给定L和R。任务是找到一个数字,该数字以二进制表示形式设置第L个和第R个索引之间的所有位,而其余位未设置。二进制表示形式为32位。
例子:
Input: L = 2, R = 5
Output: 60
Explanation: The binary representation is
0..0111100 => 60
Input: L = 1, R = 3
Output: 14
Explanation: The binary representation is
0..01110 => 14
天真的方法:天真的方法是从i = L迭代到i = R并计算2 i的所有幂的加法。
下面的程序说明了幼稚的方法:
C++
// CPP program to print the integer
// with all the bits set in range L-R
// Naive Approach
#include
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
// 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..
Python3
# 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.
PHP
C++
// CPP program to print the integer
// with all the bits set in range L-R
// Efficient Approach
#include
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
// 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.
Python3
# 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.
PHP
输出:
60
一种有效的方法是从右开始计算所有(R)个设置位的数量,然后从右开始计算所有(L-1)个位设置的数量,以获得所需的数量。
- 使用以下公式从右计算具有所有R设置位的数字。
(1 << (R+1)) - 1.
- 从右边减去所有(L-1)个设置位的数字。
(1<
因此计算((1 <<(R + 1))-1)-((1 << L)-1),我们得到的最终公式为:
(1<<(R+1))-(1<
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
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
// 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.
Python3
# 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.
的PHP
输出:
60
时间复杂度:O(1)
辅助空间:O(1)
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