// Problem: C. Labs
// Contest: Codeforces - Codeforces Round #593 (Div. 2)
// URL: https://codeforces.com/problemset/problem/1236/C
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 2022-03-03 18:58:39
//
// Powered by CP Editor (https://cpeditor.org)
#include<bits/stdc++.h>
using namespace std;
#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair
const ll mod = 1e9 + 7;
inline ll qmi (ll a, ll b) {
ll ans = 1;
while (b) {
if (b & 1) ans = ans * a%mod;
a = a * a %mod;
b >>= 1;
}
return ans;
}
inline int read () {
int x = 0, f = 0;
char ch = getchar();
while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
return f?-x:x;
}
template<typename T> void print(T x) {
if (x < 0) putchar('-'), x = -x;
if (x >= 10) print(x/10);
putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
return (a + b) %mod;
}
inline ll inv (ll a) {
return qmi(a, mod - 2);
}
int a[550][550];
int n;
void dfs (int x, int y, int num) {
if (num == n * n +1) return;
a[x][y] = num;
if (x - 1 >= 1 && !a[x - 1][y]) {
dfs(x - 1, y, num + 1);
}
else if (x + 1 <= n && !a[x + 1][y])
dfs(x + 1, y, num + 1);
else {
dfs(x, y + 1, num + 1);
}
}
void solve() {
cin >> n;
dfs(1, 1, 1);
for (int i = 1; i<= n; i ++) {
for (int j = 1; j <= n ;j ++)
cout << a[i][j] << " ";
puts("");
}
}
int main () {
int t;
t =1;
//cin >> t;
while (t --) solve();
return 0;
}
最大值为n * n / 2 我们先研究3*3的,我们想要每一组,对于另外一组需要有最多的贡献 之前试了一种思路比较接近正解。但是还不够.我们可以观察答案找出答案具有的结构