Codeforces Round #719 (Div. 3)C. Not Adjacent Matrix
给数字n, 求一个n x n的矩阵, 每两个相邻个子的大小之差不能是1.
先放单数1 3 5 7, 再放2 4 6 8 即可.
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 |
#include "bits/stdc++.h" using namespace std; // fast read const auto fr = [](){ std::ios_base::sync_with_stdio(0); std::cin.tie(0); std::cout << std::fixed << std::setprecision(12); return 1; }(); template<typename A> ostream& operator<<(ostream &cout, vector<A> const &v); template<typename A, typename B> ostream& operator<<(ostream &cout, pair<A, B> const &p) { return cout << "(" << p.first << ", " << p.second << ")"; }; template<typename A> ostream& operator<<(ostream &cout, vector<A> const &v) { cout << "["; for(int i = 0; i < v.size(); i++) {if (i) cout << ", "; cout << v[i];} return cout << "]"; } template<typename A, typename B> istream& operator>>(istream& cin, pair<A, B> &p) { cin >> p.first; return cin >> p.second; } // vars: using ll = long long; using ull = unsigned long long; using ld = long double; using vi = std::vector<int>; using vl = std::vector<ll>; using vvi = std::vector<vi>; using vvl = std::vector<vl>; using pii = std::pair<int,int>; using pil = std::pair<int,ll>; using pli = std::pair<ll,int>; using pll = std::pair<ll,ll>; using vpii = std::vector<pii>; using vvpii = std::vector<vpii>; // consts ll M = 0; // ksm (kuai su mi) ll ksm(ll a,ll p){ll res=1;while(p){if(p&1){res=res*a%M;}a=a*a%M;p>>=1;}return res;} int main() { fr; int T; cin >> T; while (T--) { int N; cin >> N; if(N == 1) { cout << 1 << endl; continue; } if(N == 2) { cout << -1 << endl; continue; } int count = 1; for (int i = 1; i <= N * N ; i+=2) { cout << i << " "; if((count ) % N == 0) { cout << "\n"; } count++; } for (int i = 2; i <= N * N ; i+=2) { cout << i << " "; if((count ) % N == 0) { cout << "\n"; } count++; } } return 0; } |