Codeforces 400C Inna and Huge Candy Matrix(模拟）

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题目链接：Codeforces 400C Inna and Huge Candy Matrix

题目大意：给出n，m，x，y，z和p，表示在一个n*m的矩阵上有p块糖果，给出p块糖果的坐标，然后将整个矩阵顺时针旋转x次，镜像翻转y次，逆时针旋转z次，然后按照顺序输出操作完后糖果的坐标。

解题思路：模拟，注意旋转完，n和m要交换，翻转不用，然后如果纯模拟肯定超时，很容易发现旋转4次等于没变，翻转2次也是没有变的。

#include <stdio.h>#include <string.h>const int N = 1e5+5;struct state {int xi, yi;}s[N];int n, m, x, y, z, p;void cw() {for (int i = 0; i < p; i++) {int r = s[i].yi;int l = n + 1 - s[i].xi;s[i].xi = r; s[i].yi = l;}int t = n;n = m; m = t;}void hr() {for (int i = 0; i < p; i++)s[i].yi = m + 1 - s[i].yi;}void ccw() {for (int i = 0; i < p; i++) {int r = m + 1 - s[i].yi;int l = s[i].xi;s[i].xi = r; s[i].yi = l;}int t = n;n = m; m = t;}int main () {scanf("%d%d%d%d%d%d", &n, &m, &x, &y, &z, &p);for (int i = 0; i < p; i++)scanf("%d%d", &s[i].xi, &s[i].yi);x %= 4;for (int i = 0; i < x; i++) cw();y %= 2;for (int i = 0; i < y; i++) hr();z %= 4;for (int i = 0; i < z; i++) ccw();for (int i = 0; i < p; i++)printf("%d %d\n", s[i].xi, s[i].yi);return 0;}

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