Ultraviolet A 11657,

Ultraviolet A 11657,

Link: Ultraviolet A 11657-Rational Billiard

Given a boundary M, N, and the position of the first ball and the second ball, the first ball moves in the direction of p and q and is rebounded when it hits the boundary, consistent with the line of light, ask if it is possible to concentrate the second ball.

Solution: refer to other people's ideas on the Internet. First, expand the corresponding multiples of the horizontal and vertical coordinates to ensure that each movement of p and q corresponds to the unit length on the new plane, then, you only need to consider the number of steps required for horizontal movement. After the number of steps required for vertical movement is reduced, whether the remaining steps meet the multiples of the cycle is enough. Consider four situations. The data for this question is very watery. At first, the coordinates were written incorrectly.

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;ll m, n, x0, y0, x1, y1, p, q;ll gcd (ll a, ll b) {    return b == 0 ? a : gcd(b, a % b);}bool judge () {    if (p == 0 || q == 0) {        ll u = p * (y1-y0) - q * (x1-x0);        return u == 0;    }    ll absq = q < 0 ? -q : q;    ll absp = p < 0 ? -p : p;    m *= absq; x0 *= absq; x1 *= absq;    n *= absp; y0 *= absp; y1 *= absp;    p /= absp; q /= absq;    ll d1 = q * (y1-y0) - p * (x1-x0);    ll d2 = q * (y1-y0) - p * (2*m-x1-x0);    ll d3 = q * (2*n-y1-y0) - p * (x1-x0);    ll d4 = q * (2*n-y1-y0) - p * (2*m-x1-x0);    ll g = gcd(2*m, 2*n);    if (d1 % g == 0) return true;    if (d2 % g == 0) return true;    if (d3 % g == 0) return true;    if (d4 % g == 0) return true;    return false;}int main () {    while (scanf("%lld%lld%lld%lld%lld%lld%lld%lld", &m, &n, &x0, &y0, &x1, &y1, &p, &q) == 8) {        if (!(m || n || x0 || y0 || x1 || y1 || p || q))            break;        printf("%s\n", judge() ? "HIT" : "MISS");    }    return 0;}