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 equation to Y=(X^6)%971 + (X^2)%971 + 355.
Below is the required implementation:
C++
// CPP implementation of above approach#include <bits/stdc++.h>using namespace std;// computing (a^b)%clong 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 codeint 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 approachclass GFG{// computing (a^b)%cstatic 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 approachn = 654654mod = 971print(((pow(n, 6, mod)+pow(n, 2, mod))% mod + 355)% mod) |
C#
// C# implementation of above approachusing System;class GFG{// computing (a^b)%cstatic 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
<?php// PHP implementation of above approach// computing (a^b)%cfunction 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?> |
Javascript
<script>// JavaScript implementation of above approach// computing (a^b)%cfunction modpow(base, exp, modulus) { base %= modulus; let result = 1; while (exp > 0) { if ((exp & 1)>0) result = (result * base) % modulus; base = (base * base) % modulus; exp >>= 1; } return result;}// driver code let n = 654654; let mod = 971; document.write(((modpow(n, 6, mod)+ modpow(n, 2, mod))% mod + 355)% mod); </script> |
Output:
450
Time Complexity:O(1), since there is no loop or recursion.
Auxiliary Space: O(1), since no extra space has been taken.
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