// Chinese remainder theorem: find z such that
// z % x[i] = a[i] for all i. Note that the solution is
// unique modulo M = lcm_i (x[i]). Return (z,M). On failure, M = -1.
pii chinese_remainder_theorem(const ve<int> &x, const ve<int> &a) {
pii ret = make_pair(a[0], x[0]);
for (int i = 1; i < x.size(); i++) {
ret = chinese_remainder_theorem(ret.second, ret.first, x[i], a[i]);
if (ret.second == -1) break;
}
return ret;
}
void printEdges(ve & edges){
for (int i = 0; i < edges.size(); i++){
edge e = edges[i];
printf("Edge to %i of length %f\n", e.first, e.second);
}
}
