C. Ray Tracing
There are k sensors located in the rectangular room of size n × m meters. The i-th sensor is located at point (xi, yi). All sensors are located at distinct points strictly inside the rectangle.
Opposite corners of the room are located at points (0, 0) and (n, m). Walls of the room are parallel to coordinate axes.
At the moment 0, from the point (0, 0) the laser ray is released in the direction of point (1, 1). The ray travels with a speed of 
meters per second. Thus, the ray will reach the point (1, 1) in exactly one second after the start.
When the ray meets the wall it‘s reflected by the rule that the angle of incidence is equal to the angle of reflection. If the ray reaches any of the four corners, it immediately stops.
For each sensor you have to determine the first moment of time when the ray will pass through the point where this sensor is located. If the ray will never pass through this point, print - 1 for such sensors.
Input
The first line of the input contains three integers n, m and k (2 ≤ n, m ≤ 100 000, 1 ≤ k ≤ 100 000) — lengths of the room‘s walls and the number of sensors.
Each of the following k lines contains two integers xi and yi (1 ≤ xi ≤ n - 1, 1 ≤ yi ≤ m - 1) — coordinates of the sensors. It‘s guaranteed that no two sensors are located at the same point.
Output
Print k integers. The i-th of them should be equal to the number of seconds when the ray first passes through the point where the i-th sensor is located, or - 1 if this will never happen.
Examples
input
3 3 41 11 22 12 2
output
1-1-12
input
3 4 61 12 11 22 21 32 3
output
1-1-125-1
input
7 4 51 32 25 15 34 3
output
13295-1 题意是一束光从(0,0)射出，光线速度为根号2的单位每秒，求一路上有k个传感仪，求各个传感仪第一次接收到光线信号的时间，如果接收不到就输出-1。 观察到光线与坐标轴的角度始终为45°，那样的话说明光线的位置可以用一个数字来表示，这个类似八皇后的时候的斜线表示。用两个vector来储存各斜线上的点。用dx，dy可以表示光线方向。
然后就可以枚举光线路径斜线上的传感仪更新答案，再更新光线撞墙的坐标和光线方向，直到光线到达角落。
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn = 2e5 + 10;
vector< pair<int, int> > va[maxn], vb[maxn];
ll ans[maxn];
int main()
{
int n, m, k;
memset(ans, 0x3f, sizeof(ans));
scanf("%d %d %d", &n, &m, &k);
for(int i = 1; i <= k; i++) {
int x, y;
scanf("%d %d", &x, &y);
va[x+y].emplace_back(i, x);
vb[x-y+m].emplace_back(i, x);
}
int x = 0, y = 0; //定义初始坐标
int dx = 1, dy = 1; //定义初始方向
ll t = 0; //定义初始时间
while(true) {
if(dx == dy) {
for(auto p:vb[x-y+m]) {
ans[p.first] = min(ans[p.first], t + abs(p.second - x));
}
}
else {
for(auto p:va[x+y]) {
ans[p.first] = min(ans[p.first], t + abs(p.second - x));
}
}
int time = 0x3f3f3f3f;
if(dx == -1) time = min(time, x);
else time = min(time, n - x);
if(dy == -1) time = min(time, y);
else time = min(time, m - y);
x = x + time * dx;
y = y + time * dy;
t = time + t;
bool flag1 = (x == 0 || x == n);
bool flag2 = (y == 0 || y == m);
if(flag1 && flag2) break;
if(flag1) dx = -dx;
if(flag2) dy = -dy;
}
for(int i = 1; i <= k; i++) {
if(ans[i] == 0x3f3f3f3f3f3f3f3f) puts("-1");
else printf("%I64d\n", ans[i]);
}
}