1105 Spiral Matrix (25 分)（模拟）

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

生词

英文	解释
Spiral Matrix	旋转矩阵
clockwise	顺时针的

题目大意：

将给定的N个正整数按非递增的顺序，填入“螺旋矩阵”_{所谓“螺旋矩阵”，是指从左上角第1个格子开始，按顺时针螺旋方向填充}要求矩阵的规模为m行n列，满足条件：m*n等于N；m>=n；且m-n取所有可能值中的最小值～

分析：

先计算行数m和列数n的值，n从根号N的整数部分开始，往前推一直到1，找到第一个满足N % n== 0的，m的值等于N/n～将N个给定的值输入数组a，并将a数组中的值按非递增排序，接着建立m行n列的数组b，填充时按层数填充，一个包裹矩阵的口字型为一层，计算螺旋矩阵的层数level，如果m的值为偶数，层数为m/2，如果m为奇数，层数为m/2+1，所以level = m / 2 + m % 2；因为是从左上角第1个格子开始，按顺时针螺旋方向填充，所以外层for循环控制层数i从0到level，内层for循环按左上到右上、右上到右下、右下到左下、左下到左上的顺序一层层填充，注意内层for循环中还要控制t <= N – 1，因为如果螺旋矩阵中所有的元素已经都填充完毕，就不能再重复填充～填充完毕后，输出整个矩阵～

原文链接：https://blog.csdn.net/liuchuo/article/details/52123228

柳神题解

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
int cmp(int a, int b) {return a > b;}
int main() {
    int N, m, n, t = 0;
    scanf("%d", &N);
    for (n = sqrt((double)N); n >= 1; n--) {
        //if语句>=2加大括号，抄错了的原因Orz
        if (N % n == 0) {
            m = N / n;
            break;
        }
    }
    vector<int> a(N);
    for (int i = 0; i < N; i++)
        scanf("%d", &a[i]);
    sort(a.begin(), a.end(), cmp);
    vector<vector<int> > b(m, vector<int>(n));
    int level = m / 2 + m % 2;
    for (int i = 0; i < level; i++) {
        for (int j = i; j <= n - 1 - i && t <= N - 1; j++)
                b[i][j] = a[t++];
        for (int j = i + 1; j <= m - 2 - i && t <= N - 1; j++)
                b[j][n - 1 - i] = a[t++];
        for (int j = n - i - 1; j >= i && t <= N - 1; j--)
                b[m - 1 - i][j] = a[t++];
        for (int j = m - 2 - i; j >= i + 1 && t <= N - 1; j--)
                b[j][i] = a[t++];
    }
    for (int i = 0; i < m; i++) {
        for (int j = 0 ; j < n; j++) {
            printf("%d", b[i][j]);
            if (j != n - 1) printf(" ");
        }
        printf("\n");
    }
    return 0;
}