两个不同素数乘积的前 N 个自然数的逐位异或
原文:https://www . geeksforgeeks . org/两个不同素数乘积的第一个 n 个自然数的按位异或/
给定一个正整数 N ，任务是计算第一个 N 数的位异或，这是两个截然不同的质数的乘积。
示例:
输入: N = 20 输出: 7 解释: 范围[1，20]中正好是两个不同素数乘积的数是{6，10，12，14，15，18，20}。 这些数字的按位异或= 6 ^ 10 ^ 12 ^ 14 ^ 15 ^ 18 ^ 20 = 7
输入: N = 50 输出: 26
天真法:最简单的方法是迭代每个数直到NT5使用质因数分解法找到每个数的质因数。发现不同质因数的计数为 2 的数字，然后计算它们与答案的异或。核对所有数字后，打印得到的答案。
时间复杂度: O(N√N)* 辅助空间: O(1)
高效途径:优化上述途径，思路是稍加修改使用厄拉多塞的筛。按照以下步骤解决问题:
- 将变量和初始化为 0 ，以存储所需的结果。
- 创建一个大小为 N+1 的整数数组、 arr[] ，并用全零初始化，其中 arr[i] 表示 i 的不同素数的个数。
- 使用变量 i 在范围【2，N】中迭代，如果 arr[i] 的值为 0 ，则使用变量 j 遍历 i 的所有倍数，并将 arr[j] 的值增加 1 ，因为 i 是【T29】的质因数
- 使用变量 i 在范围【2，N】内迭代，如果arr【I】等于 2 ，则取 i 与 ans 的异或。
- 打印和的值作为结果。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count prime factors
// using Sieve of Eratosthenes
void sieve(int arr[], int n)
{
// Iterate in the [2, N]
for (int i = 2; i <= n; i++) {
// If the current number is prime
if (arr[i] == 0)
// Iterate over all multiples of i
for (int j = 2 * i; j <= n; j += i) {
// Increment arr[j] by 1 since
// i is a prime factor of j
arr[j]++;
}
}
}
// Function to find Bitwise XOR
// of first N natural numbers
// satisfying the given condition
void findXOR(int n)
{
// arr[i]: Stores the number of
// distinct prime factors of i
int arr[n + 1] = { 0 };
// Initialize the base cases
arr[0] = arr[1] = 1;
// Function Call to fill
// the array, arr[]
sieve(arr, n);
// Store the required result
int ans = 0;
// Iterate over the range [2, N]
for (int i = 2; i <= n; i++) {
// Check if the i-th number has
// exactly two distinct prime factor
if (arr[i] == 2) {
// If true, update the answer
ans = (ans ^ i);
}
}
// Print the result
cout << ans;
}
// Driver Code
int main()
{
// Given Input
int n = 20;
// Function Call
findXOR(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
public class GFG {
// Function to count prime factors
// using Sieve of Eratosthenes
static void sieve(int arr[], int n)
{
// Iterate in the [2, N]
for (int i = 2; i <= n; i++) {
// If the current number is prime
if (arr[i] == 0)
// Iterate over all multiples of i
for (int j = 2 * i; j <= n; j += i) {
// Increment arr[j] by 1 since
// i is a prime factor of j
arr[j]++;
}
}
}
// Function to find Bitwise XOR
// of first N natural numbers
// satisfying the given condition
static void findXOR(int n)
{
// arr[i]: Stores the number of
// distinct prime factors of i
int arr[] = new int[n + 1];
// Initialize the base cases
arr[0] = arr[1] = 1;
// Function Call to fill
// the array, arr[]
sieve(arr, n);
// Store the required result
int ans = 0;
// Iterate over the range [2, N]
for (int i = 2; i <= n; i++) {
// Check if the i-th number has
// exactly two distinct prime factor
if (arr[i] == 2) {
// If true, update the answer
ans = (ans ^ i);
}
}
// Print the result
System.out.println(ans);
}
// Driver code
public static void main(String[] args)
{
// Given Input
int n = 20;
// Function Call
findXOR(n);
}
}
// This code is contributed by abhinavjain194
Python 3
# Python3 program for the above approach
# Function to count prime factors
# using Sieve of Eratosthenes
def sieve(arr, n):
# Iterate in the [2, N]
for i in range(2, n + 1, 1):
# If the current number is prime
if (arr[i] == 0):
# Iterate over all multiples of i
for j in range(2 * i, n + 1, i):
# Increment arr[j] by 1 since
# i is a prime factor of j
arr[j] += 1
# Function to find Bitwise XOR
# of first N natural numbers
# satisfying the given condition
def findXOR(n):
# arr[i]: Stores the number of
# distinct prime factors of i
arr = [0 for i in range(n + 1)]
# Initialize the base cases
arr[0] = arr[1] = 1
# Function Call to fill
# the array, arr[]
sieve(arr, n)
# Store the required result
ans = 0
# Iterate over the range [2, N]
for i in range(2, n + 1, 1):
# Check if the i-th number has
# exactly two distinct prime factor
if (arr[i] == 2):
# If true, update the answer
ans = (ans ^ i)
# Print the result
print(ans)
# Driver Code
if __name__ == '__main__':
# Given Input
n = 20
# Function Call
findXOR(n)
# This code is contributed by SURENDRA_GANGWAR
C
// C# program for the above approach
using System;
class GFG{
// Function to count prime factors
// using Sieve of Eratosthenes
static void sieve(int []arr, int n)
{
// Iterate in the [2, N]
for(int i = 2; i <= n; i++)
{
// If the current number is prime
if (arr[i] == 0)
// Iterate over all multiples of i
for(int j = 2 * i; j <= n; j += i)
{
// Increment arr[j] by 1 since
// i is a prime factor of j
arr[j]++;
}
}
}
// Function to find Bitwise XOR
// of first N natural numbers
// satisfying the given condition
static void findXOR(int n)
{
// arr[i]: Stores the number of
// distinct prime factors of i
int []arr = new int[n + 1];
// Initialize the base cases
arr[0] = arr[1] = 1;
// Function Call to fill
// the array, arr[]
sieve(arr, n);
// Store the required result
int ans = 0;
// Iterate over the range [2, N]
for(int i = 2; i <= n; i++)
{
// Check if the i-th number has
// exactly two distinct prime factor
if (arr[i] == 2)
{
// If true, update the answer
ans = (ans ^ i);
}
}
// Print the result
Console.WriteLine(ans);
}
// Driver code
public static void Main(String[] args)
{
// Given Input
int n = 20;
// Function Call
findXOR(n);
}
}
// This code is contributed by ankThon
java 描述语言
<script>
// JavaScript program for the above approach
// Function to count prime factors
// using Sieve of Eratosthenes
function sieve( arr, n)
{
// Iterate in the [2, N]
for (var i = 2; i <= n; i++) {
// If the current number is prime
if (arr[i] == 0)
// Iterate over all multiples of i
for (var j = 2 * i; j <= n; j += i) {
// Increment arr[j] by 1 since
// i is a prime factor of j
arr[j]++;
}
}
}
// Function to find Bitwise XOR
// of first N natural numbers
// satisfying the given condition
function findXOR( n)
{
// arr[i]: Stores the number of
// distinct prime factors of i
var arr = new Array(n + 1);
arr.fill(0);
// Initialize the base cases
arr[0] = arr[1] = 1;
// Function Call to fill
// the array, arr[]
sieve(arr, n);
// Store the required result
var ans = 0;
// Iterate over the range [2, N]
for (var i = 2; i <= n; i++) {
// Check if the i-th number has
// exactly two distinct prime factor
if (arr[i] == 2) {
// If true, update the answer
ans = (ans ^ i);
}
}
// Print the result
document.write( ans);
}
n = 20;
// Function Call
findXOR(n);
// This code is contributed by SoumikMondal
</script>
Output:
7
时间复杂度:O(N * log(logN)) 辅助空间: O(N)
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