单词积累
clockwise 顺时针方向的
spiral 螺旋的
思路
本质是模拟的思想，用i标记遍历到第几圈，然后分为矩形的四个边（对应下图不同颜色），分别去赋值，注意边界点即可。
一般情况
此外，还有可能出现3 7这样的数字，数组为3*1，为了不让第四次的值覆盖第二次的值，此处选取的做法是判断值有没有赋完，赋完立刻退出。
特殊考虑
代码
#include <bits/stdc++.h>
using namespace std;
const int maxn = 10002;
int N;
int m, n;
int num[maxn];
int ans[maxn][maxn];
bool cmp (int a, int b) {
return a > b;
}
int main() {
cin>>N;
for (int i = 0; i < N; i++) {
cin>>num[i];
}
sort(num, num + N, cmp);
for (int i = sqrt(N); i >= 0; i--) {
if (N % i == 0) {
n = i;
m = N / n;
break;
}
}
int indexs = 0;
int cri = 0;
if (n % 2 == 0) cri = n / 2;
else cri = n / 2 + 1;
for (int i = 0; i < cri; i++) {
for (int j = i; j < n - i; j++) {
ans[i][j] = num[indexs++];
// cout<<1<<" "<<i<<" "<<j<<" "<<ans[i][j]<<endl;
}
if (indexs >= N) break;
for (int j = i + 1; j < m - i; j++) {
ans[j][n - i - 1] = num[indexs++];
// cout<<2<<" "<<j<<" "<<n - i - 1<<" "<<ans[j][n - i - 1]<<endl;
}
if (indexs >= N) break;
for (int j = n - i - 2; j >= i; j--) {
ans[m - i - 1][j] = num[indexs++];
// cout<<3<<" "<<m - i - 1<<" "<<j<<" "<<ans[m - i - 1][j]<<endl;
}
if (indexs >= N) break;
for (int j = m - i - 2; j > i; j--) {
ans[j][i] = num[indexs++];
// cout<<4<<" "<<j<<" "<<i<<" "<<ans[j][i]<<endl;
}
if (indexs >= N) break;
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
cout<<ans[i][j];
if (j != n - 1) cout<<" ";
}
cout<<endl;
}
}
