给定数字N ，任务是对给定数字N的数字执行按位运算。按位运算包括：
- 查找给定数字N的所有数字的XOR
- 查找给定数字N的所有数字的OR
- 查找给定数字N的所有数字的AND
例子：
Input: N = 486
Output:
XOR = 10
OR = 14
AND = 0
Input: N = 123456
Output:
XOR = 10
OR = 14
AND = 0
方法：
- 取得号码
- 查找数字的数字并将其存储在数组中以进行计算。
- 现在，在此数组上一个接一个地执行各种按位运算(XOR，OR和AND)。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
int digit[100000];
// Function to find the digits
int findDigits(int n)
{
int count = 0;
while (n != 0) {
digit[count] = n % 10;
n = n / 10;
++count;
}
return count;
}
// Function to Find OR
// of all digits of a number
int OR_of_Digits(int n, int count)
{
int ans = 0;
for (int i = 0; i < count; i++) {
// Find OR of all digits
ans = ans | digit[i];
}
// return OR of digits
return ans;
}
// Function to Find AND
// of all digits of a number
int AND_of_Digits(int n, int count)
{
int ans = 0;
for (int i = 0; i < count; i++) {
// Find AND of all digits
ans = ans & digit[i];
}
// return AND of digits
return ans;
}
// Function to Find XOR
// of all digits of a number
int XOR_of_Digits(int n, int count)
{
int ans = 0;
for (int i = 0; i < count; i++) {
// Find XOR of all digits
ans = ans ^ digit[i];
}
// return XOR of digits
return ans;
}
// Driver code
void bitwise_operation(int N)
{
// Find and store all digits
int countOfDigit = findDigits(N);
// Find XOR of digits
cout << "XOR = "
<< XOR_of_Digits(N, countOfDigit)
<< endl;
// Find OR of digits
cout << "OR = "
<< OR_of_Digits(N, countOfDigit)
<< endl;
// Find AND of digits
cout << "AND = "
<< AND_of_Digits(N, countOfDigit)
<< endl;
}
// Driver code
int main()
{
int N = 123456;
bitwise_operation(N);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG{
static int []digit = new int[100000];
// Function to find the digits
static int findDigits(int n)
{
int count = 0;
while (n != 0) {
digit[count] = n % 10;
n = n / 10;
++count;
}
return count;
}
// Function to Find OR
// of all digits of a number
static int OR_of_Digits(int n, int count)
{
int ans = 0;
for (int i = 0; i < count; i++) {
// Find OR of all digits
ans = ans | digit[i];
}
// return OR of digits
return ans;
}
// Function to Find AND
// of all digits of a number
static int AND_of_Digits(int n, int count)
{
int ans = 0;
for (int i = 0; i < count; i++) {
// Find AND of all digits
ans = ans & digit[i];
}
// return AND of digits
return ans;
}
// Function to Find XOR
// of all digits of a number
static int XOR_of_Digits(int n, int count)
{
int ans = 0;
for (int i = 0; i < count; i++) {
// Find XOR of all digits
ans = ans ^ digit[i];
}
// return XOR of digits
return ans;
}
// Driver code
static void bitwise_operation(int N)
{
// Find and store all digits
int countOfDigit = findDigits(N);
// Find XOR of digits
System.out.print("XOR = "
+ XOR_of_Digits(N, countOfDigit)
+"\n");
// Find OR of digits
System.out.print("OR = "
+ OR_of_Digits(N, countOfDigit)
+"\n");
// Find AND of digits
System.out.print("AND = "
+ AND_of_Digits(N, countOfDigit)
+"\n");
}
// Driver code
public static void main(String[] args)
{
int N = 123456;
bitwise_operation(N);
}
}
// This code is contributed by sapnasingh4991Python 3
# Python 3 implementation of the approach
digit = [0]*(100000)
# Function to find the digits
def findDigits(n):
count = 0
while (n != 0):
digit[count] = n % 10;
n = n // 10;
count += 1
return count
# Function to Find OR
# of all digits of a number
def OR_of_Digits( n,count):
ans = 0
for i in range(count):
# Find OR of all digits
ans = ans | digit[i]
# return OR of digits
return ans
# Function to Find AND
# of all digits of a number
def AND_of_Digits(n, count):
ans = 0
for i in range(count):
# Find AND of all digits
ans = ans & digit[i]
# return AND of digits
return ans
# Function to Find XOR
# of all digits of a number
def XOR_of_Digits(n, count):
ans = 0
for i in range(count):
# Find XOR of all digits
ans = ans ^ digit[i]
# return XOR of digits
return ans
# Driver code
def bitwise_operation( N):
# Find and store all digits
countOfDigit = findDigits(N)
# Find XOR of digits
print("XOR = ",XOR_of_Digits(N, countOfDigit))
# Find OR of digits
print("OR = ",OR_of_Digits(N, countOfDigit))
# Find AND of digits
print("AND = ",AND_of_Digits(N, countOfDigit))
# Driver code
N = 123456;
bitwise_operation(N)
# This code is contributed by apurva rajC#
// C# implementation of the approach
using System;
class GFG{
static int []digit = new int[100000];
// Function to find the digits
static int findDigits(int n)
{
int count = 0;
while (n != 0) {
digit[count] = n % 10;
n = n / 10;
++count;
}
return count;
}
// Function to Find OR
// of all digits of a number
static int OR_of_Digits(int n, int count)
{
int ans = 0;
for (int i = 0; i < count; i++) {
// Find OR of all digits
ans = ans | digit[i];
}
// return OR of digits
return ans;
}
// Function to Find AND
// of all digits of a number
static int AND_of_Digits(int n, int count)
{
int ans = 0;
for (int i = 0; i < count; i++) {
// Find AND of all digits
ans = ans & digit[i];
}
// return AND of digits
return ans;
}
// Function to Find XOR
// of all digits of a number
static int XOR_of_Digits(int n, int count)
{
int ans = 0;
for (int i = 0; i < count; i++) {
// Find XOR of all digits
ans = ans ^ digit[i];
}
// return XOR of digits
return ans;
}
// Driver code
static void bitwise_operation(int N)
{
// Find and store all digits
int countOfDigit = findDigits(N);
// Find XOR of digits
Console.Write("XOR = "
+ XOR_of_Digits(N, countOfDigit)
+"\n");
// Find OR of digits
Console.Write("OR = "
+ OR_of_Digits(N, countOfDigit)
+"\n");
// Find AND of digits
Console.Write("AND = "
+ AND_of_Digits(N, countOfDigit)
+"\n");
}
// Driver code
public static void Main(String[] args)
{
int N = 123456;
bitwise_operation(N);
}
}
// This code is contributed by 29AjayKumar输出：
XOR = 7
OR = 7
AND = 0