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
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
Javascript
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.
Nth 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
输出:
Yes
时间复杂度: O(1)
辅助空间: O(1)