Ultraviolet A 10548-find the right changes (Extended Euclidean)

Link to the question: Ultraviolet A 10548-find the right changes

Given a, B, c, X, Y, so that Xa + Yb = C, how many solutions are there.

Solution: Expand Euclidean to ensure that X and Y are both greater than or equal to 0, and determine the value of T in the general solution.

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;typedef long long ll;const ll INF = 0x3f3f3f3f3f3f3f;ll A, B, C;void gcd (ll a, ll b, ll& d, ll& x, ll& y) {    if (b == 0) {        d = a;        x = 1;        y = 0;    } else {        gcd(b, a%b, d, y, x);        y -= (a/b) * x;    }}void solve () {    ll d, x, y;    gcd(A, B, d, x, y);    if (C % d) {        printf("Impossible\n");        return;    }    ll up = INF, lower = -INF;    if (B / d > 0)        lower = max(lower, (ll)ceil( (-1.0*x*C) / B) );    else        up = min(up, (ll)floor( (-1.0*x*C) / B) );    if (A / d > 0)        up = min(up, (ll)floor( (1.0*y*C) / A));    else        lower = max(lower, (ll)ceil( (1.0*y*C) / A));    if (up == INF || lower == -INF)        printf("Infinitely many solutions\n");    else if (up < lower)        printf("Impossible\n");    else        printf("%lld\n", up - lower + 1);}int main () {    int cas;    scanf("%d", &cas);    while (cas--) {        scanf("%lld%lld%lld", &A, &B, &C);        solve();    }    return 0;}