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.
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.
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.
3 3 4
1 1
1 2
2 1
2 2
1
-1
-1
2
3 4 6
1 1
2 1
1 2
2 2
1 3
2 3
1
-1
-1
2
5
-1
7 4 5
1 3
2 2
5 1
5 3
4 3
13
2
9
5
-1
题意：n*m的矩阵，从(0,0)沿45度角射出一束光，速度为根号2，碰到边界反射，到角上结束。k个点，分别输出光到每个点的时间.
一段一段的模拟。
每个点y-x和y+x为定值，用两个vector存点，v1[b]，v2[b],这样在模拟每一段线段的时候就不用遍历整个点的集合。只需要遍历当前k和b下的vector中的点。
#include<iostream> #include<cstring> #include<cstdio> #include<string> #include<algorithm> #include<vector> using namespace std; int n,m,t; struct Node { int num; int x,y; }; Node getPoint(Node st,int k) { Node p; if(k*(-st.x)+st.y!=st.y&&k*(-st.x)+st.y>=0&&k*(-st.x)+st.y<=m) { p.x=0; p.y=k*(-st.x)+st.y; return p; } if(k*(n-st.x)+st.y!=st.y&&k*(n-st.x)+st.y>=0&&k*(n-st.x)+st.y<=m) { p.x=n; p.y=k*(n-st.x)+st.y; return p; } if((-st.y+k*st.x)/k!=st.x&&(-st.y+k*st.x)/k>=0&&(-st.y+k*st.x)/k<=n) { p.x=(-st.y+k*st.x)/k; p.y=0; return p; } if((m-st.y+k*st.x)/k!=st.x&&(m-st.y+k*st.x)/k>=0&&(m-st.y+k*st.x)/k<=n) { p.x=(m-st.y+k*st.x)/k; p.y=m; return p; } } bool isEnd(int x,int y) { if(x==n&&y==0) return 1; if(x==n&&y==m) return 1; if(x==0&&y==m) return 1; return 0; } int getCost(int sx,int sy,int k,int tmpx,int tmpy) { int b=sy-k*sx; if(tmpy!=k*tmpx+b) return -1; return abs(tmpx-sx); } long long ans[100005]; vector<Node> v1[200005]; vector<Node> v2[200005]; int main() { scanf("%d%d%d",&n,&m,&t); /*Node a; a.x=3; a.y=3; Node b=getPoint(a,-1); cout<<b.x<<" "<<b.y<<endl;*/ int all=0; for(int i=0; i<t; i++) { Node node; node.num=i; scanf("%d%d",&node.x,&node.y); int b1=node.y-node.x+100005; int b2=node.y+node.x; v1[b1].push_back(node); v2[b2].push_back(node); all++; } //for(int i=0;i<v2[6].size();i++) // cout<<v2[6][i].num<<" "<<v2[6][i].x<<" "<<v2[6][i].y<<endl; Node st; st.x=0; st.y=0; int cnt=0; long long time=0; while(!isEnd(st.x,st.y)) { int k; if(cnt%2) k=-1; else k=1; cnt++; Node en=getPoint(st,k); int b=st.y-k*st.x; if(k==1) { b+=100005; for(int i=0; i<v1[b].size(); i++) { if(ans[v1[b][i].num]==0) { int cost=getCost(st.x,st.y,k,v1[b][i].x,v1[b][i].y); ans[v1[b][i].num]=time+cost; } } } if(k==-1) { for(int i=0; i<v2[b].size(); i++) { if(ans[v2[b][i].num]==0) { int cost=getCost(st.x,st.y,k,v2[b][i].x,v2[b][i].y); ans[v2[b][i].num]=time+cost; } } } time+=abs(en.x-st.x); st=en; } //cout<<"cnt:"<<cnt<<endl; for(int i=0; i<all; i++) if(ans[i]==0) printf("-1 "); else printf("%I64d ",ans[i]); return 0; }