检查给定位置的位是置位还是未置位
原文:https://www . geesforgeks . org/check-what-bit-given-position-set-unset/
给定两个正整数 n 和 k 。问题是检查在 n 的二进制表示中从右侧开始的位置 k 处的位是置位(' 1 ')还是未置位(' 0 ')。 约束: 1 < = k < =二进制表示的位数 n 。 示例:
Input : n = 10, k = 2
Output : Set
(10)<sub>10</sub> = (1010)2
The 2nd bit from the right is set.
Input : n = 21, k = 4
Output : Unset
进场:以下为步骤:
- 计算new _ num=(n>>(k–1))。
- 如果(new_num & 1) == 1，则该位为“置位”，否则为“未置位”。
C++
// C++ implementation to check whether the bit
// at given position is set or unset
#include <bits/stdc++.h>
using namespace std;
// function to check whether the bit
// at given position is set or unset
bool bitAtGivenPosSetOrUnset(unsigned int n,
unsigned int k)
{
int new_num = n >> (k - 1);
// if it results to '1' then bit is set,
// else it results to '0' bit is unset
return (new_num & 1);
}
// Driver program to test above
int main()
{
unsigned int n = 10, k = 2;
if (bitAtGivenPosSetOrUnset(n, k))
cout << "Set";
else
cout << "Unset";
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to
// check the set bit
// at kth position
import java.io.*;
class GFG {
// function to check whether
// the bit at given position
// is set or unset
static int bitAtGivenPosSetOrUnset
( int n, int k)
{
// to shift the kth bit
// at 1st position
int new_num = n >> (k - 1);
// Since, last bit is now
// kth bit, so doing AND with 1
// will give result.
return (new_num & 1);
}
public static void main (String[] args)
{
// K and n must be greater than 0
int n = 10, k = 2;
if (bitAtGivenPosSetOrUnset(n, k)==1)
System.out.println("Set");
else
System.out.println("Unset");
}
}
//This code is contributed by Gitanjali
Python 3
# python implementation to check
# whether the bit at given
# position is set or unset
import math
#function to check whether the bit
# at given position is set or unset
def bitAtGivenPosSetOrUnset( n, k):
new_num = n >> (k - 1)
#if it results to '1' then bit is set,
#else it results to '0' bit is unset
return (new_num & 1)
# Driver code
n = 10
k = 2
if (bitAtGivenPosSetOrUnset(n, k)):
print("Set")
else:
print("Unset")
#This code is contributed by Gitanjali
C
// C# program to check the set bit
// at kth position
using System;
class GFG {
// function to check whether
// the bit at given position
// is set or unset
static int bitAtGivenPosSetOrUnset(
int n, int k)
{
// to shift the kth bit
// at 1st position
int new_num = n >> (k - 1);
// Since, last bit is now
// kth bit, so doing AND with 1
// will give result.
return (new_num & 1);
}
// Driver code
public static void Main ()
{
// K and n must be greater
// than 0
int n = 10, k = 2;
if (bitAtGivenPosSetOrUnset(n, k)==1)
Console.Write("Set");
else
Console.Write("Unset");
}
}
// This code is contributed by Sam007.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation to check whether the bit
// at given position is set or unset
// function to check whether the bit
// at given position is set or unset
function bitAtGivenPosSetOrUnset($n, $k)
{
$new_num = $n >> ($k - 1);
// if it results to '1' then bit is set,
// else it results to '0' bit is unset
return ($new_num & 1);
}
// Driver Code
$n = 10;
$k = 2;
if (bitAtGivenPosSetOrUnset($n, $k))
echo "Set";
else
echo "Unset";
// This code is contributed by Sam007
?>
java 描述语言
<script>
// javascript program to
// check the set bit
// at kth position
// function to check whether
// the bit at given position
// is set or unset
function bitAtGivenPosSetOrUnset
(n, k)
{
// to shift the kth bit
// at 1st position
let new_num = n >> (k - 1);
// Since, last bit is now
// kth bit, so doing AND with 1
// will give result.
return (new_num & 1);
}
// Driver Function
// K and n must be greater than 0
let n = 10, k = 2;
if (bitAtGivenPosSetOrUnset(n, k)==1)
document.write("Set");
else
document.write("Unset");
// This code is contributed by susmitakundugoaldanga.
</script>
输出:
Set
时间复杂度: O(1)。
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