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

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

分析

#include<iostream>
#include<vector>
#include<algorithm>
#include<cmath>

using namespace std;

int main()
{
int n;
cin >> n;
vector<int> arr(n, 0);
// 行、列、根号n
int r = 0, c = 0, sqrtn = (int)sqrt(n);
for (int i = 1; i <= n; i++)
cin >> arr[i - 1];
sort(arr.begin(), arr.end());

// 寻找使r-c最小的r和c
for (c = sqrtn; c >= 1; c--) {
if (n%c == 0) {
r = n / c;
break;
}
}

// 结果是一个r行，c列的二维数组
vector<vector<int>> ans(r, vector<int>(c));

int i = 0, j = 0, count = n - 1, round = 1;
// 控制方向
bool left = false, right = true, down = false, up = false;

// j的取值范围为[round-1，c-round+1]，i的取值范围为[round-1, r-round+1]
while (count >= 0) {
if (left) {
ans[i][j--] = arr[count--];
// 到达左边界，向上拐
if (j < round - 1) {
left = false;
up = true;
j++;
i--;
}
}
else if (right) {
ans[i][j++] = arr[count--];
// 到达右边界，向下拐
if (j >= c - round + 1) {
right = false;
down = true;
i++;
j--;
}
}
else if (down) {
ans[i++][j] = arr[count--];
// 到达下边界，向左拐
if (i >= r - round + 1) {
down = false;
left = true;
j--;
i--;
}
}
else {
ans[i--][j] = arr[count--];
// 到达上边界，向右拐
if (i <= round - 1) {
// 修改起点
i = round;
j = round;
round++;
up = false;
right = true;
}
}
}
for (int i = 0; i < r; i++) {
for (int j = 0; j < c - 1; j++) cout << ans[i][j] << " ";
cout << ans[i][c - 1] << endl;
}
return 0;
}