結果
問題 | No.217 魔方陣を作ろう |
ユーザー |
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提出日時 | 2015-05-26 23:09:10 |
言語 | C++11 (gcc 13.3.0) |
結果 |
WA
|
実行時間 | - |
コード長 | 2,164 bytes |
コンパイル時間 | 768 ms |
コンパイル使用メモリ | 85,568 KB |
実行使用メモリ | 6,944 KB |
最終ジャッジ日時 | 2024-07-06 10:18:57 |
合計ジャッジ時間 | 1,597 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge5 |
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ファイルパターン | 結果 |
---|---|
other | AC * 15 WA * 3 |
コンパイルメッセージ
main.cpp: In function ‘int main()’: main.cpp:92:8: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result] 92 | scanf("%d",&n); | ~~~~~^~~~~~~~~
ソースコード
#include <cstdio>#include <algorithm>#include <stack>#include <queue>#include <deque>#include <vector>#include <string>#include <string.h>#include <cstdlib>#include <ctime>#include <cmath>#include <map>#include <set>#include <iostream>#include <sstream>#include <numeric>#include <cctype>#define fi first#define se second#define rep(i,n) for(int i = 0; i < n; ++i)#define rrep(i,n) for(int i = 1; i <= n; ++i)#define drep(i,n) for(int i = n-1; i >= 0; --i)#define gep(i,g,j) for(int i = g.head[j]; i != -1; i = g.e[i].next)#define each(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)#define rng(a) a.begin(),a.end()#define maxs(x,y) x = max(x,y)#define mins(x,y) x = min(x,y)#define pb push_back#define sz(x) (int)(x).size()#define pcnt __builtin_popcount#define snuke srand((unsigned)clock()+(unsigned)time(NULL));using namespace std;typedef long long int ll;typedef pair<int,int> P;typedef vector<int> vi;typedef vector<vi> vvi;inline int in() { int x; scanf("%d",&x); return x;}inline void priv(vi a) { rep(i,sz(a)) printf("%2d%c",a[i],i==sz(a)-1?'\n':' ');}const int MX = 100005, INF = 1000010000;const ll LINF = 1000000000000000000ll;const double eps = 1e-10;void odd(vvi& a, int n) {rep(i,n)rep(j,n) {int x = n/2-i+j, y = j*2-x;x = (x+n)%n;y = (y+n)%n;a[y][x] = i*n+j+1;}}const int d[3][2][2] = {{{4,1},{2,3}},{{1,4},{2,3}},{{1,4},{3,2}}};void even(vvi& a, int n) {vvi b(n/2,vi(n/2));odd(b,n/2);rep(i,n/2)rep(j,n/2) {b[i][j] = (b[i][j]-1)*4;int p = 0;if (i == n/2-1) p = 2;if (i == n/2-2) p = 1;if (i == n/2-2 && j == n/2/2) p = 0;if (i == n/2-3 && j == n/2/2) p = 1;rep(ai,2)rep(aj,2) {a[i*2+ai][j*2+aj] = b[i][j]+d[p][ai][aj];}}}void four(vvi& a, int n) {rep(i,n)rep(j,n) {if ((i-j+n)%4 == 0 || (i+j)%4 == 3) a[i][j] = i*n+j+1;else a[i][j] = n*n-(i*n+j);}}vvi gen(int n) {vvi a(n, vi(n));if (n&1) odd(a,n);else if (n%4) even(a,n);else four(a,n);return a;}int main() {int n;scanf("%d",&n);vvi a = gen(n);rep(i,n) priv(a[i]);return 0;}