检查数字的二进制等价词是否以“ 001”结尾

给定正整数N ，任务是检查该整数的等效二进制数是否以“ 001”结尾。
如果以“ 001”结尾，则打印“是”。否则，打印“否”。
例子 ：

Input: N = 9 
Output: Yes 
Explanation 
Binary of 9 = 1001, which ends with 001
Input: N = 5 
Output: No 
Binary of 5 = 101, which does not end in 001 
 

天真的方法
查找N的二进制等效项，并检查001是否为其二进制等效项的后缀。
下面是上述方法的实现：

C++

// C++ implementation of the
// above approach
 
#include 
using namespace std;
 
// Function returns true if
// s1 is suffix of s2
bool isSuffix(string s1,
              string s2)
{
    int n1 = s1.length();
    int n2 = s2.length();
    if (n1 > n2)
        return false;
    for (int i = 0; i < n1; i++)
        if (s1[n1 - i - 1]
            != s2[n2 - i - 1])
            return false;
    return true;
}
 
// Function to check if binary equivalent
// of a number ends in "001" or not
bool CheckBinaryEquivalent(int N)
{
 
    // To store the binary
    // number
    int B_Number = 0;
    int cnt = 0;
 
    while (N != 0) {
 
        int rem = N % 2;
        int c = pow(10, cnt);
        B_Number += rem * c;
        N /= 2;
 
        // Count used to store
        // exponent value
        cnt++;
    }
 
    string bin = to_string(B_Number);
    return isSuffix("001", bin);
}
 
// Driver code
int main()
{
 
    int N = 9;
    if (CheckBinaryEquivalent(N))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java

// Java implementation of the above approach
class GFG{
     
// Function returns true if
// s1 is suffix of s2
static boolean isSuffix(String s1, String s2)
{
    int n1 = s1.length();
    int n2 = s2.length();
     
    if (n1 > n2)
        return false;
             
    for(int i = 0; i < n1; i++)
       if (s1.charAt(n1 - i - 1) !=
           s2.charAt(n2 - i - 1))
           return false;
    return true;
}
     
// Function to check if binary equivalent
// of a number ends in "001" or not
static boolean CheckBinaryEquivalent(int N)
{
     
    // To store the binary
    // number
    int B_Number = 0;
    int cnt = 0;
     
    while (N != 0)
    {
     
        int rem = N % 2;
        int c = (int)Math.pow(10, cnt);
        B_Number += rem * c;
        N /= 2;
     
        // Count used to store
        // exponent value
        cnt++;
    }
    String bin = Integer.toString(B_Number);
    return isSuffix("001", bin);
}
     
// Driver code
public static void main (String[] args)
{
    int N = 9;
     
    if (CheckBinaryEquivalent(N))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the
# above approach
 
# Function returns true if
# s1 is suffix of s2
def isSuffix(s1, s2) :
 
    n1 = len(s1);
    n2 = len(s2);
    if (n1 > n2) :
        return False;
    for i in range(n1) :
        if (s1[n1 - i - 1] != s2[n2 - i - 1]) :
            return False;
    return True;
 
# Function to check if binary equivalent
# of a number ends in "001" or not
def CheckBinaryEquivalent(N) :
 
    # To store the binary
    # number
    B_Number = 0;
    cnt = 0;
 
    while (N != 0) :
 
        rem = N % 2;
        c = 10 ** cnt;
        B_Number += rem * c;
        N //= 2;
 
        # Count used to store
        # exponent value
        cnt += 1;
 
    bin = str(B_Number);
    return isSuffix("001", bin);
 
# Driver code
if __name__ == "__main__" :
 
    N = 9;
    if (CheckBinaryEquivalent(N)) :
        print("Yes");
    else :
        print("No");
     
# This code is contributed by AnkitRai01

C#

// C# implementation of the above approach
using System;
 
class GFG{
     
// Function returns true if
// s1 is suffix of s2
static bool isSuffix(string s1, string s2)
{
    int n1 = s1.Length;
    int n2 = s2.Length;
         
    if (n1 > n2)
        return false;
                 
    for(int i = 0; i < n1; i++)
       if (s1[n1 - i - 1] !=
           s2[n2 - i - 1])
           return false;
    return true;
}
         
// Function to check if binary equivalent
// of a number ends in "001" or not
static bool CheckBinaryEquivalent(int N)
{
         
    // To store the binary
    // number
    int B_Number = 0;
    int cnt = 0;
         
    while (N != 0)
    {
        int rem = N % 2;
        int c = (int)Math.Pow(10, cnt);
        B_Number += rem * c;
        N /= 2;
         
        // Count used to store
        // exponent value
        cnt++;
    }
    string bin = B_Number.ToString();
    return isSuffix("001", bin);
}
     
// Driver code
public static void Main (string[] args)
{
    int N = 9;
     
    if (CheckBinaryEquivalent(N))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed by AnkitRai01

输出：

Yes

时间复杂度： O(N)
辅助空间： O(1)
高效方法
我们可以观察到，仅当(N – 1)被8整除时，数字的二进制等价词以“ 001”结尾。

Illustration: 
The sequence 1, 9, 17, 25, 33……. has 001 as the suffix in their binary representation. 
N^th term of the above sequence is denoted by 8 * N + 1 
So the binary equivalent of a number ends in “001” only when (N – 1) % 8 == 0 
 

下面是上述方法的实现：

C++

// C++ implementation of the above
// approach
 
#include 
using namespace std;
 
// Function to check if binary
// equivalent of a number ends
// in "001" or not
bool CheckBinaryEquivalent(int N)
{
    // To check if binary equivalent
    // of a number ends in
    // "001" or not
    return (N - 1) % 8 == 0;
}
 
// Driver code
int main()
{
 
    int N = 9;
    if (CheckBinaryEquivalent(N))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java

// Java implementation of the above approach
class GFG{
     
// Function to check if binary
// equivalent of a number ends
// in "001" or not
static boolean CheckBinaryEquivalent(int N)
{
     
    // To check if binary equivalent
    // of a number ends in
    // "001" or not
    return (N - 1) % 8 == 0;
}
     
// Driver code
public static void main (String[] args)
{
    int N = 9;
     
    if (CheckBinaryEquivalent(N))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the above approach
 
# Function to check if binary
# equivalent of a number ends
# in "001" or not
def CheckBinaryEquivalent(N):
 
    # To check if binary equivalent
    # of a number ends in
    # "001" or not
    return (N - 1) % 8 == 0;
 
# Driver code
if __name__ == "__main__":
 
    N = 9;
     
    if (CheckBinaryEquivalent(N)):
        print("Yes");
    else :
        print("No");
     
# This code is contributed by AnkitRai01

C＃

// C# implementation of the above approach
using System;
 
class GFG{
         
// Function to check if binary
// equivalent of a number ends
// in "001" or not
static bool CheckBinaryEquivalent(int N)
{
         
    // To check if binary equivalent
    // of a number ends in
    // "001" or not
    return (N - 1) % 8 == 0;
}
         
// Driver code
public static void Main (string[] args)
{
    int N = 9;
         
    if (CheckBinaryEquivalent(N))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed by AnkitRai01

Java脚本

 
// Javascript implementation of the above
// approach
 
// Function to check if binary
// equivalent of a number ends
// in "001" or not
function CheckBinaryEquivalent(N)
{
    // To check if binary equivalent
    // of a number ends in
    // "001" or not
    return (N - 1) % 8 == 0;
}
 
// Driver code
var N = 9;
if (CheckBinaryEquivalent(N))
    document.write( "Yes");
else
    document.write( "No");
 

输出：

Yes

时间复杂度： O(1)
辅助空间： O(1)