给定整数N ，任务是对小于或等于N的数字进行计数，以使每个数字至少包含一个重复的数字。
例子：
Input: N = 20
Output: 1
Explanation:
Numbers containing at least one repeated digit and less than or equal to N(= 20) are {11}.
Therefore, the required output is 1.
Input: N = 100
Output: 10
Explanation:
Numbers containing at least one repeated digit and less than or equal to N(= 100) are {11, 22, 33, 44, 55, 66, 77, 88, 99, 100}.
Therefore, the required output is 10.
方法：请按照以下步骤解决问题：
- 初始化一个变量，例如X ，以将数字的总计数存储在N中。
- 初始化一个变量，例如cntNumbers ，以存储小于或等于N的唯一数字组成的总数。
- 计算包含唯一数字的X位数的总数，并更新cntNumbers 。以下是用于计算具有所有唯一数字的X位数总数的公式。
Total count of X-digit numbers with all unique digits = {(9 * factorial(9)) / factorial(10 – X)}
- 计算X -数位号码的总数小于或等于n，其通过在多个的每个可能的数字检查所有可能的值包含唯一位数小于或等于N和更新cntNumbers。
- 最后，打印(N – cntNumbers)的值。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
long factorial(int n)
{
return (n == 1 || n == 0) ? 1 :
n * factorial(n - 1);
}
// Function to count arrangements to
// select K elements from N elements
long NPR(int n, int k)
{
return factorial(n) / factorial(n - k);
}
// Function to count numbers less than or equal
// to N with at least one repeated digits
int cntNumLessThanEqualNRepeatedDigits(int N)
{
// Stores the value of N
int temp = N;
// Stores count of
// digits in N
int X = 0;
// Store all possible
// digits of N
vector nums;
// Calculate digits in N
while(temp)
{
// Update X
X += 1;
// Insert digits of N
// into nums
nums.push_back(temp % 10);
// Update temp
temp /= 10;
}
reverse(nums.begin(), nums.end());
// Count numbers of (X)-digit
// with no repeated digits
int cntNumbers = 0;
// Calculate count of numbers of less than
// (X)-digit and no repeated digits
for(int i = 1; i < X; i++)
{
// Update cntnumbers
cntNumbers += 9 * (factorial(9) /
factorial(10 - i));
}
// Stores unique digits of
// (X)-digit numbers
unordered_set vis;
// Calculate (X) digits
// numbers with no repeated digits
for(int i = 0; i < nums.size(); i++)
{
// Stores count of unique
// value at i_th digit
int k = 0;
// Stores minimum possible value of
// i_th digit of a number
int Min = 0;
// If current digit is the first
// digit of a number
if (i == 0)
{
// Update Min
Min = 1;
}
else
{
// Update Min
Min = 0;
}
// Stores maximum possible value of
// i_th digit if a number
int Max = 0;
// If current digit is the last
// digit of a number
if (i == X - 1)
{
// Update Max
Max = nums[i];
}
else
{
// Update Max
Max = nums[i] - 1;
}
// Iterate over all possible value
// of current digit
for(int j = Min; j <= Max; j++)
{
// If value of current digit
// already occurred in vis
if (vis.find(j) != vis.end())
{
continue;
}
// Update k
k += 1;
}
// Update cntNumbers
cntNumbers += k * (NPR(9 - i,
X - i - 1));
// If current value of i-th digit
// already occurred in vis
if (vis.find(nums[i]) != vis.end())
{
break;
}
// Insert val in vis
vis.insert(nums[i]);
}
// Return total count of numbers less
// than or equal to N with repetition
return (N - cntNumbers);
}
// Driver Code
int main()
{
int N = 100;
cout << cntNumLessThanEqualNRepeatedDigits(N);
return 0;
}
// This code is contributed by Manu Pathria Java
// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
class GFG{
public static int factorial(int n)
{
return (n == 1 || n == 0) ? 1 :
n * factorial(n - 1);
}
// Function to count arrangements to
// select K elements from N elements
public static int NPR(int n, int k)
{
return factorial(n) / factorial(n - k);
}
// Function to count numbers less than or equal
// to N with at least one repeated digits
public static int cntNumLessThanEqualNRepeatedDigits(int N)
{
// Stores the value of N
int temp = N;
// Stores count of
// digits in N
int X = 0;
// Store all possible
// digits of N
List nums = new ArrayList<>();
// Calculate digits in N
while (temp > 0)
{
// Update X
X += 1;
// Insert digits of N
// into nums
nums.add(temp % 10);
// Update temp
temp /= 10;
}
Collections.reverse(nums);
// Count numbers of (X)-digit
// with no repeated digits
int cntNumbers = 0;
// Calculate count of numbers of less than
// (X)-digit and no repeated digits
for(int i = 1; i < X; i++)
{
// Update cntnumbers
cntNumbers += 9 * (factorial(9) /
factorial(10 - i));
}
// Stores unique digits of
// (X)-digit numbers
Set vis=new HashSet<>();
// Calculate (X) digits
// numbers with no repeated digits
for(int i = 0; i < nums.size(); i++)
{
// Stores count of unique
// value at i_th digit
int k = 0;
// Stores minimum possible value of
// i_th digit of a number
int Min = 0;
// If current digit is the first
// digit of a number
if (i == 0)
{
// Update Min
Min = 1;
}
else
{
// Update Min
Min = 0;
}
// Stores maximum possible value of
// i_th digit if a number
int Max = 0;
// If current digit is the last
// digit of a number
if (i == X - 1)
{
// Update Max
Max = nums.get(i);
}
else
{
// Update Max
Max = nums.get(i) - 1;
}
// Iterate over all possible value
// of current digit
for(int j = Min; j <= Max; j++)
{
// If value of current digit
// already occurred in vis
if (vis.contains(j))
{
continue;
}
// Update k
k += 1;
}
// Update cntNumbers
cntNumbers += k * (NPR(9 - i,
X - i - 1));
// If current value of i-th digit
// already occurred in vis
if (vis.contains(nums.get(i)))
{
break;
}
// Insert val in vis
vis.add(nums.get(i));
}
// Return total count of numbers less
// than or equal to N with repetition
return (N - cntNumbers);
}
// Driver Code
public static void main(String[] args)
{
int N = 100;
System.out.print(
cntNumLessThanEqualNRepeatedDigits(N));
}
}
// This code is contributed by Manu Pathria Python3
# Python3 program to implement
# the above approach
import math
# Function to count arrangements to
# select K elements from N elements
def NPR(n, k):
return (math.factorial(n) //
math.factorial(n-k))
# Function to count numbers less than or equal
# to N with at least one repeated digits
def cntNumLessThanEqualNRepeatedDigits(N):
# Stores the value of N
temp = N
# Stores count of
# digits in N
X = 0;
# Store all possible
# digits of N
nums = []
# Calculate digits in N
while(temp):
# Update X
X += 1
# Insert digits of N
# into nums
nums.append(temp % 10)
# Update temp
temp //= 10
# Reverse nums
nums.reverse()
# Count numbers of (X)-digit
# with no repeated digits
cntNumbers = 0
# Calculate count of numbers of less than
# (X)-digit and no repeated digits
for i in range(1, X):
# Update cntnumbers
cntNumbers += 9 * (math.factorial(9) //
math.factorial(10 - i))
# Stores unique digits of
# (X)-digit numbers
vis = set()
# Calculate (X) digits
# numbers with no repeated digits
for i, val in enumerate(nums):
# Stores count of unique
# value at i_th digit
k = 0
# Stores minimum possible value of
# i_th digit of a number
Min = 0;
# If current digit is the first
# digit of a number
if i == 0:
#Update Min
Min = 1
else:
# Update Min
Min = 0
# Stores maximum possible value of
# i_th digit if a number
Max = 0;
# If current digit is the last
# digit of a number
if i == X - 1:
# Update Max
Max = val
else:
# Update Max
Max = val - 1;
# Iterate over all possible value
# of current digit
for j in range(Min, Max + 1):
# If value of current digit
# already occurred in vis
if (j in vis):
continue
# Update k
k += 1
# Update cntNumbers
cntNumbers += k * (NPR(9 - i,
X - i - 1))
# If current value of i-th digit
# already occurred in vis
if val in vis:
break
# Insert val in vis
vis.add(val)
# Return total count of numbers less
# than or equal to N with repetition
return (N - cntNumbers)
# Driver Code
if __name__ == '__main__':
N = 100
# Function call
print(cntNumLessThanEqualNRepeatedDigits(N))C#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG {
static int factorial(int n)
{
return (n == 1 || n == 0) ? 1 :
n * factorial(n - 1);
}
// Function to count arrangements to
// select K elements from N elements
static int NPR(int n, int k)
{
return factorial(n) / factorial(n - k);
}
// Function to count numbers less than or equal
// to N with at least one repeated digits
static int cntNumLessThanEqualNRepeatedDigits(int N)
{
// Stores the value of N
int temp = N;
// Stores count of
// digits in N
int X = 0;
// Store all possible
// digits of N
List nums = new List();
// Calculate digits in N
while (temp > 0)
{
// Update X
X += 1;
// Insert digits of N
// into nums
nums.Add(temp % 10);
// Update temp
temp /= 10;
}
nums.Reverse();
// Count numbers of (X)-digit
// with no repeated digits
int cntNumbers = 0;
// Calculate count of numbers of less than
// (X)-digit and no repeated digits
for(int i = 1; i < X; i++)
{
// Update cntnumbers
cntNumbers += 9 * (factorial(9) /
factorial(10 - i));
}
// Stores unique digits of
// (X)-digit numbers
HashSet vis = new HashSet();
// Calculate (X) digits
// numbers with no repeated digits
for(int i = 0; i < nums.Count; i++)
{
// Stores count of unique
// value at i_th digit
int k = 0;
// Stores minimum possible value of
// i_th digit of a number
int Min = 0;
// If current digit is the first
// digit of a number
if (i == 0)
{
// Update Min
Min = 1;
}
else
{
// Update Min
Min = 0;
}
// Stores maximum possible value of
// i_th digit if a number
int Max = 0;
// If current digit is the last
// digit of a number
if (i == X - 1)
{
// Update Max
Max = nums[i];
}
else
{
// Update Max
Max = nums[i] - 1;
}
// Iterate over all possible value
// of current digit
for(int j = Min; j <= Max; j++)
{
// If value of current digit
// already occurred in vis
if (vis.Contains(j))
{
continue;
}
// Update k
k += 1;
}
// Update cntNumbers
cntNumbers += k * (NPR(9 - i,
X - i - 1));
// If current value of i-th digit
// already occurred in vis
if (vis.Contains(nums[i]))
{
break;
}
// Insert val in vis
vis.Add(nums[i]);
}
// Return total count of numbers less
// than or equal to N with repetition
return (N - cntNumbers);
}
static void Main()
{
int N = 100;
Console.WriteLine(cntNumLessThanEqualNRepeatedDigits(N));
}
}
// This code is contributed by divyeshrabadiya07 输出：
10时间复杂度： O((log 10 N) 2 )
辅助空间： O(1)