Visual Cube

• 将这个立方体分块，分成上中下三个部分，利用长宽高计算行列，最后输出即可。
  每个部分都分成奇偶行来输出，总共有\(2*(b+c)+1\)行，共\(2*(a+b)+1\)列。设当前行为\(i\),划分方式:当行数小于等于\(2*b\)时，在上部，当$i > (2 * b) $ && \(i<=2*c+1\)时，在中部,其余在下部。这样的划分可以应对绝大部分情况，但是对于上部有一中情况是不包括的，当$b>c $时，如果继续之前的划分，那么在头部判定之后，它依然符合下部的判定，所以这时也被当作下部输出，这不是预期的结果，所以在上部加上额外的判定。

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
using namespace std;
int main(void) {int t, a, b, c;cin >> t;while (t-- > 0) {cin >> a >> b >> c;for (int i = 1; i <= (2 * (b + c) + 1); i++) {//if (i != 1)cout << endl;//headif (i % 2 == 1 && i <= (2 * b)) {for (int j = 1; j <= 2*(b-i/2); j++) {cout << ".";}for (int j = 2 * (b - i / 2) + 1; j <= 2 * (b - i / 2) + a * 2; j+=2) {cout << "+-";}cout << "+";if (b < c || i <= (2 * c + 1)) {for (int j = 2 * (b - i / 2) + a * 2 + 2; j <= 2 * (a + b) + 1; j += 2) {cout << ".+";}}else{cout << ".";for (int j = 1; j <= c; j++) {cout << "+.";}for (int j = 2 * (b - i / 2) + a * 2 + 2*c+3; j <= 2 * (a + b) + 1; j++) {cout << ".";}}cout << endl;continue;}if (i % 2 == 0 && i <= (2 * b)) {for (int j = 1; j <= 2 * (b - i / 2)+1; j++) {cout << ".";}for (int j = 2 * (b - i / 2) + 1; j <= 2 * (b - i / 2) + a * 2; j += 2) {cout << "/.";}if (b < c ||i<=(2*c+1)) {for (int j = 2 * (b - i / 2) + a * 2 + 2; j <= 2 * (a + b) + 1; j += 2) {cout << "/|";}}else{cout << "/";for(int j=1;j<=c;j++){cout << "|/";}for (int j = 2 * (b - i / 2) + a * 2+2*c+3; j <= 2 * (a + b) + 1; j++) {cout << ".";}}cout << endl;continue;}//middleif (i % 2 == 1 && i > (2 * b)&& i<=2*c+1) {for (int j = 1; j <= 2 * a; j+=2) {cout << "+-";}cout << "+";for (int j = 2 * (a+1); j <= 2 * (a + b) + 1; j += 2) {cout << ".+";}cout << endl;continue;}if (i % 2 == 0 && i > (2 * b) && i <= 2 * c + 1) {for (int j = 1; j <= 2 * a; j+=2) {cout << "|.";}cout << "|";for (int j = 2 * (a + 1); j <= 2 * (a + b) + 1; j += 2) {cout << "/|";}cout << endl;continue;}//bottomif (i % 2 == 1 && i > (2 * c + 1)) {for (int j = 1; j <= 2 * a; j+=2) {cout << "+-";}cout << "+";int k = (2 * (c + b) + 1 - i) / 2+1;for (int j = 1; j <= k-1; j++) {cout << ".+";}for (int j = 2 * a + 1 + 2 * k; j <= 2 * (a + b) + 2; j++) {cout << ".";}cout << endl;continue;}if (i % 2 == 0 && i > (2 * c + 1)) {for (int j = 1; j <= 2 * a; j+=2) {cout << "|.";}int k = (2 * (c + b) + 1 - i) / 2+1;for (int j = 1; j <= k; j++) {cout << "|/";}for (int j = 2 * a + 1 + 2 * k; j <= 2 * (a + b) + 1; j++) {cout << ".";}cout << endl;continue;}}}return 0;
}