求出函数Y的值=(X ^ 6 + X ^ 2 + 9894845)％971

给定一个函数，对于给定值， Y =(X ^ 6 + X ^ 2 + 9894845)％971 。任务是找到函数的值。

例子：

Input: x = 5
Output: 469

Input: x = 654654
Output: 450

解释：

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.

以下是所需的实现：

C++

// CPP implementation of above approach
#include 
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

Java

// 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;

Python3

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

C#

// 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;

PHP

 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
?>

输出：

450