分析:模拟可以过,但是好烦啊。。不会写。还有一个扩展欧几里得的方法,见下:
假设光线没有反射,而是对应的感应器镜面对称了一下的话
左下角红色的地方是原始的
假设光线没有经过反射,那么用函数可以表示为
对于一个点,光线的真实情况是可能会经过多次,这意味着一个点以及它的镜像点可能有多个在
代码:
/*****************************************************/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <map>
#include <set>
#include <ctime>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define offcin ios::sync_with_stdio(false)
#define sigma_size 26
#define lson l,m,v<<1
#define rson m+1,r,v<<1|1
#define slch v<<1
#define srch v<<1|1
#define sgetmid int m = (l+r)>>1
#define LL long long
#define ull unsigned long long
#define mem(x,v) memset(x,v,sizeof(x))
#define lowbit(x) (x&-x)
#define bits(a) __builtin_popcount(a)
#define mk make_pair
#define pb push_back
#define fi first
#define se second
const int INF = 0x3f3f3f3f;
const LL INFF = 1e18;
const double pi = acos(-1.0);
const double inf = 1e18;
const double eps = 1e-9;
const LL mod = 1e9+7;
const int maxmat = 10;
const ull BASE = 31;
/*****************************************************/
//对于不定整数方程pa+qb=c,若 c mod Gcd(a, b)=0,则该方程存在整数解,否则不存在整数解
//p = p1 + b/Gcd(a, b) * t
//q = q1 - a/Gcd(a, b) * t(其中t为任意整数,正负皆可)
LL exgcd(LL a,LL b,LL &x,LL &y) {
if(!b) {x=1; y=0; return a; }
LL r=exgcd(b,a%b,y,x);
y-=a/b*x;
return r;
}
bool solve_equ(LL a,LL b,LL ca,LL cb,LL &x,LL &y,LL &d,LL &ans) {
LL c = ca + cb;
d = exgcd(a, b, x, y);
if(c % d) return false;
LL k = c / d; LL g = b / d;
if (g < 0) g = -g;
g *= a; x *= k; y *= k;
ans = a * x - cb;
ans = (ans % g + g) % g;
return true;
}
int main(int argc, char const *argv[]) {
int N, M, K;
cin>>N>>M>>K;
LL over = INFF;
for (int i = 0; i < 3; i ++) {
LL a, b;
if (i == 0) a = N, b = M;
else if (i == 1) a = 0, b = M;
else a = N, b = 0;
LL x, y, d, res, ans = INFF;
for (int k = 0; k < 4; k ++) {
LL tmpa = a, tmpb = b;
if (k & 1) tmpa = -tmpa;
if ((k >> 1) & 1) tmpb = -tmpb;
if (solve_equ(2 * M, -2 * N, tmpa, tmpb, x, y, d, res))
ans = min(ans, res);
}
over = min(over, ans);
}
while (K --) {
LL a, b;
cin>>a>>b;
LL ans = INFF, x, y, d, res;
for (int k = 0; k < 4; k ++) {
LL tmpa = a, tmpb = b;
if (k & 1) tmpa = -tmpa;
if ((k >> 1) & 1) tmpb = -tmpb;
if (solve_equ(2 * M, -2 * N, tmpa, tmpb, x, y, d, res))
ans = min(ans, res);
}
if (ans == INFF || (over != INFF && ans > over)) puts("-1");
else cout<<ans<<endl;
}
return 0;
}