题目描述

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

思路

1.关于行数、列数的求解
最接近n的平方的两个整数一定是row-col最小的两个整数。
2.关于矩阵的填充
填充的顺序是右下左上，由于vector的初始值为0且题目保证是正数，因此在一个周期内只需按照顺序判断即可。注意在往上填充的时候右边是空的，因此需要增设一个flag，当flag为1是才判断右方向。

代码

#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
bool cmp1(int a, int b) { return a > b; }
int main() {
    int n, row, col, min = 99999999, r = 0, c = 0, cnt = 0, flag=1;
    cin >> n;
    vector<int> mat(n);
    for (int i = 0; i < n; i++) cin >> mat[i];
    for (col = sqrt((double)n); col >= 1; col--) {
        if (n % col == 0) {
            row = n / col;
            break;
        }
    }
    vector<vector<int>> ans(row, vector<int>(col));
    sort(mat.begin(), mat.end(), cmp1);
    while (cnt < n) {
        ans[r][c] = mat[cnt];
        if (c + 1 < col && ans[r][c + 1] == 0 && flag == 1) c++;
        else if (r + 1 < row && ans[r + 1][c] == 0) r++;
        else if (c - 1 >= 0 && ans[r][c - 1] == 0) c--;
        else if (r - 1 >= 0 && ans[r - 1][c] == 0) {
            r--;
            if (r - 1 >= 0 && ans[r - 1][c] == 0) flag = 0;
            else flag = 1;
        }
        cnt++;
    }
    for (int i = 0; i < row; i++) {
        for (int j = 0; j < col; j++)
            cout << (j == 0 ? "" : " ") << ans[i][j];
        cout << endl;
    }
    return 0;
}