前一个较小的整数，少一个设置位
原文:https://www . geesforgeks . org/previous-small-integer-one-less-number-set-bits/
给定一个正整数“n”，其二进制表示中有“x”个设置位。问题是找到先前的较小整数(小于 n 的最大整数)，在其二进制表示中具有(x-1)个设置位。 注:1<=nT5例:
Input : 8
Output : 0
(8)<sub>10</sub> = (1000)2
is having 1 set bit.
(0)<sub>10</sub> = (0)2
is having 0 set bit and is the previous smaller.
Input : 25
Output : 24
以下是步骤:
C++
// C++ implementation to find the previous
// smaller integer with one less number of
// set bits
#include<bits/stdc++.h>
using namespace std;
// function to find the position of
// rightmost set bit.
int getFirstSetBitPos(int n)
{
return log2(n & -n) + 1;
}
// function to find the previous smaller
// integer
int previousSmallerInteger(int n)
{
// position of rightmost set bit of n
int pos = getFirstSetBitPos(n);
// turn off or unset the bit at
// position 'pos'
return (n & ~(1 << (pos - 1)));
}
// Driver program
int main()
{
int n = 25;
cout << previousSmallerInteger(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation to find the previous
// smaller integer with one less number of
// set bits
class GFG {
// function to find the position of
// rightmost set bit.
static int getFirstSetBitPos(int n)
{
return (int)(Math.log(n & -n) / Math.log(2)) + 1;
}
// function to find the previous smaller
// integer
static int previousSmallerInteger(int n)
{
// position of rightmost set bit of n
int pos = getFirstSetBitPos(n);
// turn off or unset the bit at
// position 'pos'
return (n & ~(1 << (pos - 1)));
}
// Driver code
public static void main(String[] args)
{
int n = 25;
System.out.print("Previous smaller Integer ="
+ previousSmallerInteger(n));
}
}
// This code is contributed by Anant Agarwal.
Python 3
# Python3 implementation to find
# the previous smaller integer with
# one less number of set bits
import math
# Function to find the position
# of rightmost set bit.
def getFirstSetBitPos(n):
return (int)(math.log(n & -n) /
math.log(2)) + 1
# Function to find the
# previous smaller integer
def previousSmallerInteger(n):
# position of rightmost set bit of n
pos = getFirstSetBitPos(n)
# turn off or unset the bit
# at position 'pos'
return (n & ~(1 << (pos - 1)))
# Driver code
n = 25
print("Previous small Integer = ",
previousSmallerInteger(n))
# This code is contributed by Anant Agarwal.
C
// C# implementation to find the previous
// smaller integer with one less number of
// set bits
using System;
class GFG {
// function to find the position of
// rightmost set bit.
static int getFirstSetBitPos(int n)
{
return (int)(Math.Log(n & -n) /
Math.Log(2)) + 1;
}
// function to find the previous smaller
// integer
static int previousSmallerInteger(int n)
{
// position of rightmost set bit of n
int pos = getFirstSetBitPos(n);
// turn off or unset the bit at
// position 'pos'
return (n & ~(1 << (pos - 1)));
}
// Driver code
public static void Main()
{
int n = 25;
Console.WriteLine("Previous small Integer ="
+ previousSmallerInteger(n));
}
}
// This code is contributed by anant321.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation to find the previous
// smaller integer with one less number of
// set bits
// function to find the position of
// rightmost set bit.
function getFirstSetBitPos($n)
{
return log($n & -$n) + 1;
}
// function to find the previous
// smaller integer
function previousSmallerInteger($n)
{
// position of rightmost set bit of n
$pos = getFirstSetBitPos($n);
// turn off or unset the bit at
// position 'pos'
return ($n & ~(1 << ($pos - 1)));
}
// Driver Code
$n = 25;
echo "Previous smaller Integer = ", previousSmallerInteger($n);
// This code is contributed by Ajit
?>
java 描述语言
<script>
// Javascript implementation to find the previous
// smaller integer with one less number of
// set bits
// function to find the position of
// rightmost set bit.
function getFirstSetBitPos(n)
{
return parseInt(Math.log(n & -n)/Math.log(2)) + 1;
}
// function to find the previous smaller
// integer
function previousSmallerInteger(n)
{
// position of rightmost set bit of n
var pos = getFirstSetBitPos(n);
// turn off or unset the bit at
// position 'pos'
return (n & ~(1 << (pos - 1)));
}
// Driver program
var n = 25;
document.write("Previous smaller Integer = " + previousSmallerInteger(n));
</script>
输出:
Previous smaller integer = 24
本文由阿育什·乔哈里供稿。如果你喜欢 GeeksforGeeks 并想投稿，你也可以使用contribute.geeksforgeeks.org写一篇文章或者把你的文章邮寄到 contribute@geeksforgeeks.org。看到你的文章出现在极客博客主页上，帮助其他极客。 如果你发现任何不正确的地方，或者你想分享更多关于上面讨论的话题的信息，请写评论。
版权属于：月萌API www.moonapi.com，转载请注明出处
