
C++
// C++ program to count appearances of
// a digit 'd' in range from [0..n]
#include
using namespace std;
int getOccurence(int n, int d)
{
int result = 0; // Initialize result
// Count appearances in numbers starting
// from d.
int itr = d;
while (itr <= n)
{
// When the last digit is equal to d
if (itr%10 == d)
result++;
// When the first digit is equal to d then
if (itr != 0 && itr/10 == d)
{
// increment result as well as number
result++;
itr++;
}
// In case of reverse of number such as 12 and 21
else if (itr/10 == d-1)
itr = itr + (10 - d);
else
itr = itr+10;
}
return result;
}
// Driver code
int main(void)
{
int n = 11, d = 1;
cout << getOccurence(n, d);
return 0;
} Java
// java program to count appearances of
// a digit 'd' in range from [0..n]
import java.*;
public class GFG {
static int getOccurence(int n, int d)
{
// Initialize result
int result = 0;
// Count appearances in numbers
// starting from d.
int itr = d;
while (itr <= n)
{
// When the last digit is
// equal to d
if (itr % 10 == d)
result++;
// When the first digit is
// equal to d then
if (itr != 0 && itr/10 == d)
{
// increment result as
// well as number
result++;
itr++;
}
// In case of reverse of number
// such as 12 and 21
else if (itr/10 == d-1)
itr = itr + (10 - d);
else
itr = itr + 10;
}
return result;
}
// Driver code
public static void main (String[] args)
{
int n = 11, d = 1;
System.out.println(getOccurence(n, d) );
}
}
// This code is contributed by Sam007.Python3
# Python3 program to count appearances
# of a digit 'd' in range from [0..n]
import math;
def getOccurence(n, d):
# Initialize result
result = 0;
# Count appearances in numbers
# starting from d.
itr = d;
while(itr <= n):
# When the last digit is equal to d
if (itr % 10 == d):
result += 1;
# When the first digit is equal to d then
if (itr != 0 and math.floor(itr / 10) == d):
# increment result as well as number
result += 1;
itr += 1;
# In case of reverse of number
# such as 12 and 21
elif (math.floor(itr / 10) == d - 1):
itr = itr + (10 - d);
else:
itr = itr + 10;
return result;
# Driver code
n = 11;
d = 1;
print(getOccurence(n, d));
# This code is contributed by mitsC#
// C# program to count appearances of
// a digit 'd' in range from [0..n]
using System;
public class GFG {
static int getOccurence(int n, int d)
{
// Initialize result
int result = 0;
// Count appearances in numbers
// starting from d.
int itr = d;
while (itr <= n)
{
// When the last digit is
// equal to d
if (itr % 10 == d)
result++;
// When the first digit is
// equal to d then
if (itr != 0 && itr/10 == d)
{
// increment result as
// well as number
result++;
itr++;
}
// In case of reverse of number
// such as 12 and 21
else if (itr/10 == d-1)
itr = itr + (10 - d);
else
itr = itr + 10;
}
return result;
}
// Driver code
public static void Main()
{
int n = 11, d = 1;
Console.Write(getOccurence(n, d));
}
}
// This code is contributed by Sam007.PHP
输出：
4