给定非常大的数字n和x，我们需要找到n ^ x的数字总和，使得：
If n^x < 10
digSum(n^x) = n^x
Else
digSum(n^x) = Sum(digSum(n^x))例子：
Input : 5 4
Output : 4
We know 54 = 625
Sum of digits in 625 = 6 + 2 + 5 = 13
Sum of digits in 13 = 1 + 3 = 4
Input : 546878 56422
Output : 7先决条件：递归数字总和。
这个想法是：
每6个指数重复一次数字总和。
令SoD(n)= a
设b = x％6
然后
SoD(n ^ x)= SoD(a ^ b)，当a = 3＆6时b = 1除外
当a = 3＆6时，SoD(n ^ x)= 9 forall x> 1，
C++
// CPP Code for Sum of
// digit of n^x where
// n and x are very large
#include
using namespace std;
// function to get sum
// of digits of a number
long digSum(long n)
{
if (n == 0)
return 0;
return (n % 9 == 0)
? 9 : (n % 9);
}
// function to return sum
long PowDigSum(long n, long x)
{
// Find sum of digits in n
long sum = digSum(n);
// Find remainder of exponent
long rem = x % 6;
if ((sum == 3 || sum == 6)
&& x > 1)
return 9;
else if (x == 1)
return sum;
else if (x == 0)
return 1;
else if (rem == 0)
return digSum((long)pow(sum, 6));
else
return digSum((long)pow(sum, rem));
}
// Driver code
int main()
{
int n = 33333;
int x = 332654;
cout << PowDigSum(n, x);
return 0;
}
// This code is contributed by Gitanjali. Java
// Java Code for
// Sum of digit of
// n^x where n and
// x are very large
import java.util.*;
class GFG {
// function to get sum
// of digits of a number
static long digSum(long n)
{
if (n == 0)
return 0;
return (n % 9 == 0) ? 9 : (n % 9);
}
// function to return sum
static long PowDigSum(long n, long x)
{
// Find sum of digits in n
long sum = digSum(n);
// Find remainder of exponent
long rem = x % 6;
if ((sum == 3 || sum == 6) && x > 1)
return 9;
else if (x == 1)
return sum;
else if (x == 0)
return 1;
else if (rem == 0)
return digSum((long)Math.pow(sum, 6));
else
return digSum((long)Math.pow(sum, rem));
}
/* Driver program to test above function */
public static void main(String[] args)
{
int n = 33333;
int x = 332654;
System.out.println(PowDigSum(n, x));
}
}
// This code is contributed by Arnav Kr. Mandal.Python3
# Python3 Code for Sum
# of digit of n^x
import math
# function to get
# sum of digits of
# a number
def digSum(n):
if (n == 0):
return 0
if n % 9==0 :
return 9
else:
return (n % 9)
# function to return sum
def PowDigSum(n, x):
# Find sum of
# digits in n
sum = digSum(n)
# Find remainder
# of exponent
rem = x % 6
if ((sum == 3 or sum == 6) and x > 1):
return 9
elif (x == 1):
return sum
elif (x == 0):
return 1
elif (rem == 0):
return digSum(math.pow(sum, 6))
else:
return digSum(math.pow(sum, rem))
# Driver method
n = 33333
x = 332654
print (PowDigSum(n, x))
# This code is contributed by GitanjaliC#
// C# Code for Sum of digit of
// n^x where n and x are very large
using System;
class GFG {
// Function to get sum
// of digits of a number
static long digSum(long n)
{
if (n == 0)
return 0;
return (n % 9 == 0) ? 9 : (n % 9);
}
// Function to return sum
static long PowDigSum(long n, long x)
{
// Find sum of digits in n
long sum = digSum(n);
// Find remainder of exponent
long rem = x % 6;
if ((sum == 3 || sum == 6) && x > 1)
return 9;
else if (x == 1)
return sum;
else if (x == 0)
return 1;
else if (rem == 0)
return digSum((long)Math.Pow(sum, 6));
else
return digSum((long)Math.Pow(sum, rem));
}
// Driver Code
public static void Main()
{
int n = 33333;
int x = 332654;
Console.WriteLine(PowDigSum(n, x));
}
}
// This code is contributed by vt_m.PHP
1)
return 9;
else if ($x == 1)
return $sum;
else if ($x == 0)
return 1;
else if ($rem == 0)
return digSum(pow($sum, 6));
else
return digSum(pow($sum, $rem));
}
// Driver code
$n = 33333;
$x = 332654;
echo PowDigSum($n, $x);
// This code is contributed by aj_36.
?>Javascript
输出：
9