给定三个整数a，b，n。您的任务是打印a和b之间的数字个数，包括也有n个除数的数字。如果一个数字有n个除数，包括1和它本身，则称为n除数。
例子：
Input : a = 1, b = 7, n = 2
Output : 4
There are four numbers with 2 divisors in
range [1, 7]. The numbers are 2, 3, 5, and 7.
天真的方法：
天真的方法是检查a和b之间的所有数字，以找出n个除数中的多少个，从而找出每个数的每个除数的数量。如果等于n，则为n除数
高效方法：
任何数字都可以以素数分解的形式写成，即数字为x且p1，p2..pm是除以x的素数，因此x = p1 e1 * p2 e2 ….pm em其中e1，e2 … em为质数p1，p2….pm的指数。因此，x的除数为(e1 + 1)*(e2 + 1)… *(em + 1)。
现在第二个观察是质数大于sqrt(x)的指数不能超过1。让我们通过矛盾证明这一点，假设质数P大于sqrt(x)且其质数x的指数E大于1(E> = 2)所以P ^ E sqrt(x)所以P ^ E>(sqrt(x)) E和E> = 2所以P E总是大于x
第三个观察结果是，在x的素数分解中，大于sqrt(x)的质数个数始终小于等于1。这也可以通过上述矛盾类似地证明。
现在解决这个问题
第1步：应用erathesthenes筛网并计算平方根最大为sqrt(b)的素数。
步骤2：从a到b遍历每个数字，并通过将该数字除以质数来计算该数字中每个质数的指数，并使用公式numberofdivisors(x)=(e1 + 1)*(e2 + 1)… 。(em + 1)。
步骤3：如果在数字> 1的基础上将所有小于等于该数平方根的质数相除后，这意味着存在一个大于其平方根的质数，该质数将被除，并且其指数将始终如上文所述。
C++
// C++ program to count numbers with n divisors
#include
using namespace std;
// applying sieve of eratosthenes
void sieve(bool primes[], int x)
{
primes[1] = false;
// if a number is prime mark all its multiples
// as non prime
for (int i=2; i*i <= x; i++)
{
if (primes[i] == true)
{
for (int j=2; j*i <= x; j++)
primes[i*j] = false;
}
}
}
// function that returns numbers of number that have
// n divisors in range from a to b. x is sqrt(b) + 1.
int nDivisors(bool primes[], int x, int a, int b, int n)
{
// result holds number of numbers having n divisors
int result = 0;
// vector to hold all the prime numbers between 1
// ans sqrt(b)
vector v;
for (int i = 2; i <= x; i++)
if (primes[i] == true)
v.push_back (i);
// Traversing all numbers in given range
for (int i=a; i<=b; i++)
{
// initialising temp as i
int temp = i;
// total holds the number of divisors of i
int total = 1;
int j = 0;
// we need to use that prime numbers that
// are less than equal to sqrt(temp)
for (int k = v[j]; k*k <= temp; k = v[++j])
{
// holds the exponent of k in prime
// factorization of i
int count = 0;
// repeatedly divide temp by k till it is
// divisible and accordingly increase count
while (temp%k == 0)
{
count++;
temp = temp/k;
}
// using the formula no.of divisors =
// (e1+1)*(e2+1)....
total = total*(count+1);
}
// if temp is not equal to 1 then there is
// prime number in prime factorization of i
// greater than sqrt(i)
if (temp != 1)
total = total*2;
// if i is a ndvisor number increase result
if (total == n)
result++;
}
return result;
}
// Returns count of numbers in [a..b] having
// n divisors.
int countNDivisors(int a, int b, int n)
{
int x = sqrt(b) + 1;
// primes[i] = true if i is a prime number
bool primes[x];
// initialising each number as prime
memset(primes, true, sizeof(primes));
sieve(primes, x);
return nDivisors(primes, x, a, b, n);
}
// driver code
int main()
{
int a = 1, b = 7, n = 2;
cout << countNDivisors(a, b, n);
return 0;
} Java
// Java program to count numbers with n divisors
import java.util.*;
class GFG{
// applying sieve of eratosthenes
static void sieve(boolean[] primes, int x)
{
primes[1] = true;
// if a number is prime mark all its multiples
// as non prime
for (int i=2; i*i <= x; i++)
{
if (primes[i] == false)
{
for (int j=2; j*i <= x; j++)
primes[i*j] = true;
}
}
}
// function that returns numbers of number that have
// n divisors in range from a to b. x is sqrt(b) + 1.
static int nDivisors(boolean[] primes, int x, int a, int b, int n)
{
// result holds number of numbers having n divisors
int result = 0;
// vector to hold all the prime numbers between 1
// ans sqrt(b)
ArrayList v=new ArrayList();
for (int i = 2; i <= x; i++)
if (primes[i] == false)
v.add(i);
// Traversing all numbers in given range
for (int i=a; i<=b; i++)
{
// initialising temp as i
int temp = i;
// total holds the number of divisors of i
int total = 1;
int j = 0;
// we need to use that prime numbers that
// are less than equal to sqrt(temp)
for (int k = v.get(j); k*k <= temp; k = v.get(++j))
{
// holds the exponent of k in prime
// factorization of i
int count = 0;
// repeatedly divide temp by k till it is
// divisible and accordingly increase count
while (temp%k == 0)
{
count++;
temp = temp/k;
}
// using the formula no.of divisors =
// (e1+1)*(e2+1)....
total = total*(count+1);
}
// if temp is not equal to 1 then there is
// prime number in prime factorization of i
// greater than sqrt(i)
if (temp != 1)
total = total*2;
// if i is a ndvisor number increase result
if (total == n)
result++;
}
return result;
}
// Returns count of numbers in [a..b] having
// n divisors.
static int countNDivisors(int a, int b, int n)
{
int x = (int)Math.sqrt(b) + 1;
// primes[i] = true if i is a prime number
boolean[] primes=new boolean[x+1];
// initialising each number as prime
sieve(primes, x);
return nDivisors(primes, x, a, b, n);
}
// driver code
public static void main(String[] args)
{
int a = 1, b = 7, n = 2;
System.out.println(countNDivisors(a, b, n));
}
}
// This code is contributed by mits Python3
# Python3 program to count numbers
# with n divisors
import math;
# applying sieve of eratosthenes
def sieve(primes, x):
primes[1] = False;
# if a number is prime mark all
# its multiples as non prime
i = 2;
while (i * i <= x):
if (primes[i] == True):
j = 2;
while (j * i <= x):
primes[i * j] = False;
j += 1;
i += 1;
# function that returns numbers of number
# that have n divisors in range from a to b.
# x is sqrt(b) + 1.
def nDivisors(primes, x, a, b, n):
# result holds number of numbers
# having n divisors
result = 0;
# vector to hold all the prime
# numbers between 1 and sqrt(b)
v = [];
for i in range(2, x + 1):
if (primes[i]):
v.append(i);
# Traversing all numbers in given range
for i in range(a, b + 1):
# initialising temp as i
temp = i;
# total holds the number of
# divisors of i
total = 1;
j = 0;
# we need to use that prime numbers that
# are less than equal to sqrt(temp)
k = v[j];
while (k * k <= temp):
# holds the exponent of k in prime
# factorization of i
count = 0;
# repeatedly divide temp by k till it is
# divisible and accordingly increase count
while (temp % k == 0):
count += 1;
temp = int(temp / k);
# using the formula no.of divisors =
# (e1+1)*(e2+1)....
total = total * (count + 1);
j += 1;
k = v[j];
# if temp is not equal to 1 then there is
# prime number in prime factorization of i
# greater than sqrt(i)
if (temp != 1):
total = total * 2;
# if i is a ndivisor number
# increase result
if (total == n):
result += 1;
return result;
# Returns count of numbers in [a..b]
# having n divisors.
def countNDivisors(a, b, n):
x = int(math.sqrt(b) + 1);
# primes[i] = true if i is a prime number
# initialising each number as prime
primes = [True] * (x + 1);
sieve(primes, x);
return nDivisors(primes, x, a, b, n);
# Driver code
a = 1;
b = 7;
n = 2;
print(countNDivisors(a, b, n));
# This code is contributed by mitsC#
// C# program to count numbers with n divisors
using System.Collections;
using System;
class GFG{
// applying sieve of eratosthenes
static void sieve(bool[] primes, int x)
{
primes[1] = true;
// if a number is prime mark all its multiples
// as non prime
for (int i=2; i*i <= x; i++)
{
if (primes[i] == false)
{
for (int j=2; j*i <= x; j++)
primes[i*j] = true;
}
}
}
// function that returns numbers of number that have
// n divisors in range from a to b. x is sqrt(b) + 1.
static int nDivisors(bool[] primes, int x, int a, int b, int n)
{
// result holds number of numbers having n divisors
int result = 0;
// vector to hold all the prime numbers between 1
// ans sqrt(b)
ArrayList v=new ArrayList();
for (int i = 2; i <= x; i++)
if (primes[i] == false)
v.Add(i);
// Traversing all numbers in given range
for (int i=a; i<=b; i++)
{
// initialising temp as i
int temp = i;
// total holds the number of divisors of i
int total = 1;
int j = 0;
// we need to use that prime numbers that
// are less than equal to sqrt(temp)
for (int k = (int)v[j]; k*k <= temp; k = (int)v[++j])
{
// holds the exponent of k in prime
// factorization of i
int count = 0;
// repeatedly divide temp by k till it is
// divisible and accordingly increase count
while (temp%k == 0)
{
count++;
temp = temp/k;
}
// using the formula no.of divisors =
// (e1+1)*(e2+1)....
total = total*(count+1);
}
// if temp is not equal to 1 then there is
// prime number in prime factorization of i
// greater than sqrt(i)
if (temp != 1)
total = total*2;
// if i is a ndivisor number increase result
if (total == n)
result++;
}
return result;
}
// Returns count of numbers in [a..b] having
// n divisors.
static int countNDivisors(int a, int b, int n)
{
int x = (int)Math.Sqrt(b) + 1;
// primes[i] = true if i is a prime number
bool[] primes=new bool[x+1];
// initialising each number as prime
sieve(primes, x);
return nDivisors(primes, x, a, b, n);
}
// driver code
public static void Main()
{
int a = 1, b = 7, n = 2;
Console.WriteLine(countNDivisors(a, b, n));
}
}
// This code is contributed by mitsPHP
输出：
4