给定n个形式为[L，R]的查询。任务是在每个查询的范围内找到两个质数之间的最大差值。如果范围内没有素数，则打印0。所有范围都低于100005。
例子：
Input : Q = 3
query1 = [2, 5]
query2 = [2, 2]
query3 = [24, 28]
Output : 3
0
0
In first query, 2 and 5 are prime number
in the range with maximum difference which
is 3. In second and third queries, there
is only 1 prime number in range, so output
is 0.这个想法是使用Eratosthenes筛和一些预计算来计算素数。
以下是解决问题的步骤：
第1步：使用Eratosthenes算法的筛子查找质数。
步骤2：制作一个数组，假设prefix [] ，其中prefix [i]表示小于或等于i的最大素数。
步骤3：制作一个数组，假设suffix [] ，其中suffix [i]表示大于或等于i的最小素数。
步骤4：现在，对于每个具有[L，R]的查询，请执行以下操作：
if (prefix[R] R)
return 0;
else
return prefix[R] - suffix[L];以下是此方法的实现：
C++
// CPP program to find maximum differences between
// two prime numbers in given ranges
#include
using namespace std;
#define MAX 100005
// Declare global variables to assign heap memory and avoid
// stack overflow
bool prime[MAX];
int prefix[MAX], suffix[MAX];
// Precompute Sieve, Prefix array, Suffix array
void precompute(int prefix[], int suffix[])
{
memset(prime, true, sizeof(prime));
// Sieve of Eratosthenes
for (int i = 2; i * i < MAX; i++) {
if (prime[i]) {
for (int j = i * i; j < MAX; j += i)
prime[j] = false;
}
}
prefix[1] = 1;
suffix[MAX - 1] = 1e9 + 7;
// Precomputing Prefix array.
for (int i = 2; i < MAX; i++) {
if (prime[i])
prefix[i] = i;
else
prefix[i] = prefix[i - 1];
}
// Precompute Suffix array.
for (int i = MAX - 1; i > 1; i--) {
if (prime[i])
suffix[i] = i;
else
suffix[i] = suffix[i + 1];
}
}
// Function to solve each query
int query(int prefix[], int suffix[], int L, int R)
{
if (prefix[R] < L || suffix[L] > R)
return 0;
else
return prefix[R] - suffix[L];
}
// Driven Program
int main()
{
int q = 3;
int L[] = { 2, 2, 24 };
int R[] = { 5, 2, 28 };
precompute(prefix, suffix);
for (int i = 0; i < q; i++)
cout << query(prefix, suffix, L[i], R[i]) << endl;
return 0;
} Java
// Java program to find maximum differences between
// two prime numbers in given ranges
public class GFG {
final static int MAX = 100005;
// Precompute Sieve, Prefix array, Suffix array
static void precompute(int prefix[], int suffix[])
{
boolean prime[] = new boolean[MAX];
for (int i = 0; i < MAX; i++) {
prime[i] = true;
}
// Sieve of Eratosthenes
for (int i = 2; i * i < MAX; i++) {
if (prime[i]) {
for (int j = i * i; j < MAX; j += i) {
prime[j] = false;
}
}
}
prefix[1] = 1;
suffix[MAX - 1] = (int)1e9 + 7;
// Precomputing Prefix array.
for (int i = 2; i < MAX; i++) {
if (prime[i]) {
prefix[i] = i;
}
else {
prefix[i] = prefix[i - 1];
}
}
// Precompute Suffix array.
for (int i = MAX - 2; i > 1; i--) {
if (prime[i]) {
suffix[i] = i;
}
else {
suffix[i] = suffix[i + 1];
}
}
}
// Function to solve each query
static int query(int prefix[], int suffix[], int L,
int R)
{
if (prefix[R] < L || suffix[L] > R) {
return 0;
}
else {
return prefix[R] - suffix[L];
}
}
// Driven Program
public static void main(String[] args)
{
int q = 3;
int L[] = { 2, 2, 24 };
int R[] = { 5, 2, 28 };
int prefix[] = new int[MAX], suffix[]
= new int[MAX];
precompute(prefix, suffix);
for (int i = 0; i < q; i++) {
System.out.println(
query(prefix, suffix, L[i], R[i]));
}
}
}
/*This code is contributed by Rajput-Ji*/Python3
# Python 3 program to find maximum
# differences between two prime numbers
# in given ranges
from math import sqrt
MAX = 100005
# Precompute Sieve, Prefix array, Suffix array
def precompute(prefix, suffix):
prime = [True for i in range(MAX)]
# Sieve of Eratosthenes
k = int(sqrt(MAX))
for i in range(2, k, 1):
if (prime[i]):
for j in range(i * i, MAX, i):
prime[j] = False
prefix[1] = 1
suffix[MAX - 1] = int(1e9 + 7)
# Precomputing Prefix array.
for i in range(2, MAX, 1):
if (prime[i]):
prefix[i] = i
else:
prefix[i] = prefix[i - 1]
# Precompute Suffix array.
i = MAX - 2
while(i > 1):
if (prime[i]):
suffix[i] = i
else:
suffix[i] = suffix[i + 1]
i -= 1
# Function to solve each query
def query(prefix, suffix, L, R):
if (prefix[R] < L or suffix[L] > R):
return 0
else:
return prefix[R] - suffix[L]
# Driver Code
if __name__ == '__main__':
q = 3
L = [2, 2, 24]
R = [5, 2, 28]
prefix = [0 for i in range(MAX)]
suffix = [0 for i in range(MAX)]
precompute(prefix, suffix)
for i in range(0, q, 1):
print(query(prefix, suffix,
L[i], R[i]))
# This code is contributed by
# Surendra_GangwarC#
// C# program to find maximum differences between
// two prime numbers in given ranges
using System;
public class GFG {
static readonly int MAX = 100005;
// Precompute Sieve, Prefix array, Suffix array
static void precompute(int[] prefix, int[] suffix)
{
bool[] prime = new bool[MAX];
for (int i = 0; i < MAX; i++) {
prime[i] = true;
}
// Sieve of Eratosthenes
for (int i = 2; i * i < MAX; i++) {
if (prime[i]) {
for (int j = i * i; j < MAX; j += i) {
prime[j] = false;
}
}
}
prefix[1] = 1;
suffix[MAX - 1] = (int)1e9 + 7;
// Precomputing Prefix array.
for (int i = 2; i < MAX; i++) {
if (prime[i]) {
prefix[i] = i;
}
else {
prefix[i] = prefix[i - 1];
}
}
// Precompute Suffix array.
for (int i = MAX - 2; i > 1; i--) {
if (prime[i]) {
suffix[i] = i;
}
else {
suffix[i] = suffix[i + 1];
}
}
}
// Function to solve each query
static int query(int[] prefix, int[] suffix, int L,
int R)
{
if (prefix[R] < L || suffix[L] > R) {
return 0;
}
else {
return prefix[R] - suffix[L];
}
}
// Driven Program
public static void Main()
{
int q = 3;
int[] L = { 2, 2, 24 };
int[] R = { 5, 2, 28 };
int[] prefix = new int[MAX];
int[] suffix = new int[MAX];
precompute(prefix, suffix);
for (int i = 0; i < q; i++) {
Console.WriteLine(
query(prefix, suffix, L[i], R[i]));
}
}
}
/*This code is contributed by 29AjayKumar*/PHP
1; $i--)
{
if ($prime[$i])
$suffix[$i] = $i;
else
$suffix[$i] = $suffix[$i + 1];
}
}
// Function to solve each query
function query($prefix, $suffix, $L, $R)
{
if ($prefix[$R] < $L || $suffix[$L] > $R)
return 0;
else
return $prefix[$R] - $suffix[$L];
}
// Driver Code
$q = 3;
$L = array( 2, 2, 24 );
$R = array( 5, 2, 28 );
$prefix = array_fill(0, $MAX + 1, 0);
$suffix = array_fill(0, $MAX + 1, 0);
precompute($prefix, $suffix);
for ($i = 0; $i < $q; $i++)
echo query($prefix, $suffix,
$L[$i], $R[$i]) . "\n";
// This code is contributed by mits
?>输出：
3
0
0