给定一个整数“和”(小于10 ^ 8)，任务是找到一对和等于给定“和”的素数
在所有可能的对中，所选对之间的绝对差必须最小。
如果不能将“和”表示为两个质数之和，则打印“不能表示为两个质数之和”。
例子：
Input : Sum = 1002
Output : Primes: 499 503
Explanation
1002 can be represented as sum of many prime number pairs
such as
499 503
479 523
461 541
439 563
433 569
431 571
409 593
401 601...
But 499 and 503 is the only pair which has minimum difference
Input :Sum = 2002
Output : Primes: 983 1019
解决方案
- 我们将创建一个Eratosthenes筛子，该筛子将存储所有质数，并检查O(1)时间中的质数是否为质数。
- 现在，找到两个质数之和等于给定变量’sum’的和。我们将开始一个从sum / 2到1的循环(以最小化绝对差)，并检查循环计数器“ i”和“ sum-i”是否均为素数。
- 如果它们是素数，那么我们将打印它们并退出循环。
- 如果不能将“和”表示为两个素数的和，那么我们将打印“不能表示为两个素数的和”。
以下是上述解决方案的实现：
C++
// C++ implementation of the above approach
#include
using namespace std;
#define MAX 100000000
// stores whether a number is prime or not
bool prime[MAX + 1];
// create the 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 false if i is Not a prime, else true.
memset(prime, true, sizeof(prime));
prime[1] = false;
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 as non-prime
for (int i = p * 2; i <= MAX; i += p)
prime[i] = false;
}
}
}
// find the two prime numbers with minimum
// difference and whose sum is equal to
// variable sum
void find_Prime(int sum)
{
// start from sum/2 such that
// difference between i and sum-i will be
// minimum
for (int i = sum / 2; i > 1; i--) {
// if both 'i' and 'sum - i' are prime then print
// them and break the loop
if (prime[i] && prime[sum - i]) {
cout << i << " " << (sum - i) << endl;
return;
}
}
// if there is no prime
cout << "Cannot be represented as sum of two primes" << endl;
}
// Driver code
int main()
{
// create the sieve
SieveOfEratosthenes();
int sum = 1002;
// find the primes
find_Prime(sum);
return 0;
} Java
//Java implementation of the above approach
class GFG {
static final int MAX = 100000000;
// stores whether a number is prime or not
static boolean prime[] = new boolean[MAX + 1];
// create the 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 false if i is Not a prime, else true.
for (int i = 0; i < prime.length; i++) {
prime[i] = true;
}
prime[1] = false;
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 as non-prime
for (int i = p * 2; i <= MAX; i += p) {
prime[i] = false;
}
}
}
}
// find the two prime numbers with minimum
// difference and whose sum is equal to
// variable sum
static void find_Prime(int sum) {
// start from sum/2 such that
// difference between i and sum-i will be
// minimum
for (int i = sum / 2; i > 1; i--) {
// if both 'i' and 'sum - i' are prime then print
// them and break the loop
if (prime[i] && prime[sum - i]) {
System.out.println(i + " " + (sum - i));
return;
}
}
// if there is no prime
System.out.println("Cannot be represented as sum of two primes");
}
public static void main(String []args) {
// create the sieve
SieveOfEratosthenes();
int sum = 1002;
// find the primes
find_Prime(sum);
}
}
/*This code is contributed by 29AjayKumar*/Python3
# Python 3 implementation of the above approach
from math import sqrt
# stores whether a number is prime or not
# create the sieve of eratosthenes
def SieveOfEratosthenes():
MAX = 1000001
# Create a boolean array "prime[0..n]" and
# initialize all entries it as true. A value
# in prime[i] will finally be false if i is
# Not a prime, else true.
prime = [True for i in range(MAX + 1)]
prime[1] = False
for p in range(2, int(sqrt(MAX)) + 1, 1):
# If prime[p] is not changed,
# then it is a prime
if (prime[p] == True):
# Update all multiples of p
# as non-prime
for i in range(p * 2, MAX + 1, p):
prime[i] = False
return prime
# find the two prime numbers with minimum
# difference and whose sum is equal to
# variable sum
def find_Prime(sum):
# start from sum/2 such that difference
# between i and sum-i will be minimum
# create the sieve
prime = SieveOfEratosthenes()
i = int(sum / 2)
while(i > 1):
# if both 'i' and 'sum - i' are prime
# then print them and break the loop
if (prime[i] and prime[sum - i]):
print(i, (sum - i))
return
i -= 1
# if there is no prime
print("Cannot be represented as sum",
"of two primes")
# Driver code
if __name__ == '__main__':
sum = 1002
# find the primes
find_Prime(sum)
# This code is contributed by
# Shashank_SharmaC#
// C# implementation of the
// above approach
class GFG
{
static int MAX = 1000000;
// stores whether a number is
// prime or not
static bool[] prime = new bool[MAX + 1];
// create the 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
// false if i is Not a prime, else true.
for (int i = 0; i < prime.Length; i++)
{
prime[i] = true;
}
prime[1] = false;
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
// as non-prime
for (int i = p * 2;
i <= MAX; i += p)
{
prime[i] = false;
}
}
}
}
// find the two prime numbers with
// minimum difference and whose sum
// is equal to variable sum
static void find_Prime(int sum)
{
// start from sum/2 such that
// difference between i and sum-i
// will be minimum
for (int i = sum / 2; i > 1; i--)
{
// if both 'i' and 'sum - i'
// are prime then print
// them and break the loop
if (prime[i] && prime[sum - i])
{
System.Console.WriteLine(i + " " +
(sum - i));
return;
}
}
// if there is no prime
System.Console.WriteLine("Cannot be represented " +
"as sum of two primes");
}
// Driver Code
static void Main()
{
// create the sieve
SieveOfEratosthenes();
int sum = 1002;
// find the primes
find_Prime(sum);
}
}
// This code is contributed by mits输出：
499 503