给定三个整数N，L和R。任务是计算在[L，R](包括两个端点)范围内的自然数的数量，这些自然数与N相对质数。
例子：
Input: N = 10, L = 1, R = 25
Output: 10
Explanation:
10 natural numbers (in the range 1 to 25) are relatively prime to 10.
They are 1, 3, 7, 9, 11, 13, 17, 19, 21, 23.
Input: N = 12, L = 7, R = 38
Output: 11
Explanation:
11 natural numbers (in the range 1 to 38) are relatively prime to 12.
They are 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37.
方法：
- 首先，将数字N分解。因此，找出N的所有素数。
- 将数为N的素数存储在数组中。
- 我们可以确定不大于R且可被N的素数整除的自然数的总数。
- 假设值为y。因此，不大于R的y个自然数至少具有一个与N的公因数。
- 因此，这些y数不能相对于N素数。
- 因此，相对于N而言素数不大于R的自然数为R-y。
- 现在，类似地，我们需要找出不大于L-1的N的相对质数的数量。
- 然后，从R的答案中减去L-1的结果。
下面是上述方法的实现：
C++
// C++ code to count of natural
// numbers in range [L, R] which
// are relatively prime with N
#include
using namespace std;
#define maxN (long long)1000000000000
// container of all the primes
// up to sqrt(n)
vector prime;
// Function to calculate prime
// factors of n
void sieve(long long n)
{
// run the sieve of Eratosthenes
bool check[1000007] = { 0 };
long long i, j;
// 0(false) means prime,
// 1(true) means not prime
check[0] = 1, check[1] = 1,
check[2] = 0;
// no even number is
// prime except for 2
for (i = 4; i <= n; i += 2)
check[i] = true;
for (i = 3; i * i <= n; i += 2)
if (!check[i]) {
// all the multiples of each
// each prime numbers are
// non-prime
for (j = i * i; j <= n; j += 2 * i)
check[j] = true;
}
prime.push_back(2);
// get all the primes
// in prime vector
for (int i = 3; i <= n; i += 2)
if (!check[i])
prime.push_back(i);
return;
}
// Count the number of numbers
// up to m which are divisible
// by given prime numbers
long long count(long long a[],
int n, long long m)
{
long long parity[3] = { 0 };
// Run from i= 000..0 to i= 111..1
// or check all possible
// subsets of the array
for (int i = 1; i < (1 << n); i++) {
long long mult = 1;
for (int j = 0; j < n; j++)
if (i & (1 << j))
mult *= a[j];
// take the multiplication
// of all the set bits
// if the number of set bits
// is odd, then add to the
// number of multiples
parity[__builtin_popcount(i) & 1]
+= (m / mult);
}
return parity[1] - parity[0];
}
// Function calculates all number
// not greater than 'm' which are
// relatively prime with n.
long long countRelPrime(
long long n,
long long m)
{
long long a[20];
int i = 0, j = 0;
long long pz = prime.size();
while (n != 1 && i < pz) {
// if square of the prime number
// is greater than 'n', it can't
// be a factor of 'n'
if ((long long)prime[i]
* (long long)prime[i]
> n)
break;
// if prime is a factor of
// n then increment count
if (n % prime[i] == 0)
a[j] = (long long)prime[i], j++;
while (n % prime[i] == 0)
n /= prime[i];
i++;
}
if (n != 1)
a[j] = n, j++;
return m - count(a, j, m);
}
void countRelPrimeInRange(
long long n, long long l,
long long r)
{
sieve(sqrt(maxN));
long long result
= countRelPrime(n, r)
- countRelPrime(n, l - 1);
cout << result << "\n";
}
// Driver code
int main()
{
long long N = 7, L = 3, R = 9;
countRelPrimeInRange(N, L, R);
return 0;
} Java
// Java code to count of natural
// numbers in range [L, R] which
// are relatively prime with N
import java.util.*;
class GFG{
static int maxN = 100000000;
// Container of all the primes
// up to sqrt(n)
static Vector prime = new Vector();
// Function to calculate prime
// factors of n
static void sieve(int n)
{
// Run the sieve of Eratosthenes
boolean[] check = new boolean[1000007];
for(int i = 0; i < 1000007; i++)
check[i] = false;
int i, j;
// 0(false) means prime,
// 1(true) means not prime
check[0] = false;
check[1] = true;
check[2] = false;
// No even number is
// prime except for 2
for(i = 4; i <= n; i += 2)
check[i] = true;
for(i = 3; i * i <= n; i += 2)
if (!check[i])
{
// All the multiples of each
// each prime numbers are
// non-prime
for(j = i * i; j <= n; j += 2 * i)
check[j] = true;
}
prime.add(2);
// Get all the primes
// in prime vector
for(i = 3; i <= n; i += 2)
if (!check[i])
prime.add(i);
return;
}
static int countSetBits(int n)
{
int count = 0;
while (n > 0)
{
count += n & 1;
n >>= 1;
}
return count;
}
// Count the number of numbers
// up to m which are divisible
// by given prime numbers
static int count(int a[],
int n, int m)
{
int[] parity = new int[3];
for(int i = 0; i < 3; i++)
parity[i] = 0;
// Run from i= 000..0 to i= 111..1
// or check all possible
// subsets of the array
for(int i = 1; i < (1 << n); i++)
{
int mult = 1;
for(int j = 0; j < n; j++)
if ((i & (1 << j)) != 0)
mult *= a[j];
// Take the multiplication
// of all the set bits
// If the number of set bits
// is odd, then add to the
// number of multiples
parity[countSetBits(i) & 1] += (m / mult);
}
return parity[1] - parity[0];
}
// Function calculates all number
// not greater than 'm' which are
// relatively prime with n.
static int countRelPrime(int n, int m)
{
int[] a = new int[20];
int i = 0, j = 0;
int pz = prime.size();
while(n != 1 && i < pz)
{
// If square of the prime number
// is greater than 'n', it can't
// be a factor of 'n'
if ((int)prime.get(i) *
(int)prime.get(i) > n)
break;
// If prime is a factor of
// n then increment count
if (n % prime.get(i) == 0)
{
a[j] = (int)prime.get(i);
j++;
}
while (n % prime.get(i) == 0)
n /= prime.get(i);
i++;
}
if (n != 1)
{
a[j] = n;
j++;
}
return m - count(a, j, m);
}
static void countRelPrimeInRange(int n, int l,
int r)
{
sieve((int)Math.sqrt(maxN));
int result = countRelPrime(n, r) -
countRelPrime(n, l - 1);
System.out.println(result);
}
// Driver code
public static void main(String[] args)
{
int N = 7, L = 3, R = 9;
countRelPrimeInRange(N, L, R);
}
}
// This code is contributed by grand_master Python3
# Python3 code to count of natural
# numbers in range [L, R] which
# are relatively prime with N
from math import sqrt, floor
maxN = 1000000000000
# Container of all the primes
# up to sqrt(n)
prime = []
# Function to calculate prime
# factors of n
def sieve(n):
# Run the sieve of Eratosthenes
check = [0] * (1000007)
i, j = 0, 0
# 0(false) means prime,
# 1(True) means not prime
check[0] = 1
check[1] = 1
check[2] = 0
# No even number is
# prime except for 2
for i in range(4, n + 1, 2):
check[i] = True
for i in range(3, n + 1, 2):
if i * i > n:
break
if (not check[i]):
# All the multiples of each
# each prime numbers are
# non-prime
for j in range(2 * i, n + 1, 2 * i):
check[j] = True
prime.append(2)
# Get all the primes
# in prime vector
for i in range(3, n + 1, 2):
if (not check[i]):
prime.append(i)
return
# Count the number of numbers
# up to m which are divisible
# by given prime numbers
def count(a, n, m):
parity = [0] * 3
# Run from i = 000..0 to i = 111..1
# or check all possible
# subsets of the array
for i in range(1, 1 << n):
mult = 1
for j in range(n):
if (i & (1 << j)):
mult *= a[j]
# Take the multiplication
# of all the set bits
# If the number of set bits
# is odd, then add to the
# number of multiples
parity[bin(i).count('1') & 1] += (m // mult)
return parity[1] - parity[0]
# Function calculates all number
# not greater than 'm' which are
# relatively prime with n.
def countRelPrime(n, m):
a = [0] * 20
i = 0
j = 0
pz = len(prime)
while (n != 1 and i < pz):
# If square of the prime number
# is greater than 'n', it can't
# be a factor of 'n'
if (prime[i] * prime[i] > n):
break
# If prime is a factor of
# n then increment count
if (n % prime[i] == 0):
a[j] = prime[i]
j += 1
while (n % prime[i] == 0):
n //= prime[i]
i += 1
if (n != 1):
a[j] = n
j += 1
return m - count(a, j, m)
def countRelPrimeInRange(n, l, r):
sieve(floor(sqrt(maxN)))
result = (countRelPrime(n, r) -
countRelPrime(n, l - 1))
print(result)
# Driver code
if __name__ == '__main__':
N = 7
L = 3
R = 9
countRelPrimeInRange(N, L, R)
# This code is contributed by mohit kumar 29C#
// C# code to count of natural
// numbers in range [L, R] which
// are relatively prime with N
using System;
using System.Collections.Generic;
class GFG{
static int maxN = 100000000;
// Container of all the primes
// up to sqrt(n)
static List prime = new List();
// Function to calculate prime
// factors of n
static void sieve(int n)
{
// Run the sieve of Eratosthenes
bool[] check = new bool[1000007];
for(int I = 0; I < 1000007; I++)
check[I] = false;
int i, j;
// 0(false) means prime,
// 1(true) means not prime
check[0] = false;
check[1] = true;
check[2] = false;
// No even number is
// prime except for 2
for(i = 4; i <= n; i += 2)
check[i] = true;
for(i = 3; i * i <= n; i += 2)
if (!check[i])
{
// All the multiples of each
// each prime numbers are
// non-prime
for(j = i * i; j <= n; j += 2 * i)
check[j] = true;
}
prime.Add(2);
// Get all the primes
// in prime vector
for(i = 3; i <= n; i += 2)
if (!check[i])
prime.Add(i);
return;
}
static int countSetBits(int n)
{
int count = 0;
while (n > 0)
{
count += n & 1;
n >>= 1;
}
return count;
}
// Count the number of numbers
// up to m which are divisible
// by given prime numbers
static int count(int[] a, int n, int m)
{
int[] parity = new int[3];
for(int i = 0; i < 3; i++)
parity[i] = 0;
// Run from i= 000..0 to i= 111..1
// or check all possible
// subsets of the array
for(int i = 1; i < (1 << n); i++)
{
int mult = 1;
for(int j = 0; j < n; j++)
if ((i & (1 << j)) != 0)
mult *= a[j];
// Take the multiplication
// of all the set bits
// If the number of set bits
// is odd, then add to the
// number of multiples
parity[countSetBits(i) & 1] += (m / mult);
}
return parity[1] - parity[0];
}
// Function calculates all number
// not greater than 'm' which are
// relatively prime with n.
static int countRelPrime(int n, int m)
{
int[] a = new int[20];
int i = 0, j = 0;
int pz = prime.Count;
while (n != 1 && i < pz)
{
// If square of the prime number
// is greater than 'n', it can't
// be a factor of 'n'
if ((int)prime[i] * (int)prime[i] > n)
break;
// If prime is a factor of
// n then increment count
if (n % prime[i] == 0)
{
a[j] = (int)prime[i];
j++;
}
while (n % prime[i] == 0)
n /= prime[i];
i++;
}
if (n != 1)
{
a[j] = n;
j++;
}
return m - count(a, j, m);
}
static void countRelPrimeInRange(int n, int l,
int r)
{
sieve((int)Math.Sqrt(maxN));
int result = countRelPrime(n, r) -
countRelPrime(n, l - 1);
Console.WriteLine(result);
}
// Driver Code
static void Main()
{
int N = 7, L = 3, R = 9;
countRelPrimeInRange(N, L, R);
}
}
// This code is contributed by divyeshrabadiya07 输出：
6