给定算术级数的第一项(A)和共同差(D)，以及质数(P)。任务是找到给定AP中第一个元素的位置，该位置是给定素数P的倍数。
例子：
Input: A = 4, D = 9, P = 11
Output: 2
Explanation :
The third term of the given AP is
a multiple of prime number 11.
First Term = 4
Second Term = 4+9 = 13
Third Term = 4+2*9 = 22
Input: A = 5, D = 6, P = 7
Output: 5
Explanation :
The sixth term of the given AP is
a multiple of prime number 7.
First Term = 5
Second Term = 5+6 = 11
Third Term = 5+2*6 = 17
Fourth Term = 5+3*6 = 23
Fifth Term = 5+4*6 = 29
Sixth Term = 5+5*5 = 35
方法：
设项为A N。所以，
AN = (A + (N-1)*D)现在，假设A N是P的倍数。然后，
A + (N-1)*D = k*P
Where, k is a constant.
现在让A为(A％P)，D为(D％P)。因此，我们有(N-1)* D =(k * P – A)。
在RHS上加上和减去P，我们得到：
(N-1)*D = P(k-1) + (P-A),
Where P-A is a non-negative number
(since A is replaced by A%P which is less than P)
最终双方都接受mod：
((N-1)*D)%P = (P-A)%P
or, ((N-1)D)%P = P-A
让我们找到一个X 因此答案N为： 下面是上述方法的实现：((X*(P-A)) % P) + 1.
C++
#include
Java
// Java Program to Find First
// element in AP which is
// multiple of given prime
class GFG
{
// Iterative Function to
// calculate (x^y)%p in
// O(log y) */
static int power(int x,
int y, int p)
{
// Initialize result
int res = 1;
// Update x if it is
// more than or equal to p
x = x % p;
while (y > 0)
{
// If y is odd, multiply
// x with result
if ((y & 1) != 0)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
// function to find nearest
// element in common
static int NearestElement(int A,
int D, int P)
{
// base conditions
if (A == 0)
return 0;
else if (D == 0)
return -1;
else
{
int X = power(D, P - 2, P);
return (X * (P - A)) % P;
}
}
// Driver code
public static void main(String args[])
{
int A = 4, D = 9, P = 11;
// module both A and D
A %= P;
D %= P;
// function call
System.out.println(NearestElement(A, D, P));
}
}
// This code is contributed
// by Arnab Kundu
Python 3
# Python 3 Program to Find First
# element in AP which is
# multiple of given prime
# Iterative Function to calculate
# (x^y)%p in O(log y)
def power(x, y, p) :
# Initialize result
res = 1
# Update x if it is more than or
# equal to p
x = x % p
while y > 0 :
# If y is odd, multiply x with result
if y & 1 :
res = (res * x) % p
# y must be even now
# y = y/2
y = y >> 1
x = (x * x) % p
return res
# function to find nearest element in common
def NearestElement(A, D, P) :
# base conditions
if A == 0 :
return 0
elif D == 0 :
return -1
else :
X = power(D, P - 2, P)
return (X * (P - A)) % P
# Driver Code
if __name__ == "__main__" :
A, D, P = 4, 9, 11
# module both A and D
A %= P
D %= P
# function call
print(NearestElement(A, D, P))
# This code is contributed by ANKITRAI1
C#
// C# Program to Find First
// element in AP which is
// multiple of given prime
using System;
class GFG
{
// Iterative Function to
// calculate (x^y)%p in
// O(log y) */
static int power(int x,
int y, int p)
{
// Initialize result
int res = 1;
// Update x if it is
// more than or equal to p
x = x % p;
while (y > 0)
{
// If y is odd, multiply
// x with result
if ((y & 1) != 0)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
// function to find nearest
// element in common
static int NearestElement(int A,
int D, int P)
{
// base conditions
if (A == 0)
return 0;
else if (D == 0)
return -1;
else
{
int X = power(D, P - 2, P);
return (X * (P - A)) % P;
}
}
// Driver code
public static void Main()
{
int A = 4, D = 9, P = 11;
// module both A and D
A %= P;
D %= P;
// function call
Console.WriteLine(NearestElement(A, D, P));
}
}
// This code is contributed
// by chandan_jnu.
PHP
0)
{
// If y is odd, multiply
// x with result
if ($y & 1)
$res = ($res * $x) %$p;
// y must be even now
$y = $y >> 1; // y = y/2
$x = ($x * $x) %$p;
}
return $res;
}
// function to find nearest
// element in common
function NearestElement($A, $D, $P)
{
// base conditions
if ($A == 0)
return 0;
else if ($D == 0)
return -1;
else
{
$X = power($D, $P - 2, $P);
return ($X * ($P - $A)) %$P;
}
}
// Driver code
$A = 4; $D = 9; $P = 11;
// module both A and D
$A %= $P;
$D %= $P;
// function call
echo NearestElement($A, $D, $P);
// This code is contributed
// by chandan_jnu.
?>
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