矩形翻转

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一个矩形，如果四个角都标上数字，要得通过旋转，翻转等操作后的所有结果，只需要连续做对称翻转和上下翻转就可以了。
如题目：
http://hihocoder.com/problemset/problem/1268

#include <cstdio>#include <algorithm>#include <cstring>#define rep(i, s, e) for (int i= s; i< e; i++)int G[4][4];int org[3][3] = {4, 9, 2, 3,5,7,8,1,6};int a[9][3][3];int main(){//freopen("in.txt", "r", stdin);    rep(i, 0, 3) rep(j, 0, 3) a[0][i][j] = org[i][j];    rep(i, 1, 8)    {        if (i%2){            rep(j, 0, 3)                rep(k, 0, 3)                    a[i][j][k] = a[i-1][k][j];        }        else{            rep(j, 0, 3)                rep(k, 0, 3)                    a[i][j][k] = a[i-1][j][2-k];        }    }   rep(i, 0, 3)        rep(j, 0, 3)    {        scanf("%d", &G[i][j]);    }    int ans = -1;    int ansNum = 0;    rep(i, 0, 8)    {        bool flag = true;        rep(j, 0, 3) rep(k, 0, 3)            {                if (G[j][k] != 0 && G[j][k] != a[i][j][k])                    flag = false;            }        if (flag)        {            ansNum++;            ans = i;        }    }    if (ansNum > 1)        printf("Too Many\n");    else{        rep(i, 0, 3)        {            rep(j, 0, 3)                printf("%d ", a[ans][i][j]);            printf("\n");        }    }}

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