Problem L. Visual Cube（杭电多校2018年第三场+模拟）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6330

题目：

题意：给你长宽高，让你画出一个正方体。

思路：模拟即可，湘潭邀请赛热身赛原题，不过比那个容易很多，湘潭那个没写==，这个模拟还是很难受的，写了好久……

代码实现如下：

#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long ll;
typedef pair<ll, ll> pll;
typedef pair<int, ll> pil;;
typedef pair<int, int> pii;
typedef unsigned long long ull;

#define lson i<<1
#define rson i<<1|1
#define bug printf("*********
");
#define FIN freopen("D://code//in.txt", "r", stdin);
#define debug(x) cout<<"["<<x<<"]" <<endl;
#define IO ios::sync_with_stdio(false),cin.tie(0);

const double eps = 1e-8;
const int mod = 10007;
const int maxn = 100 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f;

int t, a, b, c, n, m;
char mp[maxn][maxn];

int main() {
    //FIN;
    scanf("%d", &t);
    while(t--) {
        scanf("%d%d%d", &a, &b, &c);
        n = 2 * c + 1 + 2 * b;
        m = 2 * a + 1 + 2 * b;
        //初始化
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= m; j++) {
                mp[i][j] = '.';
            }
        }
        //上方（正方体正面的上面那条边为界线，下同）的+-构造
        for(int i = 1; i <= 2 * b; i += 2) {
            for(int j = 2 * b - i + 2; j <= 2 * b - i + 2 + 2 * a; j++) {
                if(j & 1) mp[i][j] = '+';
                else mp[i][j] = '-';
            }
            for(int j = 2 * b - i + 2 + 2 * a + 2; j <= m; j++) {
                if(j & 1) mp[i][j] = '+';
            }
        }
        //|/构造
        for(int i = 2; i <= 2 * b; i+=2) {
            for(int j = 2 * b - i + 2; j <= m; j++) {
                if(j % 2 == 0) mp[i][j] = '/';
            }
            for(int j = m; j >= m - i + 2; j -= 2) {
                mp[i][j] = '|';
            }
        }
        for(int i = 2 * b + 2; i <= n; i += 2) {
            for(int j = 1; j <= m; j += 2) {
                mp[i][j] = '|';
            }
            for(int j = 2 * a + 2; j <= m; j+= 2) {
                mp[i][j] = '/';
            }
        }
        //下方的+-构造
        for(int i = 2 * b + 1; i <= 2 * c + 1; i += 2) {
            for(int j = 1; j <= 2 * a + 1; j++) {
                if(j & 1) mp[i][j] = '+';
                else mp[i][j] = '-';
            }
            for(int j = 2 * a + 2; j <= m; j++) {
                if(j & 1) mp[i][j] = '+';
            }
        }
        for(int i = n; i > 2 * c + 1; i-=2) {
            for(int j = 1; j <= 2 * a + 1; j++) {
                if(j & 1) mp[i][j] = '+';
                else mp[i][j] = '-';
            }
            for(int j = 2 * a + 2; j <= m; j++) {
                if(j & 1) mp[i][j] = '+';
            }
        }
        //我上面的处理会让下方一些应该为.的变成其他的，所以需要用.覆盖回来
        int y = 0;
        for(int i = 1; i < 2 * b + 1; i++) {
            for(int j = 1; j <= 2 * b + 1 - i; j++) {
                mp[i][j] = '.';
            }
        }
        for(int i = n; i > 2 * c + 1; i--) {
            for(int j = 2 * a + 2 + y; j <= m; j++) {
                mp[i][j] = '.';
            }
            y++;
        }
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= m; j++) {
                printf("%c", mp[i][j]);
            }
            printf("
");
        }
    }
    return 0;
}