Ultraviolet A 10548-find the right changes (number theory)

Ultraviolet A 10548-find the right changes

Question Link

Given A, B, and C, it indicates the value of the goods. There are several methods to determine whether there are unsolvable and unlimited types of goods A, B, and C.

Train of Thought: Extend Euclidean for a general solution, and calculate the upper limit and lower limit to determine

Code:

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const long long INF = 10000000000000000LL;int t;long long a, b, c;long long exgcd(long long a, long long b, long long &x, long long &y) {    if (!b) {x = 1; y = 0; return a;}    long long d = exgcd(b, a % b, y, x);    y -= a / b * x;    return d;}long long solve() {    long long x, y, d;    d = exgcd(a, b, x, y);    if (c % d) return 0;    long long down = -INF, up = INF;    if (b / d > 0)down = max(down, (long long)ceil((-x * 1.0 * c) / b));    elseup = min(up, (long long)floor((-x * 1.0 * c) / b));    if (a / d > 0)up = min(up, (long long )floor((y * 1.0 * c) / a));    elsedown = max(down, (long long)ceil((y * 1.0 * c) / a));    if (up == INF || down == -INF) return -1;    if (down <= up) return up - down + 1;    return 0;}int main() {    scanf("%d", &t);    while (t--) {scanf("%lld%lld%lld", &a, &b, &c);long long ans = solve();if (ans == 0) printf("Impossible\n");else if (ans < 0) printf("Infinitely many solutions\n");else printf("%lld\n", ans);    }    return 0;}