# Find Positive Integer Solution for a Given Equation in C++

Suppose we have a function f that takes two parameters (x, y). We have to return all pairs of x and y, for which f(x, y) = z. The z is given as input, and x, y are positive integers. The function is constantly increasing function. So f(x, y) < f(x + 1, y) and f(x, y) < f(x, y + 1).

To solve this we will perform straight-forward approach. Take i in range 1 to 1000, and j in range 1 to 1000, for all combinations of i,j, if f(i, j) = 0, then return true, otherwise false.

Consider the function id (should be provided) is 1 for Addition, 2 for multiplication. It also takes the z value.

## Example

Let us see the following implementation to get better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<vector<int> > v){
cout << "[";
for(int i = 0; i<v.size(); i++){
cout << "[";
for(int j = 0; j <v[i].size(); j++){
cout << v[i][j] << ", ";
}
cout << "],";
}
cout << "]"<<endl;
}
class CustomFunction {
int id;
public:
CustomFunction(int id){
this->id = id;
}
int f(int x, int y){
if(id == 1)
return y + x;
else if(id == 2)
return y * x;
return 0;
}
};
class Solution {
public:
vector<vector<int>> findSolution(CustomFunction& c, int z) {
vector  < vector <int > > ans;
for(int i = 1; i <= 1000; i++ ){
for(int j = 1; j <= 1000; j++){
if(c.f(i,j) == z){
vector <int> t;
t.push_back(i);
t.push_back(j);
ans.push_back(t);
}
}
}
return ans;
}
};
main(){
Solution ob;
CustomFunction c(1);
print_vector(ob.findSolution(c, 7));
}

## Input

1
7

## Output

[[1, 6, ],[2, 5, ],[3, 4, ],[4, 3, ],[5, 2, ],[6, 1, ],]

Updated on: 29-Apr-2020

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