这道题其实不难，只是一道非常简单的模拟题。
我们发现，每顺时针转动 \(4\) 次、镜面对称 \(2\) 次、逆时针旋转 \(4\) 次，就变回原来的样子了。
所以，在操作前，我们可以让 \(x\gets x\bmod4\)，\(y\gets y\bmod2\)，\(z\gets z\bmod4\)。
接下来，只需在草稿纸上画一画，即可知道顺时针转一次，一个点的 \(x\) 值会变为原来的 \(y\) 值，\(y\) 值会变为原来的 \(n-x+1\) 的值；镜面对称一次，一个点的 \(y\) 值会变为原来的 \(m-y+1\) 的值；逆时针转一次，一个点的 \(y\) 值会变为原来的 \(x\) 值，\(x\) 值会变为原来的 \(m-y+1\) 的值。要注意，两个旋转操作后要交换 \(n\) 和 \(m\)。
This problem is not difficult, it's just a very simple simulation problem.
We found that every \(4\) clockwise rotations, \(2\) mirror symmetry rotations, and \(4\) counterclockwise rotations, it returns to its original state.
So, before proceeding, we can set \(x\gets x\bmod4\), \(y\gets y\bmod2\), and \(z\gets z\bmod4\).
Next, just draw a picture on the draft paper to know that if you turn it clockwise once, the \(x\) value of a point will become the original \(y\) value, and the \(y\) value will become the original \(n-x+1\) value; Once the mirror is symmetrical, the \(y\) value of a point will change to the original \(m-y+1\) value; Rotate counterclockwise once, and the \(y\) value of a point will become the original \(x\) value, and the \(x\) value will become the original \(m-y+1\) value. Note that after two rotation operations, \(n\) and \(m\) need to be swapped.
AC Code：
#include<bits/stdc++.h>
using namespace std;
int n,m,x,y,z,p;
struct e{
int x,y;
}d[100005];
void fun1(int t){
while(t--){
for(int i=1;i<=p;i++){
int o=d[i].x;
d[i].x=d[i].y;
d[i].y=n-o+1;
}
swap(n,m); //记得交换
}
}
void fun2(int t){
while(t--){
for(int i=1;i<=p;i++){
d[i].y=m-d[i].y+1;
}
}
}
void fun3(int t){
while(t--){
for(int i=1;i<=p;i++){
int o=d[i].y;
d[i].y=d[i].x;
d[i].x=m-o+1;
}
swap(n,m);
}
}
int main(){
cin>>n>>m>>x>>y>>z>>p;
x%=4;y%=2;z%=4; //记得模一下
for(int i=1;i<=p;i++)cin>>d[i].x>>d[i].y;
fun1(x);fun2(y);fun3(z);
for(int i=1;i<=p;i++)cout<<d[i].x<<" "<<d[i].y<<"\n";
return 0;
}