codeforces 400C - Inna and Huge Candy Matrix

題目鏈接：http://codeforces.com/problemset/problem/400/C

題目大意：給出n，m，x，y，z，p，n*m的矩陣上有p塊糖果，給出p塊糖果的座標，輸出矩陣順時針旋轉x次，鏡像翻轉y次，逆時針旋轉z次後糖果座標。

題目分析：旋轉完n和m要交換，翻轉不用，旋轉4次和翻轉2次都是不變的。

順時針旋轉：

n=3, m=2

(1,1)->(1,3) (1,2)->(2,3)

(2,1)->(1,2) (2,2)->(2,2)

(3,1)->(1,1) (3,2)->(2,1)

得到規律：x'=y，y'=1+n-x

n=3, m=2

(1,1)->(1,2) (1,2)->(1,1)

(2,1)->(2,2) (2,2)->(2,1)
(3,1)->(3,2) (3,2)->(3,1)
得到規律：x'=x，y'=n-y

n=3, m=2

(1,1)->(2,1) (1,2)->(1,1)

(2,1)->(2,2) (2,2)->(1,2)

(3,1)->(2,3) (3,2)->(1,3)

得到規律：x'=m+1-y，y'=x

代碼參考：

#include<set>
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 1e5+9;
struct Point
{
    int x, y;
}pos[N];
int n, m, x, y, z, p;
void clockwise()
{
    int x, y;
    for(int i=0; i<p; ++i)
    {
        x = pos[i].y;
        y = 1+n-pos[i].x;
        pos[i].x = x;
        pos[i].y = y;
    }
    swap(n, m);
}
void horizontal()
{
    for(int i=0; i<p; ++i)
    {
        pos[i].y = m+1-pos[i].y;
    }
}
void counter()
{
    int x, y;
    for(int i=0; i<p; ++i)
    {
        x = m+1-pos[i].y;
        y = pos[i].x;
        pos[i].x = x;
        pos[i].y = y;
    }
    swap(n, m);
}
int main()
{
    int i, j, a, b;
    while(~scanf("%d%d%d%d%d%d", &n, &m, &x, &y, &z, &p))
    {
        for(i=0; i<p; ++i)
        {
            scanf("%d%d", &pos[i].x, &pos[i].y);
        }
        for(i=0; i<x%4; ++i) clockwise();
        for(i=0; i<y%2; ++i) horizontal();
        for(i=0; i<z%4; ++i) counter();
        for(i=0; i<p; ++i)
        {
            printf("%d %d\n", pos[i].x, pos[i].y);
        }
    }
    return 0;
}