仅在第 L 和第 R 索引之间设置位的数字
原文:https://www . geesforgeks . org/number-with-set-bits-only-介于-l-th 和-r-th-index/
给定 L 和 R，任务是找到在二进制表示中，第 L 和第 R 个索引之间的所有位都被设置，其余位不被设置的数字。二进制表示是 32 位。
示例:
输入: L = 2，R = 5 输出: 60 说明:二进制表示为 0..0111100 = > 60
输入: L = 1，R = 3 输出: 14 说明:二进制表示为 0..01110 = > 14
天真法:求数的天真法是从 i = L 迭代到 i = R，计算 2 i 的所有次方之和。 下面的程序说明了天真的方法:
C++
// CPP program to print the integer
// with all the bits set in range L-R
// Naive Approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the integer
// with all the bits set in range L-R
int getInteger(int L, int R)
{
int number = 0;
// iterate from L to R
// and add all powers of 2
for (int i = L; i <= R; i++)
number += pow(2, i);
return number;
}
// Driver Code
int main()
{
int L = 2, R = 5;
cout << getInteger(L, R);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to print the
// integer with all the bits
// set in range L-R Naive Approach
import java.io.*;
class GFG
{
// Function to return the
// integer with all the
// bits set in range L-R
static int getInteger(int L,
int R)
{
int number = 0;
// iterate from L to R
// and add all powers of 2
for (int i = L; i <= R; i++)
number += Math.pow(2, i);
return number;
}
// Driver Code
public static void main (String[] args)
{
int L = 2, R = 5;
System.out.println(getInteger(L, R));
}
}
// This code is contributed by anuj_67..
Python 3
# Python 3 program to print the integer
# with all the bits set in range L-R
# Naive Approach
from math import pow
# Function to return the integer
# with all the bits set in range L-R
def getInteger(L, R):
number = 0
# iterate from L to R
# and add all powers of 2
for i in range(L, R + 1, 1):
number += pow(2, i)
return number
# Driver Code
if __name__ == '__main__':
L = 2
R = 5
print(int(getInteger(L, R)))
# This code is contributed by
# Surendra_Gangwar
C
// C# program to print the
// integer with all the bits
// set in range L-R Naive Approach
using System;
class GFG
{
// Function to return the
// integer with all the
// bits set in range L-R
static int getInteger(int L,
int R)
{
int number = 0;
// iterate from L to R
// and add all powers of 2
for (int i = L; i <= R; i++)
number += (int)Math.Pow(2, i);
return number;
}
// Driver Code
public static void Main ()
{
int L = 2, R = 5;
Console.Write(getInteger(L, R));
}
}
// This code is contributed
// by shiv_bhakt.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to print
// the integer with all
// the bits set in range
// L-R Naive Approach
// Function to return the
// integer with all the
// bits set in range L-R
function getInteger($L, $R)
{
$number = 0;
// iterate from L to R
// and add all powers of 2
for ($i = $L; $i <= $R; $i++)
$number += pow(2, $i);
return $number;
}
// Driver Code
$L = 2;
$R = 5;
echo getInteger($L, $R);
// This code is contributed
// by shiv_bhakt.
?>
java 描述语言
<script>
// Javascript program to print the integer
// with all the bits set in range L-R
// Naive Approach
// Function to return the integer
// with all the bits set in range L-R
function getInteger(L, R)
{
var number = 0;
// iterate from L to R
// and add all powers of 2
for (var i = L; i <= R; i++)
number += Math.pow(2, i);
return number;
}
// Driver Code
var L = 2, R = 5;
document.write( getInteger(L, R));
</script>
输出:
60
一种有效的方法是计算所有(R)位从右边设置的数，然后减去所有(L-1)位从右边设置的数，得到所需的数。
1.用下面的公式计算右边所有 R 位的个数。
(1 << (R+1)) - 1.
2.从右边减去所有(L-1)设定位的数字。
(1<<L) - 1
因此，计算((1 <
(1<<(R+1))-(1<<L)
下面的程序说明了有效的方法:
C++
// CPP program to print the integer
// with all the bits set in range L-R
// Efficient Approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the integer
// with all the bits set in range L-R
int setbitsfromLtoR(int L, int R)
{
return (1 << (R + 1)) - (1 << L);
}
// Driver Code
int main()
{
int L = 2, R = 5;
cout << setbitsfromLtoR(L, R);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to print
// the integer with all
// the bits set in range
// L-R Efficient Approach
import java.io.*;
class GFG
{
// Function to return the
// integer with all the
// bits set in range L-R
static int setbitsfromLtoR(int L,
int R)
{
return (1 << (R + 1)) -
(1 << L);
}
// Driver Code
public static void main (String[] args)
{
int L = 2, R = 5;
System.out.println(setbitsfromLtoR(L, R));
}
}
// This code is contributed
// by shiv_bhakt.
Python 3
# Python3 program to print
# the integer with all the
# bits set in range L-R
# Efficient Approach
# Function to return the
# integer with all the
# bits set in range L-R
def setbitsfromLtoR(L, R):
return ((1 << (R + 1)) -
(1 << L))
# Driver Code
L = 2
R = 5
print(setbitsfromLtoR(L, R))
# This code is contributed
# by Smita
C
// C# program to print
// the integer with all
// the bits set in range
// L-R Efficient Approach
using System;
class GFG
{
// Function to return the
// integer with all the
// bits set in range L-R
static int setbitsfromLtoR(int L,
int R)
{
return (1 << (R + 1)) -
(1 << L);
}
// Driver Code
public static void Main ()
{
int L = 2, R = 5;
Console.WriteLine(setbitsfromLtoR(L, R));
}
}
// This code is contributed
// by shiv_bhakt.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to print
// the integer with all
// the bits set in range
// L-R Efficient Approach
// Function to return the
// integer with all the
// bits set in range L-R
function setbitsfromLtoR($L, $R)
{
return (1 << ($R + 1)) -
(1 << $L);
}
// Driver Code
$L = 2;
$R = 5;
echo setbitsfromLtoR($L, $R);
// This code is contributed
// by shiv_bhakt.
?>
java 描述语言
<script>
// Javascript program to print the integer
// with all the bits set in range L-R
// Efficient Approach
// Function to return the integer
// with all the bits set in range L-R
function setbitsfromLtoR(L, R)
{
return (1 << (R + 1)) - (1 << L);
}
// Driver Code
var L = 2, R = 5;
document.write( setbitsfromLtoR(L, R));
// This code is contributed by famously.
</script>
输出:
60
时间复杂度:O(1) T3】辅助空间 : O(1)
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