Split number into sum of preselected other numbers - c++

Question

I have a number (for example 301, but can be even 10^11).
n = lenght of that number
I have to break it down to sum of max n components. Those components are 0^n, 1^n, 2^n, 3^n...9^n.
How can I do that?

Answer 1

Since you have 1^n included in your options, this becomes a really simple problem solvable through Greedy Approach.
Firstly, let me clarify that the way I understand this, for an input N of length n, you want some solution to this equation:
A.1^n + B.2^n + C.3^n + ... + H.8^n + I.9^n
There are infinitely many possible solutions (just by theory of equations). One possible solution can be found as follows:
a = [x ** n for x in range(0,10)]
consts = [0] * 10
ctr = 9
while N > 0:
consts[ctr] = N // a[ctr]
N = N % a[ctr]
ctr -= 1
return consts
This consts array will have the constant values for the above equation at respective indices.
PS: I've written this in python but you can translate it to C++ as you want. I saw that tag later. If you have any confusion regarding the code, feel free to ask in the comments.

Answer 2

You could use the following to determine the number of components.
int remain = 301; // Target number
int exp = 3; // Length of number (exponent)
int total = 0; // Number of components
bool first = true; // Used to determinie if plus sign is output
for ( int comp = 9; comp > 0; --comp )
{
int count = 0; // Number of times this component is needed
while ( pow(comp, exp) <= remain )
{
++total; // Count up total number of components
++count; // Count up number of times this component is used
remain -= int(pow(comp, exp));
}
if ( count ) // If count is not zero, component is used
{
if ( first )
{
first = false;
}
else
{
printf(" + ");
}
if ( count > 1 )
{
printf("%d(%d^%d)", count, comp, exp);
}
else
{
printf("%d^%d", comp, exp);
}
}
}
if ( total == exp )
{
printf("\r\nTarget number has %d components", exp);
}
else if ( total < exp )
{
printf("\r\nTarget number has less than %d components", exp);
}
else
{
printf("\r\nTarget number has more than %d components", exp);
}
Output for 301:
6^3 + 4^3 + 2(2^3) + 5(1^3)
Target number has more than 3 components
Output for 251:
6^3 + 3^3 + 2^3
Target number has 3 components