给定范围[L，R]，我们需要找到范围[L，R]中素数总数的计数，其中0 <= L <= R <10000。请考虑存在大量查询不同的范围。
例子：
Input : Query 1 : L = 1, R = 10
Query 2 : L = 5, R = 10
Output : 4
2
Explanation
Primes in the range L = 1 to R = 10 are
{2, 3, 5, 7}. Therefore for query, answer
is 4 {2, 3, 5, 7}.
For the second query, answer is 2 {5, 7}.
一个简单的解决方案是对每个查询[L，R]进行跟踪。从L遍历到R，检查当前数字是否为质数。如果是，则增加计数。最后返回计数。
一个有效的解决方案是使用Eratosthenes筛子查找所有达到给定极限的素数。然后，我们计算一个前缀数组以存储计数，直到每个值超出限制为止。一旦有了前缀数组，就可以在O(1)时间内回答查询。我们只需要返回prefix [R] – prefix [L-1]。
C++
// CPP program to answer queries for count of
// primes in given range.
#include
using namespace std;
const int MAX = 10000;
// prefix[i] is going to store count of primes
// till i (including i).
int prefix[MAX + 1];
void buildPrefix()
{
// Create a boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
bool prime[MAX + 1];
memset(prime, true, sizeof(prime));
for (int p = 2; p * p <= MAX; p++) {
// If prime[p] is not changed, then
// it is a prime
if (prime[p] == true) {
// Update all multiples of p
for (int i = p * 2; i <= MAX; i += p)
prime[i] = false;
}
}
// Build prefix array
prefix[0] = prefix[1] = 0;
for (int p = 2; p <= MAX; p++) {
prefix[p] = prefix[p - 1];
if (prime[p])
prefix[p]++;
}
}
// Returns count of primes in range from L to
// R (both inclusive).
int query(int L, int R)
{
return prefix[R] - prefix[L - 1];
}
// Driver code
int main()
{
buildPrefix();
int L = 5, R = 10;
cout << query(L, R) << endl;
L = 1, R = 10;
cout << query(L, R) << endl;
return 0;
} Java
// Java program to answer queries for
// count of primes in given range.
import java.util.*;
class GFG {
static final int MAX = 10000;
// prefix[i] is going to store count
// of primes till i (including i).
static int prefix[] = new int[MAX + 1];
static void buildPrefix() {
// Create a boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
boolean prime[] = new boolean[MAX + 1];
Arrays.fill(prime, true);
for (int p = 2; p * p <= MAX; p++) {
// If prime[p] is not changed, then
// it is a prime
if (prime[p] == true) {
// Update all multiples of p
for (int i = p * 2; i <= MAX; i += p)
prime[i] = false;
}
}
// Build prefix array
prefix[0] = prefix[1] = 0;
for (int p = 2; p <= MAX; p++) {
prefix[p] = prefix[p - 1];
if (prime[p])
prefix[p]++;
}
}
// Returns count of primes in range
// from L to R (both inclusive).
static int query(int L, int R)
{
return prefix[R] - prefix[L - 1];
}
// Driver code
public static void main(String[] args) {
buildPrefix();
int L = 5, R = 10;
System.out.println(query(L, R));
L = 1; R = 10;
System.out.println(query(L, R));
}
}
// This code is contributed by Anant Agarwal.Python3
# Python3 program to answer queries for
# count of primes in given range.
MAX = 10000
# prefix[i] is going to
# store count of primes
# till i (including i).
prefix =[0]*(MAX + 1)
def buildPrefix():
# Create a boolean array value in
# prime[i] will "prime[0..n]". A
# finally be false if i is Not a
# prime, else true.
prime = [1]*(MAX + 1)
p = 2
while(p * p <= MAX):
# If prime[p] is not changed,
# then it is a prime
if (prime[p] == 1):
# Update all multiples of p
i = p * 2
while(i <= MAX):
prime[i] = 0
i += p
p+=1
# Build prefix array
# prefix[0] = prefix[1] = 0;
for p in range(2,MAX+1):
prefix[p] = prefix[p - 1]
if (prime[p]==1):
prefix[p]+=1
# Returns count of primes
# in range from L to
# R (both inclusive).
def query(L, R):
return prefix[R]-prefix[L - 1]
# Driver code
if __name__=='__main__':
buildPrefix()
L = 5
R = 10
print(query(L, R))
L = 1
R = 10
print(query(L, R))
# This code is contributed by mits.C#
// C# program to answer
// queries for count of
// primes in given range.
using System;
class GFG
{
static int MAX = 10000;
// prefix[i] is going
// to store count of
// primes till i (including i).
static int[] prefix = new int[MAX + 1];
static void buildPrefix()
{
// Create a boolean array
// "prime[0..n]". A value
// in prime[i] will finally
// be false if i is Not a
// prime, else true.
bool[] prime = new bool[MAX + 1];
for (int p = 2;
p * p <= MAX; p++)
{
// If prime[p] is
// not changed, then
// it is a prime
if (prime[p] == false)
{
// Update all
// multiples of p
for (int i = p * 2;
i <= MAX; i += p)
prime[i] = true;
}
}
// Build prefix array
prefix[0] = prefix[1] = 0;
for (int p = 2; p <= MAX; p++)
{
prefix[p] = prefix[p - 1];
if (prime[p] == false)
prefix[p]++;
}
}
// Returns count of primes
// in range from L to R
// (both inclusive).
static int query(int L, int R)
{
return prefix[R] -
prefix[L - 1];
}
// Driver code
public static void Main()
{
buildPrefix();
int L = 5, R = 10;
Console.WriteLine(query(L, R));
L = 1; R = 10;
Console.WriteLine(query(L, R));
}
}
// This code is contributed
// by mits.PHP
输出：
2
4