我们给了两个整数n和d，我们需要计算所有没有数字d的n个数字。
例子 ：
Input : n = 2, d = 7
Output : 72
All two digit numbers that don't have
7 as digit are 10, 11, 12, 13, 14, 15,
16, 18, .....
Input : n = 3, d = 9
Output : 648一个简单的解决方案是遍历所有d位数字。对于每个数字，请检查是否有x作为数字。
高效方法在此方法中，我们首先检查排除数字d是否为零或非零。如果为零，则我们有9个数字(1到9)作为第一个数字，否则我们有8个数字(x位数和0除外)。然后，对于所有其他数字，我们有9个选择，即(0到9，不包括d数字)。我们用n-1个数字作为第一个数字来简单地调用digitNumber函数，我们已经发现它可以是8或9并简单地乘以它。对于其他数字，我们检查是否位数是奇数或偶数，如果它是奇数我们称之为与第(n-1)/ 2位digitNumber函数和乘法它由9 oterwise我们称之为digitNumber函数具有n / 2个数字，并将它们存储在结果和取结果方块。
插图 ：
Number from 1 to 7 excluding digit 9.
digits multiple number
1 8 8
2 8*9 72
3 8*9*9 648
4 8*9*9*9 5832
5 8*9*9*9*9 52488
6 8*9*9*9*9*9 472392
7 8*9*9*9*9*9*9 4251528如我们所见，在每一步中，我们的位数是数字的一半。假设我们用7(7)位数字从main调用函数的6(7-1)位数字。当我们将一半数字留下3(6/2)个数字时。因此，我们必须将3位数字的结果与其自身相乘，以获得6位数字的结果。同样，对于3位数字，我们有奇数位，也有奇数位。我们找到1((3-1)/ 2)位数字的结果，找到结果平方并乘以9，因为我们找到d-1位数字的结果。
C++
// C++ Implementation of above method
#include
#define mod 1000000007
using namespace std;
// Finding number of possible number with
// n digits excluding a particular digit
long long digitNumber(long long n) {
// Checking if number of digits is zero
if (n == 0)
return 1;
// Checking if number of digits is one
if (n == 1)
return 9;
// Checking if number of digits is odd
if (n % 2) {
// Calling digitNumber function
// with (digit-1)/2 digits
long long temp = digitNumber((n - 1) / 2) % mod;
return (9 * (temp * temp) % mod) % mod;
} else {
// Calling digitNumber function
// with n/2 digits
long long temp = digitNumber(n / 2) % mod;
return (temp * temp) % mod;
}
}
int countExcluding(int n, int d)
{
// Calling digitNumber function
// Checking if excluding digit is
// zero or non-zero
if (d == 0)
return (9 * digitNumber(n - 1)) % mod;
else
return (8 * digitNumber(n - 1)) % mod;
}
// Driver function to run above program
int main() {
// Initializing variables
long long d = 9;
int n = 3;
cout << countExcluding(n, d) << endl;
return 0;
} Java
// Java Implementation of above method
import java.lang.*;
class GFG {
static final int mod = 1000000007;
// Finding number of possible number with
// n digits excluding a particular digit
static int digitNumber(long n) {
// Checking if number of digits is zero
if (n == 0)
return 1;
// Checking if number of digits is one
if (n == 1)
return 9;
// Checking if number of digits is odd
if (n % 2 != 0) {
// Calling digitNumber function
// with (digit-1)/2 digits
int temp = digitNumber((n - 1) / 2) % mod;
return (9 * (temp * temp) % mod) % mod;
}
else {
// Calling digitNumber function
// with n/2 digits
int temp = digitNumber(n / 2) % mod;
return (temp * temp) % mod;
}
}
static int countExcluding(int n, int d) {
// Calling digitNumber function
// Checking if excluding digit is
// zero or non-zero
if (d == 0)
return (9 * digitNumber(n - 1)) % mod;
else
return (8 * digitNumber(n - 1)) % mod;
}
// Driver function to run above program
public static void main(String[] args) {
// Initializing variables
int d = 9;
int n = 3;
System.out.println(countExcluding(n, d));
}
}
// This code is contributed by Anant Agarwal.Python3
# Python Implementation
# of above method
mod=1000000007
# Finding number of
# possible number with
# n digits excluding
# a particular digit
def digitNumber(n):
# Checking if number
# of digits is zero
if (n == 0):
return 1
# Checking if number
# of digits is one
if (n == 1):
return 9
# Checking if number
# of digits is odd
if (n % 2!=0):
# Calling digitNumber function
# with (digit-1)/2 digits
temp = digitNumber((n - 1) // 2) % mod
return (9 * (temp * temp) % mod) % mod
else:
# Calling digitNumber function
# with n/2 digits
temp = digitNumber(n // 2) % mod
return (temp * temp) % mod
def countExcluding(n,d):
# Calling digitNumber function
# Checking if excluding digit is
# zero or non-zero
if (d == 0):
return (9 * digitNumber(n - 1)) % mod
else:
return (8 * digitNumber(n - 1)) % mod
# Driver code
d = 9
n = 3
print(countExcluding(n, d))
# This code is contributed
# by Anant Agarwal.C#
// C# Implementation of above method
using System;
class GFG {
static int mod = 1000000007;
// Finding number of possible number with
// n digits excluding a particular digit
static int digitNumber(long n) {
// Checking if number of digits is zero
if (n == 0)
return 1;
// Checking if number of digits is one
if (n == 1)
return 9;
// Checking if number of digits is odd
if (n % 2 != 0) {
// Calling digitNumber function
// with (digit-1)/2 digits
int temp = digitNumber((n - 1) / 2) % mod;
return (9 * (temp * temp) % mod) % mod;
}
else {
// Calling digitNumber function
// with n/2 digits
int temp = digitNumber(n / 2) % mod;
return (temp * temp) % mod;
}
}
static int countExcluding(int n, int d) {
// Calling digitNumber function
// Checking if excluding digit is
// zero or non-zero
if (d == 0)
return (9 * digitNumber(n - 1)) % mod;
else
return (8 * digitNumber(n - 1)) % mod;
}
// Driver function to run above program
public static void Main() {
// Initializing variables
int d = 9;
int n = 3;
Console.WriteLine(countExcluding(n, d));
}
}
// This code is contributed by vt_m.PHP
Javascript
输出：
648时间复杂度： O(log n)。