Codeforces 724C Ray Tracing 扩展欧几里得

标签： 解题报告 数学

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原题见CF 724C

n*m的矩形内有k个点，四周有围墙围起来。从(0,0)45度发射小球，速度为2√每次遇到墙都正常反弹，直到射到顶点被吸收。问每个点第一次被经过的时刻。

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分析

把矩形对称展开，最后小球在横纵坐标均为maxx=mn/gcd(m,n)处被吸收。
原坐标为(x,y)的小球经过轴对称展开，其坐标为(2kn±x,2sm±y),k,s为整数.要使得在吸收前经过点，则坐标必须在线段(0, 0)到(maxx, mxx)之间。
即要解方程2kn±x=2sm±y，求为正最小的2kn±x。利用扩展欧几里得解方程。

求ax+by=c最小正整数解，坑点：ran<0时不搞成正数会出错！

LL equation(LL a, LL b, LL c, LL &x, LL &y){    LL g = extend_Euclid(a, b, x, y);    if(c % g) return -1;    LL ran = b / g;    x *= c/g;    if(ran < 0) ran = -ran; //wa point    x = (x%ran + ran) % ran;    return 0;}

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代码

/*-------------------------------------------- * File Name: CF 724C * Author: Danliwoo * Mail: [e-mail] * Created Time: 2016-10-08 23:37:05--------------------------------------------*/#include <bits/stdc++.h>using namespace std;#define LL long longLL n, m;LL extend_Euclid(LL a, LL b, LL &x, LL &y){    if(b==0){        x = 1; y = 0;        return a;    }    LL r = extend_Euclid(b, a%b, y, x);    y -= a/b*x;    return r;}LL equation(LL a, LL b, LL c, LL &x, LL &y){    LL g = extend_Euclid(a, b, x, y);    if(c % g) return -1;    LL ran = b / g;    x *= c/g;    if(ran < 0) ran = -ran;    x = (x%ran + ran) % ran;    return 0;}LL gao(LL dx, LL dy, LL M) {    LL k, s;    if(equation(2*n, -2*m, -dx+dy, k, s) == -1)        return M + 1;    LL tx = 2 * k * n + dx;    if(tx < 0 || tx > M) return M + 1;    return tx;}LL minL(LL a, LL b) {    return a < b ? a : b;}LL solve(LL x, LL y) {    LL g = __gcd(n, m);    LL maxx = 1LL * m / g * n;    LL ans = maxx + 1;    ans = minL(ans, gao(-x, -y, maxx));    ans = minL(ans, gao(-x, y, maxx));    ans = minL(ans, gao(x, -y, maxx));    ans = minL(ans, gao(x, y, maxx));    if(ans == maxx + 1) return -1;    return ans;}int main() {    int k;    while(~scanf("%I64d%I64d%d", &n, &m, &k)) {        for(int i = 0;i < k;i++) {            LL x, y;            scanf("%I64d%I64d", &x, &y);            printf("%I64d\n", solve(x, y));        }    }    return 0;}