PAT (Advanced Level) Practise 1105 Spiral Matrix (25)

1105. Spiral Matrix (25)

时间限制

150 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

This time your job is to fill a sequence of N positive integers into a spiral matrix in non-increasing order. A spiral matrix is filled in from the first element at the upper-left corner, then move in a clockwise spiral. The matrix has m rows and ncolumns, where m and n satisfy the following: m*n must be equal to N; m>=n; and m-n is the minimum of all the possible values.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N. Then the next line contains N positive integers to be filled into the spiral matrix. All the numbers are no more than 10⁴. The numbers in a line are separated by spaces.

Output Specification:

For each test case, output the resulting matrix in m lines, each contains n numbers. There must be exactly 1 space between two adjacent numbers, and no extra space at the end of each line.

Sample Input:

12
37 76 20 98 76 42 53 95 60 81 58 93

Sample Output:

98 95 93
42 37 81
53 20 76
58 60 76
简单的排列一下数字。
#include<cstdio>
#include<vector>
#include<queue>
#include<string>
#include<map>
#include<cmath>
#include<iostream>
#include<cstring>
#include<functional>
#include<algorithm>
using namespace std;
typedef long long LL;
const int INF = 0x7FFFFFFF;
const int maxn = 1e4 + 10;
int n, a[maxn], x, y, vis[maxn];
vector<int> c[maxn];
int u[4] = { 0, 1, 0, -1 };
int v[4] = { 1, 0, -1, 0 };

int main()
{
	scanf("%d", &n);
	for (int i = 0; i < n; i++) scanf("%d", &a[i]);
	for (int i = 1; i*i <= n; i++)
	{
		if (n % i == 0)
		{
			x = n / i;	y = i;
		}
	}
	for (int i = 1; i <= x; i++)
	{
		for (int j = 0; j <= y; j++) c[i].push_back(0);
	}
	sort(a, a + n, greater<int>());
	int vec = 0, X = 1, Y = 1;
	for (int i = 0; i < n; i++)
	{
		c[X][Y] = a[i];	vis[(X - 1)*y + Y - 1] = 1;
		if (X + u[vec]>x || X + u[vec] < 1 || Y + v[vec]< 1 || Y + v[vec]>y) vec = (vec + 1) % 4;
		else if (vis[(X + u[vec] - 1) * y + Y + v[vec] - 1]) vec = (vec + 1) % 4;
		X += u[vec];	Y += v[vec];
	}
	for (int i = 1; i <= x; i++)
	{
		for (int j = 1; j <= y; j++)
		{
			printf("%d%s", c[i][j], j == y ? "\n" : " ");
		}
	}
	return 0;
}