二进制数(二进制中没有连续的 1)
原文:https://www . geesforgeks . org/https www-geesforgeks-orgfibinary-numbers/
给定 N，检查该数是否为二进制数。Fibbinary 数是整数，其二进制表示不包含连续的 1。
示例:
Input : 10
Output : YES
Explanation: 1010 is the binary representation
of 10 which does not contains any
consecutive 1's.
Input : 11
Output : NO
Explanation: 1011 is the binary representation
of 11, which contains consecutive
1's
这样做的想法是右移数字，直到 n！=0.对于 1 的每个二进制表示，检查找到的最后一位是否为 1。通过做 a(n&1)得到整数的二进制表示的最后一位。如果二进制表示的最后一位是 1，右移之前的前一位也是 1，我们会遇到连续的 1。因此我们得出结论，它不是一个二进制数。 前几个数字是:
0, 2, 4, 8, 10, 16, 18, 20.......
卡片打印处理机（Card Print Processor 的缩写）
// CPP program to check if a number
// is fibinnary number or not
#include <iostream>
using namespace std;
// function to check if binary
// representation of an integer
// has consecutive 1s
bool checkFibinnary(int n)
{
// stores the previous last bit
// initially as 0
int prev_last = 0;
while (n)
{
// if current last bit and
// previous last bit is 1
if ((n & 1) && prev_last)
return false;
// stores the last bit
prev_last = n & 1;
// right shift the number
n >>= 1;
}
return true;
}
// Driver code to check above function
int main()
{
int n = 10;
if (checkFibinnary(n))
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 program to check if a number
// is fibinnary number or not
class GFG {
// function to check if binary
// representation of an integer
// has consecutive 1s
static boolean checkFibinnary(int n)
{
// stores the previous last bit
// initially as 0
int prev_last = 0;
while (n != 0)
{
// if current last bit and
// previous last bit is 1
if ((n & 1) != 0 && prev_last != 0)
return false;
// stores the last bit
prev_last = n & 1;
// right shift the number
n >>= 1;
}
return true;
}
// Driver code to check above function
public static void main(String[] args)
{
int n = 10;
if (checkFibinnary(n) == true)
System.out.println("YES");
else
System.out.println("NO");
}
}
// This code is contributed by
// Smitha Dinesh Semwal
Python 3
# Python 3 program to check if a
# number is fibinnary number or
# not
# function to check if binary
# representation of an integer
# has consecutive 1s
def checkFibinnary(n):
# stores the previous last bit
# initially as 0
prev_last = 0
while (n):
# if current last bit and
# previous last bit is 1
if ((n & 1) and prev_last):
return False
# stores the last bit
prev_last = n & 1
# right shift the number
n >>= 1
return True
# Driver code
n = 10
if (checkFibinnary(n)):
print("YES")
else:
print("NO")
# This code is contributed by Smitha Dinesh Semwal
C
// C# program to check if a number
// is fibinnary number or not
using System;
class GFG {
// function to check if binary
// representation of an integer
// has consecutive 1s
static bool checkFibinnary(int n)
{
// stores the previous last bit
// initially as 0
int prev_last = 0;
while (n != 0)
{
// if current last bit and
// previous last bit is 1
if ((n & 1) != 0 && prev_last != 0)
return false;
// stores the last bit
prev_last = n & 1;
// right shift the number
n >>= 1;
}
return true;
}
// Driver code to check above function
public static void Main()
{
int n = 10;
if (checkFibinnary(n) == true)
Console.WriteLine("YES");
else
Console.WriteLine("NO");
}
}
// This code is contributed by vt_m.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to check if a number
// is fibinnary number or not
// function to check if binary
// representation of an integer
// has consecutive 1s
function checkFibinnary($n)
{
// stores the previous last bit
// initially as 0
$prev_last = 0;
while ($n)
{
// if current last bit and
// previous last bit is 1
if (($n & 1) && $prev_last)
return false;
// stores the last bit
$prev_last = $n & 1;
// right shift the number
$n >>= 1;
}
return true;
}
// Driver code
$n = 10;
if (checkFibinnary($n))
echo "YES";
else
echo "NO";
// This code is contributed by mits
?>
java 描述语言
<script>
// javascript program to check if a number
// is fibinnary number or not
// function to check if binary
// representation of an integer
// has consecutive 1s
function checkFibinnary(n) {
// stores the previous last bit
// initially as 0
var prev_last = 0;
while (n != 0) {
// if current last bit and
// previous last bit is 1
if ((n & 1) != 0 && prev_last != 0)
return false;
// stores the last bit
prev_last = n & 1;
// right shift the number
n >>= 1;
}
return true;
}
// Driver code to check above function
var n = 10;
if (checkFibinnary(n) == true)
document.write("YES");
else
document.write("NO");
// This code contributed by Rajput-Ji
</script>
输出:
YES
时间复杂度: O( log(n))
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