P1498 南蛮图腾题解

用分治很简单. 每次向下复制一下,向右复制一下,再向上复制一下.

   /\
  /__\
 /\  /\
/__\/__\

向下和向右:(顺便把原本的清掉)

   /\      /\
  /__\    /__\
 /\  /\  /\  /\
/__\/__\/__\/__\

再向上:

       /\
      /__\
     /\  /\
    /__\/__\
   /\      /\
  /__\    /__\
 /\  /\  /\  /\
/__\/__\/__\/__\

#include <bits/stdc++.h>

using namespace std;
//https://www.luogu.com.cn/blog/froggy/solution-p1498
const int N = 3010;//开大点
char a[N][N];
int n;
int h = 2, w = 4;//h是高,w是宽
void init() {
    memset(a, ' ', sizeof(a));
    a[1][2] = a[2][1] = '/';
    a[1][3] = a[2][4] = '\\';
    a[2][2] = a[2][3] = '_';
}

int main() {
    //初始化
    init();
    //读入
    cin >> n;
    //迭代n-1次，因为第一个三角形已经写死了~
    for (int i = 1; i < n; i++) {
        //向下和向右
        for (int j = 1; j <= h; j++) {
            for (int k = 1; k <= w; k++) {
                a[j + h][k] = a[j + h][k + w] = a[j][k];
                a[j][k] = ' ';//把上面的清掉
            }
        }
        //向上
        for (int j = 1; j <= h; j++)
            for (int k = 1; k <= w; k++)
                a[j][k + w / 2] = a[j + h][k];

        //双倍扩展
        w *= 2, h *= 2;
        //刷新完成一次
    }

    //输出大吉
    for (int i = 1; i <= h; i++) { //高度就是行
        for (int j = 1; j <= w; j++) //宽度就是列
            cout << a[i][j];
        cout << endl;
    }
    return 0;
}

标签：__,int,题解,init,复制,一下,南蛮,P1498,向下
来源： https://www.cnblogs.com/littlehb/p/15033788.html