first n digits of an exponentiation

Question

How do i determine the first n digits of an exponentiation (a^b).

eg: for a = 12, b = 13 & n = 4, the first 4 digits are 1069.

Answer 1

For this case - with magic numbers 12,13,4 in place:

#include <sstream>
#include <iomanip>
#include <cmath>

double a = 12;
int b = 13;
double result = std::pow(a,b);

std::stringstream strVal;
strVal.setf( ios::fixed, ios::floatfield );
strVal << result;
std::string output(strVal.str().substr(0,4));

output = "1069"

std::stringstream intStr(output);
int intVal;
intStr >> intVal;

intVal = 1069

EDIT: This should work for any combination where result does not overflow double.

Answer 2

The easiest way programatically to do this is to use a stringstream to convert the result of the exponentation to a string and then take the n most significant (i.e. left) characters.

if you want a way without strings then this will work:

#include <iostream>
#include <sstream>
#include <math.h>

using namespace std;

double nDigExp( double a, double b, int n )
{
    stringstream ss;
    ss.setf( ios::fixed, ios::floatfield );
    ss << pow(a,b);
    double ret;
    for ( int i = 0; i < n; ++i) ret = (10 * ret) + (ss.get() - '0'); // Yeuch!!
    return ret;
}

int main( )
{
    double result = nDigExp( 12, 13, 4 );

    cout << result << endl;

    return 0;
}

But it's hardly the most elegent code. I'm sure you can improve it.

Answer 3

Another solution, using log10:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main(int argc, char **argv) {
       int a = 12;
       int b = 13;
       int n = 4;
       double x, y;

       x = b * log10(a);
       y = floor(pow(10, x - floor(x) + n - 1));
       printf("Result: %d\n", (int)y);

       return EXIT_SUCCESS;
}