题目链接:
https://ac.nowcoder.com/acm/contest/19306/1026
题目分析
这种题,画个图,模拟就对啦。
TLE代码
#include <iostream>
using namespace std;
int n, cnt;
int w[25][25];
int main()
{
cin >> n;
// 构建框架
int dx = 1, dy = 1;
while (1)
{
while (dy <= n) w[dx][dy ++ ] = ++ cnt;
-- dy;
while (dx <= n) w[ ++ dx][dy] = ++ cnt;
-- dx;
while (dy >= 1) w[dx][ -- dy] = cnt ++ ;
++ dy, -- dx, -- cnt;
while (w[dx][dy] == 0) w[dx -- ][dy] = cnt ++ ;
break;
}
// cout << dx << ' ' << dy << ' ' << cnt << endl;
++ dx; ++ dy, -- cnt;
while (1)
{
if (w[dx][dy] && w[dx - 1][dy]
&& w[dx + 1][dy] && w[dx][dy - 1]
&& w[dx][dy + 1]) break;
// >
while (!w[dx][dy]) w[dx][dy ++ ] = ++ cnt;
// v
-- dy;
while (!w[ ++ dx][dy]) w[dx][dy] = ++ cnt;
// <
-- dx;
while (!w[dx][ -- dy]) w[dx][dy] = ++ cnt;
// ^
++ dy;
while (!w[ -- dx][dy]) w[dx][dy] = ++ cnt;
++ dx, ++ dy;
}
for (int i = 1; i <= n; ++ i)
{
for (int j = 1; j <= n; ++ j)
cout << w[i][j] << ' ';
cout << endl;
}
}
标签:AC,TLE,++,cnt,--,while,回行,dx,dy
From: https://www.cnblogs.com/ClockParadox43/p/17437986.html