pat乙级1050

pat乙级1050

这道题目刚开始没想出怎么做螺旋输出，后来借鉴了柳神的思路（原文链接：https://blog.youkuaiyun.com/liuchuo/article/details/52123209）

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;


bool cmp(int a, int b)
{
	return a > b;
}


int main()
{
	int N, m, n;
	cin >> N;
	vector<int> a(N);
	for (int i = 0; i < N; i++)
		cin >> a[i];
	sort(a.begin(), a.end(), cmp);

	//找出符合条件的m和n;
	for (int j = sqrt(N); j >= 1; j--)
	{
		if (N%j == 0)
		{
			m = N / j;
			n = j;
			break;
		}
	}

	vector<vector<int>> b(m, vector<int>(n));//初始化二维数组全部赋值为0
	int t = 0;//用来记录输出a数组的第几个
	int level = m / 2 + m % 2;
	for (int i = 0; i < level; i++)//外层for循环
	{
		for (int j = i; j <= n - 1 - i && t <= N - 1; j++)//输出内层第一个for循环
			b[i][j] = a[t++];
		for (int j = i + 1; j < m - 1 - i && t <= N - 1; j++)//输出内层第二个for循环
			b[j][n - 1 - i] = a[t++];
		for (int j = n - 1 - i; j >= i && t <= N - 1; j--)//输出内层第三个for循环
			b[m - 1 - i][j] = a[t++];
		for (int j = m - 2 - i; j >= i + 1 && t <= N - 1; j--)//输出内层第四个for循环
			b[j][i] = a[t++];
	}

	for (int i = 0; i < m; i++)
	{
		for (int j = 0; j < n; j++)
		{
			cout << b[i][j];
			if (j != n - 1)
				cout << " ";
		}
		cout << endl;
	}

	return 0;
}