https://www.acwing.com/problem/content/description/758/

#include<iostream>
#include<cmath>
using namespace std;
int res[100][100];
bool st[100][100];
int main(void)
{
	int n,m;
	cin>>n>>m;
	int dx[4]={-1,0,1,0},dy[4]={0,1,0,-1};
	for(int x=0,y=0,d=1,k=1;k<=n*m;k++)
	{
		res[x][y]=k;
		st[x][y]=true;//标记填过了 
		int a=x+dx[d],b=y+dy[d];
		if(a<0||a>=n||b<0||b>=m||st[a][b])
		{
			d=(d+1)%4;
			a=x+dx[d],b=y+dy[d];
		}
		x=a,y=b;
	}
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<m;j++)
		{
			cout<<res[i][j]<<' ';
		}
		puts("");
	}
	return 0;
}

容易理解的方式:
利用 left right top bottom 四个变量 来表示 这个矩形的边界。
链接：https://www.acwing.com/solution/content/8007/ 来源：AcWing

#include <iostream>

using namespace std;
const int N = 105;

int a[N][N];
int n, m;

int main() {
    cin >> n >> m;

    int left = 0, right = m - 1, top = 0, bottom = n - 1;
    int k = 1;
    while (left <= right && top <= bottom) {
        for (int i = left ; i <= right; i ++) {
            a[top][i] = k ++;
        }
        for (int i = top + 1; i <= bottom; i ++) {
            a[i][right] = k ++;
        }
        for (int i = right - 1; i >= left && top < bottom; i --) {
            a[bottom][i] = k ++;
        }
        for (int i = bottom - 1; i > top && left < right; i --) {
            a[i][left] = k ++;
        }
        left ++, right --, top ++, bottom --;
    }
    for (int i = 0; i < n; i ++) {
        for (int j = 0; j < m; j ++) {
            cout << a[i][j] << " ";
        }
        cout << endl;
    }
    return 0;
}