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📜  具有后缀N的不超过M的所有可能数字的计数

📅  最后修改于: 2021-04-22 07:00:16             🧑  作者: Mango

给定两个正整数NM ,任务是找到后缀为N的[1,M]范围内所有可能数字的计数。

例子:

天真的方法:最简单的方法是遍历[1,M]范围内的所有整数,并检查后缀是否为N。
时间复杂度: O(M)
辅助空间: O(1)

高效方法:为了优化上述方法,需要进行以下观察:

因此,为了计算在[1,M]范围内的可能数字的计数,需要对以下表达式求值:

下面是上述方法的实现:

C++
// C++ Program to implement
// the above approach
#include 
 
using namespace std;
 
// Function to count the
// no. of digits of N
int digitsOf(int num)
{
    return to_string(num).size();
}
 
// Function to count all possible
// numbers having Suffix as N
int count(int a, int tn)
{
    // Difference of the A.P
    int diff = pow(10, digitsOf(a));
 
    // Count of the number of terms
    return ((tn - a) / diff) + 1;
}
 
// Driver Code
int main()
{
 
    int n, m;
    n = 25, m = 4500;
    cout << count(n, m);
 
    return 0;
}


Java
// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
// Function to count the
// no. of digits of N
static int digitsOf(int num)
{
    return Integer.toString(num).length();
}
 
// Function to count all possible
// numbers having Suffix as N
static int count(int a, int tn)
{
     
    // Difference of the A.P
    int diff = (int)Math.pow(10, digitsOf(a));
 
    // Count of the number of terms
    return ((tn - a) / diff) + 1;
}
 
// Driver code
public static void main (String[] args)
{
    int n = 25, m = 4500;
     
    System.out.println(count(n, m));
}
}
 
// This code is contributed by offbeat


Python3
# Python3 program to implement
# the above approach
 
# Function to count the
# no. of digits of N
def digitsOf(num):
    return len(str(num));
 
# Function to count all possible
# numbers having Suffix as N
def count(a, tn):
 
    # Difference of the A.P
    diff = int(pow(10, digitsOf(a)));
 
    # Count of the number of terms
    return ((tn - a) / diff) + 1;
 
# Driver code
if __name__ == '__main__':
    n = 25; m = 4500;
 
    print(int(count(n, m)));
 
# This code is contributed by sapnasingh4991


C#
// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to count the
// no. of digits of N
static int digitsOf(int num)
{
    return num.ToString().Length;
}
 
// Function to count all possible
// numbers having Suffix as N
static int count(int a, int tn)
{
     
    // Difference of the A.P
    int diff = (int)Math.Pow(10, digitsOf(a));
 
    // Count of the number of terms
    return ((tn - a) / diff) + 1;
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 25, m = 4500;
     
    Console.WriteLine(count(n, m));
}
}
 
// This code is contributed by PrinciRaj1992


输出:
45




时间复杂度: O(1)
辅助空间: O(1)