给定非常大的数字n和x，我们需要求n^x的数字之和，这样：
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If n^x < 10 digSum(n^x) = n^xElse digSum(n^x) = Sum(digSum(n^x))
例如：
Input : 5 4Output : 4We know 54 = 625Sum of digits in 625 = 6 + 2 + 5 = 13Sum of digits in 13 = 1 + 3 = 4Input : 546878 56422Output : 7
先决条件： 递归数字和 . 这个想法是： 每6个指数后重复的数字之和。 设SoD（n）=a 设b=x%6 然后 SoD（n^x）=SoD（a^b），a=3和6时b=1除外 当a=3和6时，所有x>1的SoD（n^x）=9
C++
// CPP Code for Sum of // digit of n^x where // n and x are very large #include <bits/stdc++.h> 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 Gitanjali |
C#
// 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
<?php // PHP Code for Sum of // digit of n^x where // function to get sum // of digits of a number function digSum( $n ) { if ( $n == 0) return 0; return ( $n % 9 == 0) ? 9 : ( $n % 9); } // function to return sum function PowDigSum( $n , $x ) { // Find sum of digits in n $sum = digSum( $n ); // Find remainder of exponent $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(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
<script> // JavaScript Code for Sum of // digit of n^x where // n and x are very large // function to get sum // of digits of a number function digSum(n) { if (n == 0) return 0; return (n % 9 == 0) ? 9 : (n % 9); } // function to return sum function PowDigSum(n, x) { // Find sum of digits in n let sum = digSum(n); // Find remainder of exponent let 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(Math.pow(sum, 6)); else return digSum(Math.pow(sum, rem)); } // Driver code let n = 33333; let x = 332654; document.write(PowDigSum(n, x)); </script> |
输出：
9
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