We begin by focusing on the simplest case where $a,b>0$. This will be impossible unless $c,d>0$, so we further focus on this case. Note that this is exactly the Transforming Pairs task given in the silver division. When all coordinates are positive, our key insight is that it is easier to work backwards than working forwards. Suppose $c<d$, then it impossible that the last operation added to the first coordinate, as otherwise the first coordinate would have been negative immediately before the last operation. Since this is impossible because $a,b>0$, then our only possibility to consider is that the last operation added to the second coordinate, and we now consider the problem where $a'=a$, $b'=b$, $c'=c$, and $d'=d-c$. Working backwards with this logic will solve this case, however it could naively take $O(\text{MAX})$ time, where $\textrm{MAX}$ is the largest allowable absolute value of an input ($10^{18}$ for all tests). This can be simulated more efficiently by observing that, say when $c<d$, we will continue subtracting from the second coordinate until $d'=d\%c$. This modification will run in only $O(\log(\text{MAX}))$ time as desired.
Additionally, observe that the number of cases is smaller than it may initially seem. It does not change the answer to negate all the coordinates or to swap the first and second coordinates. For example, we may handle the case where $a,b,c,d<0$ by negating all the coordinates.
Moreover, we can sidestep cases where at least one of $a,b,c,d$ are $0$. If both $a$ and $b$ are $0$, then this is impossible unless both $c$ and $d$ are $0$. If one of $a$ or $b$ are 0, then the first move must add the nonzero value to the zero-valued coordinate. Similar logic holds for $c$ and $d$, noting that if one of $c$ or $d$ are $0$, then the last move must have added the non-zero value to produce the zero-valued coordinate.
With these symmetries in mind, we conclude it is sufficient to consider the following remaining cases:
Case 1 will follow more simply. Note that the sign of the coordinate may never change (or be zero) or it will be impossible. Hence, if $|a|>|b|$, our first move must add $b$ to the first coordinate, and versa when $|a|>|b|$. This can be simulated efficiently just like in the all-positive case discussed.
Case 2 will require some more complexity. Here, the first coordinate must stay positive, and at some point the second coordinate must go from negative to positive. It will be productive to consider the sequence of operations in two parts: (i) the operations where the second coordinate remains non-positive, and (ii) the operations where the second coordinate is positive. For (i), the only option is to proceed as done in Case 1: when $|a|>|b|$ we add $b$ to the first coordinate, and vice versa when $|a|<|b|$. For (ii), we will consider these operations backwards: since $c,d>0$ we work backwards as done initially in the all-positive case. With this perspective, it is clear that a prefix of the operations will follow the simulation of part (i), and a suffix will follow the simulation of part (ii). What remains is to consider the choice of the operation that brings the second coordinate from negative to non-negative, and check if it is consistent with a suffix. Very naive simluation of this would be to create the lists for parts (i) and (ii) in $O(\text{MAX})$ time, and then consider all possible combinations.
Of course, we will instead generate faster lists in $O(\log(\text{MAX}))$ time with methods similar to the aforementioned logic. Let us focus on the forward direction, part (i), for which we will generate a list of options that result from a sequence of operations where the second coordinate is negative, followed by one operation that makes the second coordinate non-negative. To generate this list, when $|a|<|b|$, then we must add $a$ to the second coordinate. If $|a|=|b|$, we will add $a$ to the second coordinate twice, and add this as an option for our list (we added $a$ twice so that the second coordinate will be positive instead of $0$). Otherwise, when $|a| > |b|$, it is possible to either add $a$ to the second coordinate (then it will become non-negative), or add $b$ to $a$. We will consider all sequences of the latter followed by one operation of the former: so $(a,a+b)$, $(a+b,a+2b)$, and so on, and add this to our list. We can simulate such a compressed list in $O(\log(\text{MAX}))$ time similar to before. For the backwards direction part (ii), we will generate the compressed list of the simulation just like in the all-positive case, where an example element of the list may contain an entry corresponding to $(x,y) \rightarrow (x,y-x) \rightarrow \dots \rightarrow (x,y\%x)$. Both lists will be of size $O(\log(\text{MAX}))$. Finally, for each pair of items in different lists, we will consider if they share a common element so we may piece their sequences together. With some casework, this may be computed for any pair in $O(1)$ time, and in total would take $O(\log^2(\text{MAX}))$ time to consider all pairs. This may be sped up to $O(\log(\text{MAX}))$ time by observing that the sequences are monotonic in the sum of the absolute values of the coordinates, so we may use a two-pointer approach instead of comparing all pairs.
Spencer's C++ Code:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pb push_back
bool sgn(ll x) { return (x > 0); }
class Obj {
public:
bool fixed;
ll cost, inc, st, en;
Obj(ll x, ll y) {
fixed = true;
st = x;
cost = y;
inc = 0LL;
en = st;
}
Obj(ll a, ll b, ll c, ll d) {
fixed = false;
st = a;
inc = b;
en = c;
if (st == en) {
fixed = true;
inc = 0;
}
cost = d;
}
};
ll compat(Obj a, Obj b, ll try_val = -1LL) {
if (a.fixed != b.fixed) {
if (!a.fixed) {
swap(a, b);
}
}
if (a.fixed && b.fixed) {
if (try_val >= 0 && try_val != a.st) {
return -1LL;
}
if (a.st == b.st) {
return a.st;
}
return -1LL;
} else if (a.fixed || b.fixed) {
bool didSwap = false;
if (b.fixed) {
didSwap = true;
swap(a, b);
}
if (try_val >= 0LL && a.st != try_val) {
return -1LL;
}
ll l = min(b.st, b.en);
ll r = max(b.st, b.en);
ll ret;
if (a.st >= l && a.st <= r &&
((a.st % llabs(b.inc)) == (l % llabs(b.inc)))) {
ret = a.st;
} else {
ret = -1LL;
}
if (didSwap) {
swap(a, b);
}
return ret;
} else {
assert(!a.fixed && !b.fixed);
ll a_l = min(a.st, a.en);
ll a_r = max(a.st, a.en);
ll b_l = min(b.st, b.en);
ll b_r = max(b.st, b.en);
if (try_val >= 0LL) {
if (a_l % llabs(a.inc) != (try_val % llabs(a.inc))) {
return -1LL;
}
if (b_l % llabs(b.inc) != (try_val % llabs(b.inc))) {
return -1LL;
}
if (try_val < a_l || try_val < b_l || try_val > a_r ||
try_val > b_r) {
return -1LL;
}
assert(a.inc < 0LL);
return try_val;
} else {
if (llabs(a.inc) != llabs(b.inc)) {
return -1LL;
}
if ((a_l % llabs(a.inc)) != (b_l % llabs(a.inc))) {
return -1LL;
}
try_val = min(a_r, b_r);
if (try_val < min(a_l, b_l)) {
return -1LL;
}
return try_val;
}
}
}
ll solve(vector<ll> x, vector<ll> y) {
ll cnt = 0;
bool done = true;
for (int i = 0; i < 2; i++) {
if (x[i] != y[i]) {
done = false;
}
}
if (done) {
return 0;
}
for (int i = 0; i < 2; i++) {
if (x[i] == 0) {
x[i] += x[1 - i];
cnt++;
}
}
if (cnt == 2) {
return -1;
}
assert((x[0] != 0) && (x[1] != 0));
done = true;
for (int i = 0; i < 2; i++) {
if (x[i] != y[i]) {
done = false;
}
}
if (done) {
return cnt;
}
int tmpY = 0;
for (int i = 0; i < 2; i++) {
if (y[i] == 0) {
y[i] -= y[1 - i];
cnt++;
tmpY++;
}
}
if (tmpY == 2) {
return -1;
}
assert((y[0] != 0) && (y[1] != 0));
done = true;
for (int i = 0; i < 2; i++) {
if (x[i] != y[i]) {
done = false;
}
}
if (done) {
return cnt;
}
// Case 1: same sign
if ((x[0] > 0) == (x[1] > 0)) {
if (x[0] < 0) {
for (int i = 0; i < 2; i++) {
x[i] = -x[i];
y[i] = -y[i];
}
}
if (min(y[0], y[1]) <= 0) {
return -1;
}
while (y[0] >= x[0] && y[1] >= x[1] &&
((x[0] + x[1]) != (y[0] + y[1])) && min(y[0], y[1]) > 0) {
if (y[0] == y[1]) {
break;
} else if (y[0] > y[1]) {
ll can = (y[0] - max(x[0], y[1]) - 1) / y[1] + 1;
y[0] -= y[1] * can;
cnt += can;
} else {
ll can = (y[1] - max(x[1], y[0]) - 1) / y[0] + 1;
y[1] -= y[0] * can;
cnt += can;
}
}
if (x[0] == y[0] && x[1] == y[1]) {
return cnt;
} else {
return -1;
}
} else {
if (sgn(x[0]) != sgn(y[0]) && sgn(x[1]) != sgn(y[1])) {
return -1;
}
if (sgn(x[0]) == sgn(y[0]) && sgn(x[1]) == sgn(y[1])) {
if (x[0] < 0) {
for (int i = 0; i < 2; i++) {
x[i] = -x[i];
y[i] = -y[i];
}
}
// x0=+, y0=+, x1=-, y1=-
assert(x[0] > 0LL && y[0] > 0LL && x[1] < 0LL && y[1] < 0LL);
x[1] = llabs(x[1]);
y[1] = llabs(y[1]);
while (y[0] <= x[0] && y[1] <= x[1] &&
(x[0] + x[1]) != (y[0] + y[1])) {
if (x[0] == x[1]) {
break;
} else if (x[0] > x[1]) {
ll can =
max(0LL, x[0] - max(x[1], y[0]) - 1LL) / x[1] + 1LL;
x[0] -= x[1] * can;
cnt += can;
} else {
ll can =
max(0LL, x[1] - max(x[0], y[1]) - 1LL) / x[0] + 1LL;
x[1] -= x[0] * can;
cnt += can;
}
}
if (x[0] == y[0] && x[1] == y[1]) {
return cnt;
} else {
return -1;
}
} else {
if (sgn(x[0]) != sgn(y[0])) {
swap(x[0], x[1]);
swap(y[0], y[1]);
}
if (x[0] < 0) {
for (int i = 0; i < 2; i++) {
x[i] = -x[i];
y[i] = -y[i];
}
}
assert(x[0] > 0 && y[0] > 0 && x[1] < 0 && y[1] > 0);
vector<pair<Obj, Obj>> pos_li;
while (y[0] > 0 && y[1] > 0) {
if (y[0] == y[1]) {
Obj l = Obj(y[0], cnt);
Obj r = Obj(y[1], cnt);
pos_li.pb(make_pair(l, r));
break;
} else if (y[0] > y[1]) {
ll can = (y[0] - y[1] - 1LL) / y[1] + 1LL;
Obj l = Obj(y[0], -y[1], y[0] - y[1] * (can - 1LL), cnt);
Obj r = Obj(y[1], cnt);
pos_li.pb(make_pair(l, r));
y[0] -= y[1] * can;
cnt += can;
} else {
ll can = (y[1] - y[0] - 1LL) / y[0] + 1LL;
Obj l = Obj(y[0], cnt);
Obj r = Obj(y[1], -y[0], y[1] - y[0] * (can - 1LL), cnt);
pos_li.pb(make_pair(l, r));
y[1] -= y[0] * can;
cnt += can;
}
}
vector<pair<Obj, Obj>> mix_li;
cnt = 0LL;
// x0=+, y0=+, x1=-, y1=+
while (x[0] > 0LL) {
if (x[1] >= 0) {
if (x[1] == 0) {
cnt++;
x[1] = x[0];
}
Obj l = Obj(x[0], cnt);
Obj r = Obj(x[1], cnt);
mix_li.pb(make_pair(l, r));
break;
}
if (llabs(x[0]) == llabs(x[1])) {
x[1] = x[0];
cnt += 2;
Obj l = Obj(x[0], cnt);
Obj r = Obj(x[1], cnt);
mix_li.pb(make_pair(l, r));
break;
}
if (x[0] > llabs(x[1])) {
ll can = (x[0] - llabs(x[1]) - 1LL) / llabs(x[1]) + 1;
Obj l =
Obj(x[0], x[1], x[0] + x[1] * (can - 1LL), cnt + 1LL);
Obj r = Obj(x[1] + x[0], x[1],
x[1] + x[0] + x[1] * (can - 1LL), cnt + 1LL);
x[0] += x[1] * can;
cnt += can;
mix_li.pb(make_pair(l, r));
} else {
ll can = (llabs(x[1]) - x[0] - 1LL) / x[0] + 1LL;
x[1] += x[0] * can;
cnt += can;
}
}
ll INF = (ll)(2e18) + 10LL;
ll ans = INF;
int point = 0;
for (int i = 0; i < pos_li.size(); i++) {
for (int j = point; j < mix_li.size(); j++) {
ll pos_min = pos_li[i].first.en + pos_li[i].second.en;
ll pos_max = pos_li[i].first.st + pos_li[i].second.st;
ll mix_min = mix_li[j].first.en + mix_li[j].second.en;
ll mix_max = mix_li[j].first.st + mix_li[j].second.st;
if (pos_min > mix_max) {
break;
}
if (j == point && (mix_min > pos_max)) {
point++;
continue;
}
auto pos = pos_li[i];
auto mix = mix_li[j];
int do_first = 0;
if (pos.second.fixed) {
do_first = 1;
}
ll meet = -1LL;
if (do_first == 1) {
swap(pos.first, pos.second);
swap(mix.first, mix.second);
}
assert(pos.first.fixed);
ll got = compat(pos.first, mix.first);
vector<ll> meets(2);
if (got == -1LL) {
if (do_first == 1) {
swap(pos.first, pos.second);
swap(mix.first, mix.second);
}
continue;
}
meets[0] = pos.first.st;
if (!mix.first.fixed) {
ll num_iters =
(meets[0] - mix.first.st) / mix.first.inc;
meet = mix.second.st + num_iters * mix.second.inc;
got = compat(pos.second, mix.second, meet);
if (got == -1LL) {
if (do_first == 1) {
swap(pos.first, pos.second);
swap(mix.first, mix.second);
}
continue;
}
meets[1] = meet;
} else {
meet = compat(pos.second, mix.second);
if (meet == -1LL) {
if (do_first == 1) {
swap(pos.first, pos.second);
swap(mix.first, mix.second);
}
continue;
}
meets[1] = meet;
}
ll cur_cost = pos.first.cost + mix.first.cost;
vector<ll> move_amt(2);
if (pos.first.inc != 0) {
move_amt[0] = (meets[0] - pos.first.st) / pos.first.inc;
} else if (pos.second.inc != 0) {
move_amt[0] =
(meets[1] - pos.second.st) / pos.second.inc;
}
if (mix.first.inc != 0) {
move_amt[1] = (meets[0] - mix.first.st) / mix.first.inc;
} else if (mix.second.inc != 0) {
move_amt[1] =
(meets[1] - mix.second.st) / mix.second.inc;
}
cur_cost += move_amt[0] + move_amt[1];
ans = min(ans, cur_cost);
if (do_first == 1) {
swap(pos.first, pos.second);
swap(mix.first, mix.second);
}
}
}
if (ans == INF) {
ans = -1LL;
}
return ans;
}
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int Q;
cin >> Q;
for (int q = 1; q <= Q; q++) {
vector<ll> x(2);
vector<ll> y(2);
cin >> x[0] >> x[1] >> y[0] >> y[1];
cout << solve(x, y) << "\n";
}
}