# 1105 Spiral Matrix （25 分）

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 10​4​​. 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
``````

## Code:

``````#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#pragma warning(disable:4996)
using namespace std;
int main()
{
int n;
scanf("%d", &n);
vector<int> vec(n);
for (int i = 0; i < n; i++)
scanf("%d", &vec[i]);
sort(vec.begin(), vec.end(), [](const int n1, const int n2) {return n1 > n2; });
int col = sqrt(n), row;
while (n % col != 0) col--;
row = n / col;
vector<vector<int>> matrix(row, vector<int>(col));
int level = row / 2 + row % 2;
int t = 0;
for (int i = 0; i < level; i++) {
for (int j = i; j <= col - 1 - i && t <= n - 1; j++)
matrix[i][j] = vec[t++];
for (int j = i + 1; j <= row - 2 - i && t <= n - 1; j++)
matrix[j][col - 1 - i] = vec[t++];
for (int j = col - i - 1; j >= i && t <= n - 1; j--)
matrix[row - 1 - i][j] = vec[t++];
for (int j = row - 2 - i; j >= i + 1 && t <= n - 1; j--)
matrix[j][i] = vec[t++];
}
for (int i = 0; i < matrix.size(); i++)
for (int j = 0; j < matrix[i].size(); j++)
printf("%d%c", matrix[i][j], (j == matrix[i].size() - 1 ? '\n' : ' '));
return 0;
}
``````

## 思路：

https://blog.csdn.net/liuchuo/article/details/52123228

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