a ^ n中的位数之和，直到一位数

给定两个数字a和n，任务是找到a ^ n(pow(a，n))的单个数字之和。在一位数总和中，我们将继续对位求和，直到剩下一位数为止。
例子：

Input : a = 5, n = 4
Output : 4
5^4 = 625 = 6+2+5 = 13
Since 13 has two digits, we
sum again 1 + 3 = 4.

Input : a = 2, n = 8
Output : 4
2^8=256 = 2+5+6 = 13 = 1+3 = 4

天真的方法是先找到a ^ n，然后使用此处讨论的方法找到a ^ n中的数字总和。
上述方法可能会导致溢出。一个更好的解决方案是基于以下观察。

int res = 1;
for (int i=1; i<=n; i++)
{
     res = res*a;
     res = digSum(res);
}

Here digSum() finds single digit sum 
of res. Please refer this for details
of digSum().

上面的伪代码说明：

For example, let a = 5, n = 4. 
After first iteration, 
res = 5
After second iteration, 
res = 7 (Note : 2 + 5 = 7)
After third iteration, 
res = 8 (Note : 3 + 5 = 8)
After 4th iteration, 
res = 4 (Note : 4 + 0 = 4)

我们可以写类似于快速模幂函数来评估digSum(一^ N)，其评价这在日志(n)的步骤。
下面是上述方法的实现：

C++

// CPP program to find single digit
// sum of a^n.
#include 
using namespace std;
 
// This function finds single digit
// sum of n.
int digSum(int n)
{
    if (n == 0)
    return 0;
    return (n % 9 == 0) ? 9 : (n % 9);
}
 
// Returns single digit sum of a^n.
// We use modular exponentiation technique.
int powerDigitSum(int a, int n)
{
    int res = 1;
    while (n) {
        if (n % 2 == 1) {
            res = res * digSum(a);
            res = digSum(res);
        }
        a = digSum(digSum(a) * digSum(a));
        n /= 2;
    }
 
    return res;
}
 
// Driver code
int main()
{
    int a = 9, n = 4;
    cout << powerDigitSum(a, n);
    return 0;
}

Java

// Java program to find single digit
// sum of a^n.
 
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG{
     
     
// This function finds single digit
// sum of n.
static int digSum(int n)
{
    if (n == 0)
    return 0;
    return (n % 9 == 0) ? 9 : (n % 9);
}
 
// Returns single digit sum of a^n.
// We use modular exponentiation technique.
static int powerDigitSum(int a, int n)
{
    int res = 1;
    while (n>0) {
        if (n % 2 == 1) {
            res = res * digSum(a);
            res = digSum(res);
        }
        a = digSum(digSum(a) * digSum(a));
        n /= 2;
    }
 
    return res;
}
 
// Driver code
public static void main(String args[])
{
    int a = 9, n = 4;
    System.out.print(powerDigitSum(a, n));
}
}

Python 3

# Python 3 Program to find single digit
# sum of a^n.
 
# This function finds single digit
# sum of n.
def digSum(n) :
 
    if n == 0 :
        return 0
 
    elif n % 9 == 0 :
        return 9
 
    else :
        return n % 9
 
# Returns single digit sum of a^n.
# We use modular exponentiation technique.
def powerDigitSum(a, n) :
 
    res = 1
    while(n) :
 
        if n %2 == 1 :
            res = res * digSum(a)
            res = digSum(res)
 
        a = digSum(digSum(a) * digSum(a))
        n //= 2
 
    return res
 
 
# Driver Code
if __name__ == "__main__" :
 
    a, n = 9, 4
    print(powerDigitSum(a, n))
 
# This code is contributed by ANKITRAI1

C#

// C# program to find single
// digit sum of a^n.
class GFG
{
 
// This function finds single
// digit sum of n.
static int digSum(int n)
{
    if (n == 0)
    return 0;
    return (n % 9 == 0) ?
                      9 : (n % 9);
}
 
// Returns single digit sum of a^n.
// We use modular exponentiation
// technique.
static int powerDigitSum(int a, int n)
{
    int res = 1;
    while (n > 0)
    {
        if (n % 2 == 1)
        {
            res = res * digSum(a);
            res = digSum(res);
        }
        a = digSum(digSum(a) * digSum(a));
        n /= 2;
    }
 
    return res;
}
 
// Driver code
static void Main()
{
    int a = 9, n = 4;
    System.Console.WriteLine(powerDigitSum(a, n));
}
}
 
// This Code is contributed by mits

Javascript

 
// javascript program to find single digit
// sum of a^n.
 
// This function finds single digit
// sum of n.
function digSum( n)
{
    if (n == 0)
    return 0;
    return (n % 9 == 0) ? 9 : (n % 9);
}
 
// Returns single digit sum of a^n.
// We use modular exponentiation technique.
function powerDigitSum( a,  n)
{
    let res = 1;
    while (n) {
        if (n % 2 == 1) {
            res = res * digSum(a);
            res = digSum(res);
        }
        a = digSum(digSum(a) * digSum(a));
        n /= 2;
    }
 
    return res;
}
 
// Driver code
 
    let a = 9, n = 4;
    document.write( powerDigitSum(a, n));
     
// This code is contributed by todaysgaurav
 
 

输出：

9