Find the value of the function Y = (X^6 + X^2 + 9894845) % 971

Given a function, Y = (X^6 + X^2 + 9894845) % 971 for a given value. The task is to find the value of the function.

Examples:

Input: x = 5
Output: 469

Input: x = 654654
Output: 450

Explanation:



Y = (X^6 + X^2 + 9894845) % 971.
If we break down the equation we get Y = (X^6)%971 + (X^2)%971 +(9894845)%971
and we can reduce the eqution to Y=(X^6)%971 + (X^2)%971 + 355.

Below is the required implementation:

C++

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// CPP implementation of above approach
#include <bits/stdc++.h>
using namespace std;
  
// computing (a^b)%c
long long int modpow(long long int base, long long int exp, long long int modulus) {
base %= modulus;
long long int result = 1;
while (exp > 0) {
    if (exp & 1) result = (result * base) % modulus;
    base = (base * base) % modulus;
    exp >>= 1;
}
return result;
}
  
// Driver code
int main(){
    long long int n = 654654, mod = 971;
    cout<<(((modpow(n, 6, mod)+modpow(n, 2, mod))% mod + 355)% mod);
  
    return 0;
}
// This code is contributed by Sanjit_Prasad

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Java

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// Java implementation of above approach
  
class GFG
{
  
// computing (a^b)%c
static long modpow(long base, long exp, long modulus) 
{
    base %= modulus;
    long result = 1;
    while (exp > 0) {
        if ((exp & 1)>0) result = (result * base) % modulus;
            base = (base * base) % modulus;
            exp >>= 1;
    }
    return result;
}
  
    public static void main(String[] args)
    {
        long n = 654654;
        long mod = 971;
        System.out.println(((modpow(n, 6, mod)+modpow(n, 2, mod))% mod + 355)% mod);
    }
}
// This code is contributed by mits;

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Python3

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# Python implementation of above approach
  
n = 654654
mod = 971
print(((pow(n, 6, mod)+pow(n, 2, mod))% mod + 355)% mod)

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C#

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// C# implementation of above approach
using System;
class GFG
{
  
// computing (a^b)%c
static long modpow(long base1, long exp, long modulus) 
{
    base1 %= modulus;
    long result = 1;
    while (exp > 0) {
        if ((exp & 1)>0) result = (result * base1) % modulus;
            base1 = (base1 * base1) % modulus;
            exp >>= 1;
    }
    return result;
}
  
    public static void Main()
    {
        long n = 654654;
        long mod = 971;
        Console.WriteLine(((modpow(n, 6, mod)+modpow(n, 2, mod))% mod + 355)% mod);
    }
}
// This code is contributed by mits;

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PHP

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<?php
// PHP implementation of above approach
  
// computing (a^b)%c
function modpow($base, $exp, $modulus)
{
    $base %= $modulus;
    $result = 1;
    while ($exp > 0) 
    {
        if ($exp & 1) $result = ($result * $base) % 
                                        $modulus;
        $base = ($base * $base) % $modulus;
        $exp >>= 1;
    }
    return $result;
}
  
// Driver code
$n = 654654;
$mod = 971;
echo (((modpow($n, 6, $mod) +
        modpow($n, 2, $mod)) % 
        $mod + 355) % $mod);
  
// This code is contributed by mits
?>

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Output:

450


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