今晚ABC436
2025-12-13 21:08:58
发布于：江西
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#include <bits/stdc++.h> using namespace std; int a[105][105]; int main() { int n; cin >> n; a[0][(n - 1) / 2] = 1; int r = 0, c = (n - 1) / 2, k = 1; for (int i = 0; i < n * n - 1; i++) { if (a[(r - 1) % n][(c + 1) % n] == 0) { a[(r - 1) % n][(c + 1) % n] = k + 1; r = (r - 1) % n; c = (c + 1) % n; k = k + 1; } else { a[(r + 1) % n][c] = k + 1; r = (r + 1) % n; k = k + 1; } } for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { cout << a[i][j] << " "; } cout << endl; } return 0; }2025-12-14 来自 江西
0主要问题是负数下标访问，把所有在
[]里的全拿出来先算再填就行AC Code：
#include<bits/stdc++.h> using namespace std; int a[105][105]; int main(){ int n; cin >> n; int r = 0; int c = (n - 1) / 2; a[r][c] = 1; for(int k = 2;k <= n * n;k++){ int nr = r - 1; int nc = c + 1; if(nr < 0) nr += n; if(nc >= n) nc -= n; if(a[nr][nc] == 0){ r = nr; c = nc; }else{ r = r + 1; if(r >= n) r -= n; } a[r][c] = k; } for(int i = 0;i < n;i++){ for(int j = 0;j < n;j++){ if(j) cout << " "; cout << a[i][j]; } cout << "\n"; } return 0; }2025-12-14 来自 北京
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