November 2013 Bronze (Correction) Problem 1: Combination Lock

Result: 10/10 **********

Upon review, I beat the solution! Though N is no larger than 100, my solution runs in O(1) time and theirs runs in O(N^3) time.

combo2.cpp

#include <iostream>
#include <fstream>

#define loopi(x) loop_start(i,x,0)
#define loopj(x) loop_start(j,x,0)
#define loop_start(i, x,s) for(int i = (s); i < (x); ++i)
#define loop_rev(a,x,stop) for(int a = (x); a >= (stop); --a)

using namespace std;

int N;

int fixrange(int n){
 --n;
 while(n < 0) n += N;
 return (n % N) + 1;
}

int overlap(int a, int b){
 // I tried with modular arithmetic but couldn't get it working
 // return 5 - min(abs(a - b) % N, N - (abs(a - b) % N));
 int a1 = fixrange(a - 2),
  a2 = fixrange(a - 1),
  a3 = a,
  a4 = fixrange(a + 1),
  a5 = fixrange(a + 2);
 int b1 = fixrange(b - 2),
  b2 = fixrange(b - 1),
  b3 = b,
  b4 = fixrange(b + 1),
  b5 = fixrange(b + 2);
 int total = 0;
 if(a1 == b1 || a1 == b2 || a1 == b3 || a1 == b4 || a1 == b5)
  ++total;
 if(a2 == b1 || a2 == b2 || a2 == b3 || a2 == b4 || a2 == b5)
  ++total;
 if(a3 == b1 || a3 == b2 || a3 == b3 || a3 == b4 || a3 == b5)
  ++total;
 if(a4 == b1 || a4 == b2 || a4 == b3 || a4 == b4 || a4 == b5)
  ++total;
 if(a5 == b1 || a5 == b2 || a5 == b3 || a5 == b4 || a5 == b5)
  ++total;
 return total;
}

#define NUMCOMBO 3
int main() {
 ofstream fout ("combo.out");
 ifstream fin ("combo.in");
 // get input
 int combo1[NUMCOMBO];
 int combo2[NUMCOMBO];
 fin >> N;
 loopi(NUMCOMBO)
  fin >> combo1[i];
 loopi(NUMCOMBO)
  fin >> combo2[i];
 // calculate:
 // Special case: there is always N overlap if N <= 5
 if(N <= 5) // 2 * N ** 3 - N ** 3 = N ** 3
  // write output
  fout << (N * N * N);
 else{
  int overlap_a = overlap(combo1[0], combo2[0]);
  int overlap_b = overlap_a ? overlap(combo1[1], combo2[1]) : 0;
  int overlap_c = overlap_b ? overlap(combo1[2], combo2[2]) : 0;
  // 2 * 5 ** 3, but subtract the overlap
  // write output
  fout << (250 - overlap_a * overlap_b * overlap_c);
 }
 return 0;
}