HDU 6330--Visual Cube(构造，计算)

标签：clu   http   visual   方式   a+b   head   --   结果   头部   

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;
            //head
            if (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;
            }
            //middle
            if (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;
            }
            //bottom
            if (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;
}

HDU 6330--Visual Cube(构造，计算)

标签：clu   http   visual   方式   a+b   head   --   结果   头部   

原文地址：https://www.cnblogs.com/FlyerBird/p/9403338.html