All possible combinations(with repetition) as values in array using recursion - c++

Question

I'm trying to solve a problem in which I need to insert math operations(+/- in this case) between digits or merge them to get a requested number.
For ex.: 123456789 => 123+4-5+6-7+8-9 = 120
My concept is basically generating different combinations of operation codes in array and calculating the expression until it equals some number.
The problem is I can't think of a way to generate every possible combination of math operations using recursion.
Here's the code:
#include <iostream>
#include <algorithm>
using namespace std;
enum {noop,opplus,opminus};//opcodes: 0,1,2
int applyOp(int opcode,int x, int y);
int calculate(int *digits,int *opcodes, int length);
void nextCombination();
int main()
{
int digits[9] = {1,2,3,4,5,6,7,8,9};
int wantedNumber = 100;
int length = sizeof(digits)/sizeof(digits[0]);
int opcodes[length-1];//math symbols
fill_n(opcodes,length-1,0);//init
while(calculate(digits,opcodes,length) != wantedNumber)
{
//recursive combination function here
}
return 0;
}
int applyOp(int opcode,int x, int y)
{
int result = x;
switch(opcode)
{
case noop://merge 2 digits together
result = x*10 + y;
break;
case opminus:
result -= y;
break;
case opplus:
default:
result += y;
break;
}
return result;
}
int calculate(int *digits,int *opcodes, int length)
{
int result = digits[0];
for(int i = 0;i < length-1; ++i)//elem count
{
result = applyOp(opcodes[i],result,digits[i+1]);//left to right, no priority
}
return result;
}

Answer 1

The key is backtracking. Each level of recursion handles
a single digit; in addition, you'll want to stop the recursion
one you've finished.
The simplest way to do this is to define a Solver class, which
keeps track of the global information, like the generated string
so far and the running total, and make the recursive function
a member. Basically something like:
class Solver
{
std::string const input;
int const target;
std::string solution;
int total;
bool isSolved;
void doSolve( std::string::const_iterator pos );
public:
Solver( std::string const& input, int target )
: input( input )
, target( target )
{
}
std::string solve()
{
total = 0;
isSolved = false;
doSolve( input.begin() );
return isSolved
? solution
: "no solution found";
}
};
In doSolve, you'll have to first check whether you've finished
(pos == input.end()): if so, set isSolved = total == target
and return immediately; otherwise, try the three possibilities,
(total = 10 * total + toDigit(*pos), total += toDigit(*pos),
and total -= toDigit(*pos)), each time saving the original
total and solution, adding the necessary text to
solution, and calling doSolve with the incremented pos.
On returning from the recursive call, if ! isSolved, restore
the previous values of total and solution, and try the next
possibility. Return as soon as you see isSolved, or when all
three possibilities have been solved.