Evaluate the N Power Sum of N: 1 ^ 1 + 2 ^ 2 + 3 ^ 3 + ...... + N ^ n

Programming for computing1 ^ 1 + 2 ^ 2 + 3 ^ 3 + ...... + N ^ nWhere N is an arbitrary integer. (Note: consider the case where the result may be out of the long range)

 

Note: This method usesInteger array elementsStores each digit of a large number.

 

 

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# Include <iostream>
Using namespace STD;

Const int n = 10000; // defines the maximum number of digits that can be stored, which can be changed as needed

 

/* Addition of large numbers: the final result is placed in the sum array, and the result array stores the K ^ K value sequence.

The rule for storing large numbers in an array is: when reading an array element with a small subscript, the low part of a large number is read backwards, that is, the highest part is read first, the last element that reads the subscript zero */

Void bigadd (int * sum, Int & height_sum, int * result, Int & height_res)
{
Int I, J;
Int tempsum;
Int carry = 0;
For (I = 0, j = 0; I {
Tempsum = sum [I] + result [J] + carry;
Sum [I] = tempsum % 10;
Carry = tempsum/10;
}
While (carry)
{
Tempsum = sum [I] + carry;
Sum [I] = tempsum % 10;
Carry = tempsum/10;
I ++;
}
Height_sum = I;
}

 

Void n_nand (int n)
{
Int sum [N] = {0}; // final result
Sum [0] = 1; // set the percentile bit to 1.
Int height_sum = 1; // The number of digits of the sum.
Int height_res;

// Calculate n-1 times and
For (int K = 1; k <= N; k ++)
{
Int result [N]; // stores the result of each power.
Result [0] = 1;
Height_res = 1; // number of digits of K ^ K
// Calculate k ^ K
For (INT I = 1; I <= K; I ++)
{
Int res = 0;
For (Int J = 0; j {
Int Buf = Result [J] * k + Res;
Result [J] = Buf % 10;
Res = Buf/10;
}
While (RES)
{
Result [height_res ++] = res % 10;
Res/= 10;
}

}
If (k> 1 & K <= N)
Bigadd (sum, height_sum, result, height_res );
}
// Output result
Int I;
For (I = N-1; I> = 0; I --)
If (sum [I]! = 0)
Break;
For (int K = I; k> = 0; k --)
Cout <sum [k];
Cout <Endl <"total length:" <I + 1 <Endl;
}

Int main ()
{
Int N;
Char CH = 'y ';
While ('y' = CH)
{
Cout <"input a number :";
Cin> N;
If (n> 10000 | n <0)
Cout <"an invalid number is entered! ";
Else
N_nand (N );
Cout <"continue? (Y/N ):";
Cin> CH;
}
Return 1;
}