更新一下码风，发一篇草履虫的做法题解

#include <iostream>
using namespace std;
int n, m, arr[100][100], i, l, r, u, d;
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    cin >> n >> m;
    for (i = 0; i < n; ++i)
        for (int j = 0; j < m; ++j) 
            cin >> arr[i][j];
    l = u = 0, r = m - 1, d = n - 1;
    while (1) {
        for (i = l; i <= r; ++i)
            cout << arr[u][i] << ' ';
        if (++u > d) break;
        for (i = u; i <= d; ++i)
            cout << arr[i][r] << ' ';
        if (l > --r) break;
        for (i = r; i >= l; --i)
            cout << arr[d][i] << ' ';
        if (u > --d) break;
        for (i = d; i >= u; --i)
            cout << arr[i][l] << ' ';
        if (++l > r) break;
    }
}

假的，我有更优雅的写法

#include <iostream>
using namespace std;
int n, m, arr[100][100], l, i, j, f, dir[4][2] = {0, 1, 1, 0, 0, -1, -1, 0};
bool vis[100][100];
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    cin >> n >> m;
    for (i = 0; i < n; ++i)
        for (j = 0; j < m; ++j) 
            cin >> arr[i][j];
    l = n * m;
    i = j = f = 0;
    for (int _ = 0, ti, tj; _ < l; ++_) {
        cout << arr[i][j] << ' ', vis[i][j] = 1;
        ti = i + dir[f][0], tj = j + dir[f][1];
        if (ti < 0 || ti >= n || tj < 0 || tj >= m || vis[ti][tj])
            f = (f + 1) % 4;
        i += dir[f][0], j += dir[f][1];
    }
}

#include<iostream>
using namespace std;
int A[110][110];
int main(){
	int n,m;
	cin >> n >> m;
	for (int i = 0; i < n; i ++){
		for (int j = 0; j < m; j ++){
			cin >> A[i][j]; 
		}
	}
	int a, b, c, d;
	a = 0, b = 0, c = n - 1, d = m - 1;
	while(a <= c && b <= d){
		for (int i = b; i <= d; i ++){
			cout << A[a][i] << " ";
		}
		for (int i = a + 1; i <= c; i ++){
			cout << A[i][d] << " ";
		}
		if(a == c || b == d) break;
		for (int i = d-1; i >= b; i --){
			cout << A[c][i] << " ";
		}
		for (int i = c - 1; i >= a + 1; i --){
			cout << A[i][b] << " ";
		}
		a++, b++, c--, d--;
	}
	return 0;
}