给定三个整数L ， R和M ，任务是找到[L，R]范围内的数字可被M整除的欧拉Totient函数的概率。
Euler’s Totient function is the count of numbers in {1, 2, 3, …, N} that are relatively prime to N, i.e., the numbers whose GCD (Greatest Common Divisor) with N is 1.
例子：
Input: L = 1, R = 5, M = 2
Output: 0.6
Explanation:
Euler’s Totient Function for N = 1, 2, 3, 4 and 5 is {1, 1, 2, 2, 4} respectively.
Count of Euler’s Totient Function divisible by M(= 2) is 3.
Therefore, the required probability is 3/5 = 0.6
Input: L = 1, R = 7, M = 4
Output: 0.142
Explanation:
Euler’s Totient Function for N = 1, 2, 3, ….7 is {1, 1, 2, 2, 4, 2, 6} respectively.
Count of Euler’s Totient Function divisible by M(= 4) is 1.
Therefore, the required probability is 1/7 = 0.142
方法：想法是预先计算Euler的Totient函数，并在给定范围内进行迭代，并计算被M整除的数字以计算概率。
For the computation of the Euler’s Totient Function, use the Euler’s Product Formula :
where pi is the prime factor of N.
对于N的每个素数i (L <= n <= R) ，执行以下步骤：
- 从[1，N]中减去i的所有倍数。
- 通过将N反复除以i来更新N。
- 如果减少的N大于1 ，则从结果中除去N的所有倍数。
要计算主要因子，请使用Eratosthenes方法的筛网。给定范围内的概率将为count /(L – R + 1) 。
下面是上述方法的实现：
C++
// C++ Program to implement
// the above approach
#include
using namespace std;
#define size 1000001
// Seieve of Erotosthenes
// to compute all primes
void seiveOfEratosthenes(int* prime)
{
prime[0] = 1, prime[1] = 0;
for (int i = 2; i * i < 1000001; i++) {
// If prime
if (prime[i] == 0) {
for (int j = i * i; j < 1000001;
j += i) {
// Mark all its multiples
// as non-prime
prime[j] = 1;
}
}
}
}
// Function to find the probability of
// Euler's Totient Function in a given range
float probabiltyEuler(int* prime, int L,
int R, int M)
{
int* arr = new int[size]{ 0 };
int* eulerTotient = new int[size]{ 0 };
int count = 0;
// Initializing two arrays
// with values from L to R
// for Euler's totient
for (int i = L; i <= R; i++) {
// Indexing from 0
eulerTotient[i - L] = i;
arr[i - L] = i;
}
for (int i = 2; i < 1000001; i++) {
// If the current number is prime
if (prime[i] == 0) {
// Checking if i is prime factor
// of numbers in range L to R
for (int j = (L / i) * i; j <= R;
j += i) {
if (j - L >= 0) {
// Update all the numbers
// which has prime factor i
eulerTotient[j - L]
= eulerTotient[j - L]
/ i * (i - 1);
while (arr[j - L] % i == 0) {
arr[j - L] /= i;
}
}
}
}
}
// If number in range has a
// prime factor > sqrt(number)
for (int i = L; i <= R; i++) {
if (arr[i - L] > 1) {
eulerTotient[i - L]
= (eulerTotient[i - L] / arr[i - L])
* (arr[i - L] - 1);
}
}
for (int i = L; i <= R; i++) {
// Count those which are divisible by M
if ((eulerTotient[i - L] % M) == 0) {
count++;
}
}
// Return the result
return (1.0 * count / (R + 1 - L));
}
// Driver Code
int main()
{
int* prime = new int[size]{ 0 };
seiveOfEratosthenes(prime);
int L = 1, R = 7, M = 3;
cout << probabiltyEuler(prime, L, R, M);
return 0;
} Java
// Java Program to implement
// the above approach
import java.util.*;
class GFG{
static final int size = 1000001;
// Seieve of Erotosthenes
// to compute all primes
static void seiveOfEratosthenes(int []prime)
{
prime[0] = 1;
prime[1] = 0;
for (int i = 2; i * i < 1000001; i++)
{
// If prime
if (prime[i] == 0)
{
for (int j = i * i; j < 1000001; j += i)
{
// Mark all its multiples
// as non-prime
prime[j] = 1;
}
}
}
}
// Function to find the probability of
// Euler's Totient Function in a given range
static float probabiltyEuler(int []prime, int L,
int R, int M)
{
int[] arr = new int[size];
int []eulerTotient = new int[size];
int count = 0;
// Initializing two arrays
// with values from L to R
// for Euler's totient
for (int i = L; i <= R; i++)
{
// Indexing from 0
eulerTotient[i - L] = i;
arr[i - L] = i;
}
for (int i = 2; i < 1000001; i++)
{
// If the current number is prime
if (prime[i] == 0)
{
// Checking if i is prime factor
// of numbers in range L to R
for (int j = (L / i) * i; j <= R; j += i)
{
if (j - L >= 0)
{
// Update all the numbers
// which has prime factor i
eulerTotient[j - L] = eulerTotient[j - L] /
i * (i - 1);
while (arr[j - L] % i == 0)
{
arr[j - L] /= i;
}
}
}
}
}
// If number in range has a
// prime factor > Math.sqrt(number)
for (int i = L; i <= R; i++)
{
if (arr[i - L] > 1)
{
eulerTotient[i - L] = (eulerTotient[i - L] / arr[i - L]) *
(arr[i - L] - 1);
}
}
for (int i = L; i <= R; i++)
{
// Count those which are divisible by M
if ((eulerTotient[i - L] % M) == 0)
{
count++;
}
}
// Return the result
return (float) (1.0 * count / (R + 1 - L));
}
// Driver Code
public static void main(String[] args)
{
int []prime = new int[size];
seiveOfEratosthenes(prime);
int L = 1, R = 7, M = 3;
System.out.print(probabiltyEuler(prime, L, R, M));
}
}
// This code is contributed by sapnasingh4991Python3
# Python3 program to implement
# the above approach
size = 1000001
# Seieve of Erotosthenes
# to compute all primes
def seiveOfEratosthenes(prime):
prime[0] = 1
prime[1] = 0
i = 2
while(i * i < 1000001):
# If prime
if (prime[i] == 0):
j = i * i
while(j < 1000001):
# Mark all its multiples
# as non-prime
prime[j] = 1
j = j + i
i += 1
# Function to find the probability of
# Euler's Totient Function in a given range
def probabiltyEuler(prime, L, R, M):
arr = [0] * size
eulerTotient = [0] * size
count = 0
# Initializing two arrays
# with values from L to R
# for Euler's totient
for i in range(L, R + 1):
# Indexing from 0
eulerTotient[i - L] = i
arr[i - L] = i
for i in range(2, 1000001):
# If the current number is prime
if (prime[i] == 0):
# Checking if i is prime factor
# of numbers in range L to R
for j in range((L // i) * i, R + 1, i):
if (j - L >= 0):
# Update all the numbers
# which has prime factor i
eulerTotient[j - L] = (eulerTotient[j - L] //
i * (i - 1))
while (arr[j - L] % i == 0):
arr[j - L] = arr[j - L] // i
# If number in range has a
# prime factor > Math.sqrt(number)
for i in range(L, R + 1):
if (arr[i - L] > 1):
eulerTotient[i - L] = ((eulerTotient[i - L] //
arr[i - L]) *
(arr[i - L] - 1))
for i in range(L, R + 1):
# Count those which are divisible by M
if ((eulerTotient[i - L] % M) == 0):
count += 1
# Return the result
return (float)(1.0 * count / (R + 1 - L))
# Driver code
prime = [0] * size
seiveOfEratosthenes(prime)
L, R, M = 1, 7, 3
print(probabiltyEuler(prime, L, R, M))
# This code is contributed by divyeshrabadiya07C#
// C# Program to implement
// the above approach
using System;
class GFG{
static readonly int size = 1000001;
// Seieve of Erotosthenes
// to compute all primes
static void seiveOfEratosthenes(int []prime)
{
prime[0] = 1;
prime[1] = 0;
for (int i = 2; i * i < 1000001; i++)
{
// If prime
if (prime[i] == 0)
{
for (int j = i * i; j < 1000001; j += i)
{
// Mark all its multiples
// as non-prime
prime[j] = 1;
}
}
}
}
// Function to find the probability of
// Euler's Totient Function in a given range
static float probabiltyEuler(int []prime, int L,
int R, int M)
{
int[] arr = new int[size];
int []eulerTotient = new int[size];
int count = 0;
// Initializing two arrays
// with values from L to R
// for Euler's totient
for (int i = L; i <= R; i++)
{
// Indexing from 0
eulerTotient[i - L] = i;
arr[i - L] = i;
}
for (int i = 2; i < 1000001; i++)
{
// If the current number is prime
if (prime[i] == 0)
{
// Checking if i is prime factor
// of numbers in range L to R
for (int j = (L / i) * i; j <= R; j += i)
{
if (j - L >= 0)
{
// Update all the numbers
// which has prime factor i
eulerTotient[j - L] = eulerTotient[j - L] /
i * (i - 1);
while (arr[j - L] % i == 0)
{
arr[j - L] /= i;
}
}
}
}
}
// If number in range has a
// prime factor > Math.Sqrt(number)
for (int i = L; i <= R; i++)
{
if (arr[i - L] > 1)
{
eulerTotient[i - L] = (eulerTotient[i - L] / arr[i - L]) *
(arr[i - L] - 1);
}
}
for (int i = L; i <= R; i++)
{
// Count those which are divisible by M
if ((eulerTotient[i - L] % M) == 0)
{
count++;
}
}
// Return the result
return (float) (1.0 * count / (R + 1 - L));
}
// Driver Code
public static void Main(String[] args)
{
int []prime = new int[size];
seiveOfEratosthenes(prime);
int L = 1, R = 7, M = 3;
Console.Write(probabiltyEuler(prime, L, R, M));
}
}
// This code is contributed by sapnasingh4991输出：
0.142857
时间复杂度： O(Nlog(N))
辅助空间： O(size)，其中size表示计算Sieve的数字。