给定一个由N 个整数组成的数组arr[]和一个范围L到R ,任务是找到数组中从索引L 到 R满足以下条件的元素总数:
where F(x) is Euler’s Totient Function.
例子:
Input: arr[] = {2, 4, 5, 8}, L = 1, R = 3
Output: 2
Explanation:
Here in the given range, F(2) = 1 and (2 – 1) = 1. Similarly, F(5) = 4 and (5 – 1) = 4.
So the total count of indices which satisfy the condition is 2.
Input: arr[] = {9, 3, 4, 6, 8}, L = 3, R = 5
Output: 0
Explanation :
In the given range there is no element that satisfies the given condition.
So the total count is 0.
朴素的方法:解决这个问题的朴素的方法是迭代数组的所有元素,并检查当前元素的欧拉Totient值是否比它本身小1。如果是,则增加计数。
时间复杂度: O(N * sqrt(N))
辅助空间: O(1)
有效的方法:
Euler’s Totient function F(n) for an input n is the count of numbers in {1, 2, 3, …, n} that are relatively prime to n, i.e., the numbers whose Greatest Common Divisor with n is 1.
- 如果我们观察,我们可以注意到上面给定的条件只有质数才能满足。
- 所以,我们需要做的就是计算给定范围内素数的总数。
- 我们将使用埃拉托色尼筛法来有效地计算素数。
- 此外,我们将预先计算计数数组中素数的个数。
下面是上述方法的实现。
C++
// C++ program for the above approach
#include
using namespace std;
long long prime[1000001] = { 0 };
// Seiev of Erotosthenes method to
// compute all primes
void seiveOfEratosthenes()
{
for (int i = 2; i < 1000001;
i++) {
prime[i] = 1;
}
for (int i = 2; i * i < 1000001;
i++) {
// If current number is
// marked prime then mark
// its multiple as non-prime
if (prime[i] == 1) {
for (int j = i * i;
j < 1000001; j += i) {
prime[j] = 0;
}
}
}
}
// Function to count the number
// of element satisfying the condition
void CountElements(int arr[],
int n, int L,
int R)
{
seiveOfEratosthenes();
long long countPrime[n + 1]
= { 0 };
// Compute the number of primes
// in count prime array
for (int i = 1; i <= n; i++) {
countPrime[i] = countPrime[i - 1]
+ prime[arr[i - 1]];
}
// Print the number of elements
// satisfying the condition
cout << countPrime[R]
- countPrime[L - 1]
<< endl;
return;
}
// Driver Code
int main()
{
// Given array
int arr[] = { 2, 4, 5, 8 };
// Size of the array
int N = sizeof(arr) / sizeof(int);
int L = 1, R = 3;
// Function Call
CountElements(arr, N, L, R);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
static int prime[] = new int[1000001];
// Seiev of Erotosthenes method to
// compute all primes
static void seiveOfEratosthenes()
{
for(int i = 2; i < 1000001; i++)
{
prime[i] = 1;
}
for(int i = 2; i * i < 1000001; i++)
{
// If current number is
// marked prime then mark
// its multiple as non-prime
if (prime[i] == 1)
{
for(int j = i * i;
j < 1000001; j += i)
{
prime[j] = 0;
}
}
}
}
// Function to count the number
// of element satisfying the condition
static void CountElements(int arr[], int n,
int L, int R)
{
seiveOfEratosthenes();
int countPrime[] = new int[n + 1];
// Compute the number of primes
// in count prime array
for(int i = 1; i <= n; i++)
{
countPrime[i] = countPrime[i - 1] +
prime[arr[i - 1]];
}
// Print the number of elements
// satisfying the condition
System.out.print(countPrime[R] -
countPrime[L - 1] + "\n");
return;
}
// Driver Code
public static void main(String[] args)
{
// Given array
int arr[] = { 2, 4, 5, 8 };
// Size of the array
int N = arr.length;
int L = 1, R = 3;
// Function Call
CountElements(arr, N, L, R);
}
}
// This code is contributed by amal kumar choubey
Python3
# Python3 program for the above approach
prime = [0] * (1000001)
# Seiev of Erotosthenes method to
# compute all primes
def seiveOfEratosthenes():
for i in range(2, 1000001):
prime[i] = 1
i = 2
while(i * i < 1000001):
# If current number is
# marked prime then mark
# its multiple as non-prime
if (prime[i] == 1):
for j in range(i * i, 1000001, i):
prime[j] = 0
i += 1
# Function to count the number
# of element satisfying the condition
def CountElements(arr, n, L, R):
seiveOfEratosthenes()
countPrime = [0] * (n + 1)
# Compute the number of primes
# in count prime array
for i in range(1, n + 1):
countPrime[i] = (countPrime[i - 1] +
prime[arr[i - 1]])
# Print the number of elements
# satisfying the condition
print(countPrime[R] -
countPrime[L - 1])
return
# Driver Code
# Given array
arr = [ 2, 4, 5, 8 ]
# Size of the array
N = len(arr)
L = 1
R = 3
# Function call
CountElements(arr, N, L, R)
# This code is contributed by sanjoy_62
C#
// C# program for the above approach
using System;
class GFG{
static int []prime = new int[1000001];
// Seiev of Erotosthenes method to
// compute all primes
static void seiveOfEratosthenes()
{
for(int i = 2; i < 1000001; i++)
{
prime[i] = 1;
}
for(int i = 2; i * i < 1000001; i++)
{
// If current number is
// marked prime then mark
// its multiple as non-prime
if (prime[i] == 1)
{
for(int j = i * i;
j < 1000001; j += i)
{
prime[j] = 0;
}
}
}
}
// Function to count the number
// of element satisfying the condition
static void CountElements(int []arr, int n,
int L, int R)
{
seiveOfEratosthenes();
int []countPrime = new int[n + 1];
// Compute the number of primes
// in count prime array
for(int i = 1; i <= n; i++)
{
countPrime[i] = countPrime[i - 1] +
prime[arr[i - 1]];
}
// Print the number of elements
// satisfying the condition
Console.Write(countPrime[R] -
countPrime[L - 1] + "\n");
return;
}
// Driver Code
public static void Main(String[] args)
{
// Given array
int []arr = { 2, 4, 5, 8 };
// Size of the array
int N = arr.Length;
int L = 1, R = 3;
// Function Call
CountElements(arr, N, L, R);
}
}
// This code is contributed by sapnasingh4991
Javascript
2
时间复杂度: O(N * log(logN))
辅助空间: O(N)
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