problem 39

/*f p is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exactly three solutions for p = 120.

{20,48,52}, {24,45,51}, {30,40,50}

For which value of p ≤ 1000, is the number of solutions maximised?
*/
#include<iostream>
#include<cmath>
#include<algorithm> 
#include<cstring>
#include<vector>
#include<map>
using namespace std;

void sides_less_than_1000(double side_3, int perimeter, map<int, int> & solutions){
   if (side_3 == int(side_3) && perimeter < 1000){
     solutions[perimeter] += 1;
     //cout << solutions[perimeter] <<endl;
   }
}

void fill_zeros(map<int, int> &solutions){
  for(int i = 0; i < 1000; i++){
    solutions.insert(pair<int,int>(i,0));
  }
}

int main() {
  map<int, int> solutions;
  fill_zeros(solutions);
  for(int side_1 = 1; side_1 < 1000; side_1++){
    for(int side_2 = side_1 + 1; side_2 < 1000; side_2++){
      double side_3 = pow(pow(side_2, 2) + pow(side_1, 2) , 0.5);
      int perimeter = side_3 + side_2 + side_1;
      sides_less_than_1000(side_3, perimeter, solutions);
    }
  }
  int max_1 = 1, max_2;
  for(int i = 0; i < solutions.size(); i++){
    if(solutions[i] > max_1){
      max_1 = solutions[i];
      max_2 = i;
    }
  }
  cout << max_1 << " : " << max_2 << endl;
   return 0;
}