POJ 1061 青蛙的约会

//POJ 1061//扩展欧几里得算法/*求解ax + by = c, 可先解得ax + by = gcd(a, b)与等式右边无关，可得结论x = x0 + kb/gcd(a, b).为了求解最小非负解，可以先取模在加上模值再取模得到*/void exgcd(LL a, LL b, LL &d, LL &x, LL &y) {if (!b) d = a, x = 1, y = 0;else exgcd(b, a%b, d, y, x), y -= x*(a / b);}int main() {LL x, y, m, n, L, X, Y, d;while (cin >> x >> y >> m >> n >> L) {if (n < m) swap(n, m), swap(x, y);exgcd(n - m, L, d, X, Y);if ((x - y) % d) printf("Impossible\n");else {LL r = L / d;LL ans = (X*(x - y) / d%r + r) % r;//printf("X = %lld r = %lld\n", X, r);printf("%lld\n", ans);}}return 0;}