1105 Spiral Matrix (25point(s)) - C語言 PAT 甲級

1105 Spiral Matrix (25point(s))

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 n columns, 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

題目大意：

1050 螺旋矩陣 (25point(s))

設計思路：

1050 螺旋矩陣（C語言）

• 總體設計方向是一層一層填充 m x n 的矩陣

1. 確定 m 和 n 的值，n 從 sqrt(N) 趨於 0，第一個 N % n == 0 成立即找到
2. 確定矩陣有幾層，level = n / 2 + n % 2
3. 每一層有四個方向

• 注意：如何保存矩陣？一種方法是直接利用一維數組模擬二維，但操作相對繁瑣，本文采用變長數組進行保存，較爲方便。不要預設矩陣爲 10000 x 10000 大小後進行操作，提交後最後一個測試點會返回各種不同的結果，個人推測是因爲計算過程中要對二維數組進行大量的尋址，造成大量時間空間的浪費，如下極端情況：N 爲 9973 素數時矩陣形狀
  9973 0 ……
  . 0 0 0 ……
  . 0 0 0 ……
  . 0 0 0 ……
  3 0 0 0 ……
  2 0 0 0 ……
  1 0 0 0 ……
  所有 0 的位置均無實際數據，但因二維數組在內存中是“一維”的，每次存取均要經過很多“0”才能到達正確地址

編譯器：C (gcc)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int comp(const void *a, const void *b)
{
    return *(int*)b - *(int*)a;
}


int main()
{
    int N, m, n, num[10000];
    int i, j, k,level;

    scanf("%d", &N);
    for(i = 0; i < N; i++)
        scanf("%d", num + i);

    qsort(num, N, sizeof(int), comp);
    for(n = sqrt((double)N); !(n > 0 && N % n == 0); n--)
        continue;
    m = N / n;

    int matrix[m][n];
    level = n / 2 + n % 2;
    for(i = 0, k = 0; i < level; i++){
        for(j = i; k <= N - 1 && j <= n - i - 1; j++)
            matrix[i][j] = num[k++];
        for(j = i + 1; k <= N - 1 && j <= m - i - 2; j++)
            matrix[j][n - i - 1] = num[k++];
        for(j = n - i - 1; k <= N - 1 && j >= i; j--)
            matrix[m - i -1][j] = num[k++];
        for(j = m - 2 - i; k <= N - 1 && j >= i + 1; j--)
            matrix[j][i] = num[k++];
    }

    for(i = 0; i < m; i++)
        for(j = 0; j < n; j++)
            printf("%d%c", matrix[i][j], j != n -1 ? ' ' : '\n');

    return 0;
}