检查给定范围内的位是否处于交替模式
原文:https://www . geesforgeks . org/check-bit-in-alternate-pattern-in-给定范围/
给定一个非负数 n 和两个值 l 和 r 。问题是检查 n 在 l 到 r 范围内的二进制表示中是否有交替模式。这里，交替模式意味着设置位和未设置位以交替顺序出现。这些位从右向左编号，即最低有效位被认为在第一个位置。 约束: 1 < = l < = r < =二进制表示中的位数 n 。 举例:
Input : n = 18, l = 1, r = 3
Output : Yes
(18)<sub>10</sub> = (10010)2
The bits in the range 1 to 3 in the
binary representation of 18 are in
alternate order.
Input : n = 22, l = 2, r = 4
Output : No
(22)<sub>10</sub> = (10110)2
The bits in the range 2 to 4 in the
binary representation of 22 are not in
alternate order.
算法:
bitsAreInAltPatrnInGivenTRange(n, l, r)
// Shift bits of given number n by
// (l-1).
num = n >> (l - 1)
// Find first bit of range
prev = num & 1
// Traverse through remaining
// bits. For every bit, check if
// current bit is same as previous
// or not.
num = num >> 1
for i = 1 to (r - l)
curr = num & 1
if curr == prev then
return false
prev = curr
num = num >> 1
return true;
C++
// C++ implementation to check whether bits are in
// alternate pattern in the given range
#include <bits/stdc++.h>
using namespace std;
// function to check whether bits are in
// alternate pattern in the given range
bool bitsAreInAltPatrnInGivenTRange(unsigned int n,
unsigned int l,
unsigned int r)
{
unsigned int num, prev, curr;
// right shift n by (l - 1) bits
num = n >> (l - 1);
// get the bit at the last position in 'num'
prev = num & 1;
// right shift 'num' by 1
num = num >> 1;
// loop until there are bits in the given range
for (int i = 1; i <= (r - l); i++) {
// get the bit at the last position in 'num'
curr = num & 1;
// if true, then bits are not in alternate
// pattern
if (curr == prev)
return false;
// update 'prev'
prev = curr;
// right shift 'num' by 1
num = num >> 1;
}
// bits are in alternate pattern in the
// given range
return true;
}
// Driver program to test above
int main()
{
unsigned int n = 18;
unsigned int l = 1, r = 3;
if (bitsAreInAltPatrnInGivenTRange(n, l, r))
cout << "Yes";
else
cout << "No";
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation to check
// whether bits are in alternate
// pattern in the given range
class GFG
{
// function to check whether
// bits are in alternate
// pattern in the given range
static boolean bitsAreInAltPatrnInGivenTRange(int n,
int l, int r)
{
int num, prev, curr;
// right shift n by (l - 1) bits
num = n >> (l - 1);
// get the bit at the
// last position in 'num'
prev = num & 1;
// right shift 'num' by 1
num = num >> 1;
// loop until there are
// bits in the given range
for (int i = 1; i <= (r - l); i++)
{
// get the bit at the
// last position in 'num'
curr = num & 1;
// if true, then bits are
// not in alternate pattern
if (curr == prev)
return false;
// update 'prev'
prev = curr;
// right shift 'num' by 1
num = num >> 1;
}
// bits are in alternate
// pattern in the given range
return true;
}
// Driver Code
public static void main(String[] args)
{
int n = 18;
int l = 1, r = 3;
if (bitsAreInAltPatrnInGivenTRange(n, l, r))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by mits
Python 3
# Python3 implementation to check
# whether bits are in alternate
# pattern in the given range
# function to check whether
# bits are in alternate
# pattern in the given range
def bitsAreInAltPatrnInGivenTRange(n, l, r):
# right shift n by (l - 1) bits
num = n >> (l - 1);
# get the bit at the
# last position in 'num'
prev = num & 1;
# right shift 'num' by 1
num = num >> 1;
# loop until there are
# bits in the given range
for i in range(1,(r - l)):
# get the bit at the
# last position in 'num'
curr = num & 1;
# if true, then bits are
# not in alternate pattern
if (curr == prev):
return False;
# update 'prev'
prev = curr;
# right shift 'num' by 1
num = num >> 1;
# bits are in alternate
# pattern in the given range
return True;
# Driver Code
if __name__ == "__main__":
n = 18;
l = 1;
r = 3;
if(bitsAreInAltPatrnInGivenTRange(n, l, r)):
print("Yes");
else:
print("No");
# This Code is contributed by mits
C
// C# implementation to check
// whether bits are in alternate
// pattern in the given range
using System;
class GFG
{
// function to check whether
// bits are in alternate
// pattern in the given range
static bool bitsAreInAltPatrnInGivenTRange(int n,
int l, int r)
{
int num, prev, curr;
// right shift n by (l - 1) bits
num = n >> (l - 1);
// get the bit at the
// last position in 'num'
prev = num & 1;
// right shift 'num' by 1
num = num >> 1;
// loop until there are
// bits in the given range
for (int i = 1; i <= (r - l); i++)
{
// get the bit at the
// last position in 'num'
curr = num & 1;
// if true, then bits are
// not in alternate pattern
if (curr == prev)
return false;
// update 'prev'
prev = curr;
// right shift 'num' by 1
num = num >> 1;
}
// bits are in alternate
// pattern in the given range
return true;
}
// Driver Code
static void Main()
{
int n = 18;
int l = 1, r = 3;
if (bitsAreInAltPatrnInGivenTRange(n, l, r))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by mits
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation to check
// whether bits are in alternate
// pattern in the given range
// function to check whether
// bits are in alternate
// pattern in the given range
function bitsAreInAltPatrnInGivenTRange($n, $l, $r)
{
// right shift n by (l - 1) bits
$num = $n >> ($l - 1);
// get the bit at the
// last position in 'num'
$prev = $num & 1;
// right shift 'num' by 1
$num = $num >> 1;
// loop until there are
// bits in the given range
for ($i = 1; $i <= ($r - $l); $i++)
{
// get the bit at the
// last position in 'num'
$curr = $num & 1;
// if true, then bits are
// not in alternate pattern
if ($curr == $prev)
return false;
// update 'prev'
$prev = $curr;
// right shift 'num' by 1
$num = $num >> 1;
}
// bits are in alternate
// pattern in the given range
return true;
}
// Driver Code
$n = 18;
$l = 1;
$r = 3;
if (bitsAreInAltPatrnInGivenTRange($n, $l, $r))
echo "Yes";
else
echo "No";
// This Code is contributed by mits
?>
java 描述语言
<script>
// Javascript implementation to check whether bits are in
// alternate pattern in the given range
// function to check whether bits are in
// alternate pattern in the given range
function bitsAreInAltPatrnInGivenTRange(n, l, r)
{
var num, prev, curr;
// right shift n by (l - 1) bits
num = n >> (l - 1);
// get the bit at the last position in 'num'
prev = num & 1;
// right shift 'num' by 1
num = num >> 1;
// loop until there are bits in the given range
for (var i = 1; i <= (r - l); i++) {
// get the bit at the last position in 'num'
curr = num & 1;
// if true, then bits are not in alternate
// pattern
if (curr == prev)
return false;
// update 'prev'
prev = curr;
// right shift 'num' by 1
num = num >> 1;
}
// bits are in alternate pattern in the
// given range
return true;
}
// Driver program to test above
var n = 18;
var l = 1, r = 3;
if (bitsAreInAltPatrnInGivenTRange(n, l, r))
document.write( "Yes");
else
document.write( "No");
// This code is contributed by rrrtnx.
</script>
输出:
Yes
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