对给定范围内的数字进行计数，这些数字的二进制表示具有相同的第一位和最后一位数字

原文:https://www . geesforgeks . org/count-numbers-from-a-给定范围-具有相同的二进制表示中的第一个和最后一个数字/

给定两个整数 L 和 R ，任务是找出在【L，R】范围内的数字的计数，它们在二进制表示中的第一个和最后一个数字是相同的。

示例:

输入: L = 1，R = 5 输出: 3 解释: (1)10=(1)2T13】(2)10=(10)2 (3)10=(11)2T23

输入: L = 6，R = 30 T3】输出: 12

天真方法:最简单的方法是迭代范围 L 到 R 找到所有数字的二进制表示并检查每个数字，如果它们的二进制表示中的第一个和最后一个数字相同或不相同。 时间复杂度: O((R-L)日志(R-L)) 辅助空间: O(1)

有效方法:给定的问题可以基于以下观察来解决:

按照以下步骤解决问题:

下面是上述方法的实现:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to count numbers in range
// [L, R] having same first and last
// digit in the binary representation
void Count_numbers(int L, int R)
{
    int count = (R - L) / 2;

    if (R % 2 != 0 || L % 2 != 0)
        count += 1;

    cout << count << endl;
}

// Drivers code
int main()
{

    // Given range [L, R]
    int L = 6, R = 30;

    // Function Call
    Count_numbers(L, R);
}

Java 语言(一种计算机语言，尤用于创建网站)

// Java program to implement
// the above approach
import java.util.*;
class GFG
{

// Function to count numbers in range
// [L, R] having same first and last
// digit in the binary representation
static void Count_numbers(int L, int R)
{
    int count = (R - L) / 2;
    if (R % 2 != 0 || L % 2 != 0)
        count += 1;
    System.out.print(count);
}

// Driver code
public static void main(String[] args)
{
    // Given range [L, R]
    int L = 6, R = 30;

    // Function Call
    Count_numbers(L, R);
}
}

// This code is contributed by sanjoy_62

Python 3

# Python program for the above approach

# Function to count numbers in range
# [L, R] having same first and last
# digit in the binary representation
def Count_numbers(L, R) :  
    count = (R - L) // 2
    if (R % 2 != 0 or L % 2 != 0) :
        count += 1
    print(count)

# Drivers code

# Given range [L, R]
L = 6
R = 30

# Function Call
Count_numbers(L, R)

# This code is contributed by code_hunt.

C

// C# program to implement
// the above approach
using System;
class GFG
{

// Function to count numbers in range
// [L, R] having same first and last
// digit in the binary representation
static void Count_numbers(int L, int R)
{
    int count = (R - L) / 2;
    if (R % 2 != 0 || L % 2 != 0)
        count += 1;
    Console.Write(count);
}

// Driver code
public static void Main(String[] args)
{

    // Given range [L, R]
    int L = 6, R = 30;

    // Function Call
    Count_numbers(L, R);
}
}

// This code is contributed by 29AjayKumar

java 描述语言

<script>
// javascript program to implement
// the above approach

    // Function to count numbers in range
    // [L, R] having same first and last
    // digit in the binary representation
    function Count_numbers(L , R)
    {
        var count = (R - L) / 2;
        if (R % 2 != 0 || L % 2 != 0)
            count += 1;
        document.write(count);
    }

    // Driver code

        // Given range [L, R]
        var L = 6, R = 30;

        // Function Call
        Count_numbers(L, R);

// This code is contributed by Rajput-Ji
</script>

Output: 

12

时间复杂度:O(1) T5辅助空间:** O(1)

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