给定Q个查询，其中每个查询由整数范围[L，R]组成，任务是从给定范围内找到其除数为质数的整数之和。
例子：
Input: Q[][] = {{2, 4}}
Output:
9
All the numbers in the range have only 2 divisors
which is prime.
(2 + 3 + 4) = 9.
Input: Q[][] = {{15, 17}, {2, 12}}
Output:
33
41
方法：使用Eratosthenes筛，找到每个素数的所有质数和除数，直到N为止。现在，创建一个前缀求和数组sum [] ，其中sum [i]将使用先前创建的Sieve数组存储范围为[0，i]的元素之和为除数的素数。现在，每个查询都可以在O(1)中以sum [r] – sum [l – 1]的形式回答。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
const int N = 100000;
// prime[i] stores 1 if i is prime
int prime[N];
// divi[i] stores the count of
// divisors of i
int divi[N];
// sum[i] will store the sum of all
// the integers from 0 to i whose
// count of divisors is prime
int sum[N];
// Function for Sieve of Eratosthenes
void SieveOfEratosthenes()
{
// Create a boolean array "prime[0..n]" and initialize
// all entries it as true. A value in prime[i] will
// finally be 0 if i is Not a prime, else true.
for (int i = 0; i < N; i++)
prime[i] = 1;
// 0 and 1 is not prime
prime[0] = prime[1] = 0;
for (int p = 2; p * p < N; p++) {
// If prime[p] is not changed, then it is a prime
if (prime[p] == 1) {
// Update all multiples of p greater than or
// equal to the square of it
// numbers which are multiple of p and are
// less than p^2 are already been marked.
for (int i = p * p; i < N; i += p)
prime[i] = 0;
}
}
}
// Function to count the divisors
void DivisorCount()
{
// For each number i we will go to each of
// the multiple of i and update the count
// of divisor of its multiple j as i is one
// of the factor of j
for (int i = 1; i < N; i++) {
for (int j = i; j < N; j += i) {
divi[j]++;
}
}
}
// Function for pre-computation
void pre()
{
for (int i = 1; i < N; i++) {
// If count of divisors of i is prime
if (prime[divi[i]] == 1) {
sum[i] = i;
}
}
// taking prefix sum
for (int i = 1; i < N; i++)
sum[i] += sum[i - 1];
}
// Driver code
int main()
{
int l = 5, r = 8;
// Find all the prime numbers till N
SieveOfEratosthenes();
// Update the count of divisors
// of all the numbers till N
DivisorCount();
// Precomputation for the prefix sum array
pre();
// Perform query
cout << sum[r] - sum[l - 1];
return 0;
} Java
//Java implementation of above approach
import java.util.*;
class GFG
{
static int N = 100000;
// prime[i] stores 1 if i is prime
static int prime[] = new int[N];
// divi[i] stores the count of
// divisors of i
static int divi[] = new int[N];
// sum[i] will store the sum of all
// the integers from 0 to i whose
// count of divisors is prime
static int sum[] = new int[N];
// Function for Sieve of Eratosthenes
static void SieveOfEratosthenes()
{
// Create a boolean array "prime[0..n]" and
// initialize all entries it as true.
// A value in prime[i] will finally be 0
// if i is Not a prime, else true.
for (int i = 0; i < N; i++)
prime[i] = 1;
// 0 and 1 is not prime
prime[0] = prime[1] = 0;
for (int p = 2; p * p < N; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p] == 1)
{
// Update all multiples of p greater than or
// equal to the square of it
// numbers which are multiple of p and are
// less than p^2 are already been marked.
for (int i = p * p; i < N; i += p)
prime[i] = 0;
}
}
}
// Function to count the divisors
static void DivisorCount()
{
// For each number i we will go to each of
// the multiple of i and update the count
// of divisor of its multiple j as i is one
// of the factor of j
for (int i = 1; i < N; i++)
{
for (int j = i; j < N; j += i)
{
divi[j]++;
}
}
}
// Function for pre-computation
static void pre()
{
for (int i = 1; i < N; i++)
{
// If count of divisors of i is prime
if (prime[divi[i]] == 1)
{
sum[i] = i;
}
}
// taking prefix sum
for (int i = 1; i < N; i++)
sum[i] += sum[i - 1];
}
// Driver code
public static void main(String args[])
{
int l = 5, r = 8;
// Find all the prime numbers till N
SieveOfEratosthenes();
// Update the count of divisors
// of all the numbers till N
DivisorCount();
// Precomputation for the prefix sum array
pre();
// Perform query
System.out.println( sum[r] - sum[l - 1]);
}
}
// This code is contributed by Arnab KunduPython3
# Python3 implementation of the approach
from math import sqrt
N = 100000;
# Create a boolean array "prime[0..n]" and
# initialize all entries it as true.
# A value in prime[i] will finally be 0 if
# i is Not a prime, else true.
# prime[i] stores 1 if i is prime
prime = [1] * N;
# divi[i] stores the count of
# divisors of i
divi = [0] * N;
# sum[i] will store the sum of all
# the integers from 0 to i whose
# count of divisors is prime
sum = [0] * N;
# Function for Sieve of Eratosthenes
def SieveOfEratosthenes() :
for i in range(N) :
prime[i] = 1;
# 0 and 1 is not prime
prime[0] = prime[1] = 0;
for p in range(2, int(sqrt(N)) + 1) :
# If prime[p] is not changed,
# then it is a prime
if (prime[p] == 1) :
# Update all multiples of p greater than or
# equal to the square of it
# numbers which are multiple of p and are
# less than p^2 are already been marked.
for i in range(p * p, N, p) :
prime[i] = 0;
# Function to count the divisors
def DivisorCount() :
# For each number i we will go to each of
# the multiple of i and update the count
# of divisor of its multiple j as i is one
# of the factor of j
for i in range(1, N) :
for j in range(i, N , i) :
divi[j] += 1;
# Function for pre-computation
def pre() :
for i in range(1, N) :
# If count of divisors of i is prime
if (prime[divi[i]] == 1) :
sum[i] = i;
# taking prefix sum
for i in range(1, N) :
sum[i] += sum[i - 1];
# Driver code
if __name__ == "__main__" :
l = 5; r = 8;
# Find all the prime numbers till N
SieveOfEratosthenes();
# Update the count of divisors
# of all the numbers till N
DivisorCount();
# Precomputation for the prefix sum array
pre();
# Perform query
print(sum[r] - sum[l - 1]);
# This code is contributed by AnkitRai01C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
static int N = 100000;
// prime[i] stores 1 if i is prime
static int []prime = new int[N];
// divi[i] stores the count of
// divisors of i
static int []divi = new int[N];
// sum[i] will store the sum of all
// the integers from 0 to i whose
// count of divisors is prime
static int []sum = new int[N];
// Function for Sieve of Eratosthenes
static void SieveOfEratosthenes()
{
// Create a boolean array "prime[0..n]"
// and initialize all entries it as true.
// A value in prime[i] will finally be 0
// if i is Not a prime, else true.
for (int i = 0; i < N; i++)
prime[i] = 1;
// 0 and 1 is not prime
prime[0] = prime[1] = 0;
for (int p = 2; p * p < N; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p] == 1)
{
// Update all multiples of p greater than
// or equal to the square of it
// numbers which are multiple of p and are
// less than p^2 are already been marked.
for (int i = p * p; i < N; i += p)
prime[i] = 0;
}
}
}
// Function to count the divisors
static void DivisorCount()
{
// For each number i we will go to each of
// the multiple of i and update the count
// of divisor of its multiple j as i is one
// of the factor of j
for (int i = 1; i < N; i++)
{
for (int j = i; j < N; j += i)
{
divi[j]++;
}
}
}
// Function for pre-computation
static void pre()
{
for (int i = 1; i < N; i++)
{
// If count of divisors of i is prime
if (prime[divi[i]] == 1)
{
sum[i] = i;
}
}
// taking prefix sum
for (int i = 1; i < N; i++)
sum[i] += sum[i - 1];
}
// Driver code
public static void Main(String []args)
{
int l = 5, r = 8;
// Find all the prime numbers till N
SieveOfEratosthenes();
// Update the count of divisors
// of all the numbers till N
DivisorCount();
// Precomputation for the prefix sum array
pre();
// Perform query
Console.WriteLine( sum[r] - sum[l - 1]);
}
}
// This code is contributed by 29AjayKumar输出：
12