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📜  在给定范围内所有数字相同的数字计数

📅  最后修改于: 2021-04-29 11:12:50             🧑  作者: Mango

给定两个整数L和R表示一个范围的开始和结束值，任务是计算该范围内所有数字相同的所有数字，例如1、22、444、3333等。

例子：

天真的方法：遍历从L到R的所有数字，对于每个数字，检查其所有数字是否相同。如果是，则增加所需的计数。最后打印此计数。

高效的方法：这个想法基于以下事实：1、11、111等的倍数(1到9)的所有数字都相同。
例如： 

1 times 1 = 1 (All digits are same)
2 times 1 = 2 (All digits are same)
3 times 1 = 3 (All digits are same)
.
.
9 times 1 = 9 (All digits are same)

Similarly
1 times 11 = 11 (All digits are same)
2 times 11 = 22 (All digits are same)
3 times 11 = 33 (All digits are same)
.
.
9 times 11 = 99 (All digits are same)

Same is the case for 111, 1111, etc.

因此，这些步骤可以定义为：

  1. 找到R中的位数。这将确定要创建的连续1的长度，直到我们必须检查该长度。
    例如，如果R = 100，则length(R)=3。因此，我们只需要检查1、11和111的倍数。
  2. 对于从1到length(R)的每个连续1s长度：
    • 将它们与2到9的所有值相乘
    • 检查它是否在[L，R]范围内。
    • 如果是，则增加所需数字的数量。
  3. 打印所需的数字计数。

      下面的代码是上述方法的实现：

      C++
      // C++ program to count the
// total numbers in the range
// L and R which have all the
// digit same
  
#include 
using namespace std;
  
// Function that count the
// total numbersProgram between L
// and R which have all the
// digit same
int count_same_digit(int L, int R)
{
    int tmp = 0, ans = 0;
  
    // length of R
    int n = log10(R) + 1;
  
    for (int i = 0; i < n; i++) {
  
        // tmp has all digits as 1
        tmp = tmp * 10 + 1;
  
        // For each multiple
        // of tmp in range 1 to 9,
        // check if it present
        // in range [L, R]
        for (int j = 1; j <= 9; j++) {
  
            if (L <= (tmp * j)
                && (tmp * j) <= R) {
  
                // Increment the required count
                ans++;
            }
        }
    }
    return ans;
}
  
// Driver Program
int main()
{
    int L = 12, R = 68;
  
    cout << count_same_digit(L, R)
         << endl;
    return 0;
}

      Java
      // Java program to count the
// total numbers in the range
// L and R which have all the
// digit same
import java.util.*;
  
class GFG{
  
// Function that count the total 
// numbersProgram between L and 
// R which have all the digit same
static int count_same_digit(int L, int R)
{
    int tmp = 0, ans = 0;
  
    // Length of R
    int n = (int)Math.log10(R) + 1;
  
    for(int i = 0; i < n; i++)
    {
         
       // tmp has all digits as 1
       tmp = tmp * 10 + 1;
         
       // For each multiple of tmp 
       // in range 1 to 9, check if
       // it present in range [L, R]
       for(int j = 1; j <= 9; j++)
       {
          if (L <= (tmp * j) && (tmp * j) <= R) 
          {
              // Increment the required count
              ans++;
          }
       }
    }
    return ans;
}
  
// Driver code
public static void main(String[] args)
{
    int L = 12, R = 68;
  
    System.out.println(count_same_digit(L, R));
}
}
  
// This code is contributed by offbeat

      Python3
      # Python3 program to count the
# total numbers in the range
# L and R which have all the
# digit same
import math
  
# Function that count the
# total numbersProgram between L
# and R which have all the
# digit same
def count_same_digit(L, R):
  
    tmp = 0; ans = 0;
  
    # length of R
    n = int(math.log10(R) + 1);
  
    for i in range(0, n):
  
        # tmp has all digits as 1
        tmp = tmp * 10 + 1;
  
        # For each multiple
        # of tmp in range 1 to 9,
        # check if it present
        # in range [L, R]
        for j in range(1, 9):
  
            if (L <= (tmp * j) and (tmp * j) <= R):
  
                # Increment the required count
                ans += 1;
              
    return ans;
  
# Driver Code
L = 12; R = 68;
  
print(count_same_digit(L, R))
  
# This code is contributed by Nidhi_biet

      C#
      // C# program to count the
// total numbers in the range
// L and R which have all the
// digit same
using System;
  
class GFG{
  
// Function that count the total 
// numbersProgram between L and 
// R which have all the digit same
static int count_same_digit(int L, int R)
{
    int tmp = 0, ans = 0;
  
    // Length of R
    int n = (int)Math.Log10(R) + 1;
  
    for(int i = 0; i < n; i++)
    {
          
        // tmp has all digits as 1
        tmp = tmp * 10 + 1;
              
        // For each multiple of tmp 
        // in range 1 to 9, check if
        // it present in range [L, R]
        for(int j = 1; j <= 9; j++)
        {
            if (L <= (tmp * j) && (tmp * j) <= R) 
            {
                // Increment the required count
                ans++;
            }
        }
    }
    return ans;
}
  
// Driver code
public static void Main()
{
    int L = 12, R = 68;
  
    Console.Write(count_same_digit(L, R));
}
}
  
// This code is contributed by Code_Mech

      输出： 
      5

      时间复杂度： O(length(R))