给定两个整数X和Y ,其中X> Y ,任务是检查是否存在质数P ,使得如果从X重复减去P则得出Y。
例子:
Input: X = 100, Y = 98
Output: Yes
(100 – (2 * 1) = 98)
Input: X = 45, Y = 31
Output: Yes
(45 – (7 * 2)) = 31
天真的方法:对从2到x的每个整数运行一个循环。如果当前数字是质数,并且满足问题中给出的条件,则它是必需的数字。
有效方法:请注意,对于有效质数p , x – k * p = y或x – y = k * p 。假设p = 2,则(x – y)= 2,4,6,… (所有偶数)。这意味着,如果(x – y)是偶数,则答案始终是正确的。如果(X – Y)是除1以外的奇数,它总是有一个主要因素。它本身就是素数,或者是较小素数和其他一些整数的乘积。因此,对于除1以外的所有奇数,答案为True。
如果(x – y)= 1怎么办,它既不是素数也不是合成数。因此,这是答案为假的唯一情况。
下面是该方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
// Function that returns true if any
// prime number satisfies
// the given conditions
bool isPossible(int x, int y)
{
// No such prime exists
if ((x - y) == 1)
return false;
return true;
}
// Driver code
int main()
{
int x = 100, y = 98;
if (isPossible(x, y))
cout << "Yes";
else
cout << "No";
return 0;
}
Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function that returns true if any
// prime number satisfies
// the given conditions
static boolean isPossible(int x, int y)
{
// No such prime exists
if ((x - y) == 1)
return false;
return true;
}
// Driver code
public static void main(String[] args)
{
int x = 100, y = 98;
if (isPossible(x, y))
System.out.print("Yes");
else
System.out.print("No");
}
}
// This code is contributed by Rajput-Ji
Python3
# Python3 implementation of the approach
# Function that returns true if any
# prime number satisfies
# the given conditions
def isPossible(x, y):
# No such prime exists
if ((x - y) == 1):
return False
return True
# Driver code
x = 100
y = 98
if (isPossible(x, y)):
print("Yes")
else:
print("No")
# This code is contributed by Mohit Kumar
C#
// C# implementation of the approach
using System;
class GFG
{
// Function that returns true if any
// prime number satisfies
// the given conditions
static bool isPossible(int x, int y)
{
// No such prime exists
if ((x - y) == 1)
return false;
return true;
}
// Driver code
public static void Main(String[] args)
{
int x = 100, y = 98;
if (isPossible(x, y))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by 29AjayKumar
输出:
Yes