给定两个整数L和R ，任务是查找范围内的双素数个数。
A number N is called double prime when the count of prime numbers in the range 1 to N (excluding 1 and including N) is also prime.
例子：
Input: L = 3, R = 10
Output: 4
Explanation:
For 3, we have the range 1, 2, 3, and count of prime number is 2 (which is also a prime no.)
For 4, we have the range 1, 2, 3, 4, and count of a prime number is 2 (which is also a prime no.)
For 5, we have the range 1, 2, 3, 4, 5, and count of a prime number is 3 (which is also a prime no.)
For 6, we have the range 1, 2, 3, 4, 5, 6, and count of prime numbers is 3 (which is also a prime no.)
For 7, we have the range 1, 2, 3, 4, 5, 6, 7, and count of prime numbers is 4 which is nonprime.
Similarly for other numbers till R = 10, the count of prime numbers is nonprime. Hence the count of double prime numbers is 4.
Input: L = 4, R = 12
Output: 5
Explanation:
For the given range there are total 5 double prime numbers.
方法：
为了解决上述问题，我们将使用Sieve的概念来生成质数。
- 生成0到10 6的所有素数并存储在数组中。
- 初始化变量计数以保持质数从1到第ith个位置的跟踪。
- 然后，对于每个素数，我们将增加计数，并设置dp [count] = 1(其中dp是存储双素数的数组)，指示从1到第ith个素数位的数量。
- 最后，找到dp数组的累加和，因此答案将为dp [R] – dp [L – 1] 。
下面是上述方法的实现：
C++
// C++ program to find the count
// of Double Prime numbers
// in the range L to R
#include
using namespace std;
// Array to make Sieve
// where arr[i]=0 indicates
// non prime and arr[i] = 1
// indicates prime
int arr[1000001];
// Array to find double prime
int dp[1000001];
// Function to find the number
// double prime numbers in range
void count()
{
int maxN = 1000000, i, j;
// Assume all numbers as prime
for (i = 0; i < maxN; i++)
arr[i] = 1;
arr[0] = 0;
arr[1] = 0;
for (i = 2; i * i <= maxN; i++) {
// Check if the number is prime
if (arr[i] == 1) {
// check for multiples of i
for (j = 2 * i; j <= maxN; j += i) {
// Make all multiples of
// ith prime as non-prime
arr[j] = 0;
}
}
}
int cnt = 0;
for (i = 0; i <= maxN; i++) {
// Check if number at ith position
// is prime then increment count
if (arr[i] == 1)
cnt++;
if (arr[cnt] == 1)
// Indicates count of numbers
// from 1 to i that are
// also prime and
// hence double prime
dp[i] = 1;
else
// If number is not a double prime
dp[i] = 0;
}
for (i = 1; i <= maxN; i++)
// finding cumulative sum
dp[i] += dp[i - 1];
}
// Driver code
int main()
{
int L = 4, R = 12;
count();
cout << dp[R] - dp[L - 1];
return 0;
} Java
// Java program to find the count
// of Double Prime numbers
// in the range L to R
import java.util.*;
import java.lang.*;
class GFG{
// Array to make Sieve
// where arr[i]=0 indicates
// non prime and arr[i] = 1
// indicates prime
static int[] arr = new int[1000001];
// Array to find double prime
static int[] dp = new int[1000001];
// Function to find the number
// double prime numbers in range
static void count()
{
int maxN = 1000000, i, j;
// Assume all numbers as prime
for (i = 0; i < maxN; i++)
arr[i] = 1;
arr[0] = 0;
arr[1] = 0;
for (i = 2; i * i <= maxN; i++)
{
// Check if the number is prime
if (arr[i] == 1)
{
// check for multiples of i
for (j = 2 * i; j <= maxN; j += i)
{
// Make all multiples of
// ith prime as non-prime
arr[j] = 0;
}
}
}
int cnt = 0;
for (i = 0; i <= maxN; i++)
{
// Check if number at ith position
// is prime then increment count
if (arr[i] == 1)
cnt++;
if (arr[cnt] == 1)
// Indicates count of numbers
// from 1 to i that are
// also prime and
// hence double prime
dp[i] = 1;
else
// If number is not a double prime
dp[i] = 0;
}
for (i = 1; i <= maxN; i++)
// finding cumulative sum
dp[i] += dp[i - 1];
}
// Driver code
public static void main(String[] args)
{
int L = 4, R = 12;
count();
System.out.println(dp[R] - dp[L - 1]);
}
}
// This code is contributed by offbeatPython3
# Python3 program to find the count
# of Double Prime numbers
# in the range L to R
# Array to make Sieve
# where arr[i]=0 indicates
# non prime and arr[i] = 1
# indicates prime
arr = [0] * 1000001
# Array to find double prime
dp = [0] * 1000001
# Function to find the number
# double prime numbers in range
def count():
maxN = 1000000
# Assume all numbers as prime
for i in range(0, maxN):
arr[i] = 1
arr[0] = 0
arr[1] = 0
i = 2
while(i * i <= maxN):
# Check if the number is prime
if (arr[i] == 1):
# Check for multiples of i
for j in range(2 * i, maxN + 1, i):
# Make all multiples of
# ith prime as non-prime
arr[j] = 0
i += 1
cnt = 0
for i in range(0, maxN + 1):
# Check if number at ith position
# is prime then increment count
if (arr[i] == 1):
cnt += 1
if (arr[cnt] == 1):
# Indicates count of numbers
# from 1 to i that are
# also prime and
# hence double prime
dp[i] = 1
else:
# If number is not a double prime
dp[i] = 0
for i in range(0, maxN + 1):
# Finding cumulative sum
dp[i] += dp[i - 1]
# Driver code
L = 4
R = 12
count()
print(dp[R] - dp[L - 1])
# This code is contributed by sanjoy_62C#
// C# program to find the count
// of Double Prime numbers
// in the range L to R
using System;
class GFG{
// Array to make Sieve
// where arr[i]=0 indicates
// non prime and arr[i] = 1
// indicates prime
static int[] arr = new int[1000001];
// Array to find double prime
static int[] dp = new int[1000001];
// Function to find the number
// double prime numbers in range
static void count()
{
int maxN = 1000000, i, j;
// Assume all numbers as prime
for (i = 0; i < maxN; i++)
arr[i] = 1;
arr[0] = 0;
arr[1] = 0;
for (i = 2; i * i <= maxN; i++)
{
// Check if the number is prime
if (arr[i] == 1)
{
// check for multiples of i
for (j = 2 * i; j <= maxN; j += i)
{
// Make all multiples of
// ith prime as non-prime
arr[j] = 0;
}
}
}
int cnt = 0;
for (i = 0; i <= maxN; i++)
{
// Check if number at ith position
// is prime then increment count
if (arr[i] == 1)
cnt++;
if (arr[cnt] == 1)
// Indicates count of numbers
// from 1 to i that are
// also prime and
// hence double prime
dp[i] = 1;
else
// If number is not a double prime
dp[i] = 0;
}
for (i = 1; i <= maxN; i++)
// finding cumulative sum
dp[i] += dp[i - 1];
}
// Driver code
public static void Main()
{
int L = 4, R = 12;
count();
Console.Write(dp[R] - dp[L - 1]);
}
}
// This code is contributed by Code_MechJavascript
输出：
5