結果

問題 No.401 数字の渦巻き
ユーザー tancahn2380
提出日時 2019-07-03 17:38:20
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 21 ms / 2,000 ms
コード長 2,214 bytes
コンパイル時間 2,248 ms
コンパイル使用メモリ 192,628 KB
最終ジャッジ日時 2025-01-07 05:52:01
ジャッジサーバーID
(参考情報)
judge3 / judge4
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ファイルパターン 結果
other AC * 30
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ソースコード

diff #
プレゼンテーションモードにする

# include "bits/stdc++.h"
using namespace std;
using LL = long long;
using ULL = unsigned long long;
const double PI = acos(-1);
template<class T>constexpr T INF() { return ::std::numeric_limits<T>::max(); }
template<class T>constexpr T HINF() { return INF<T>() / 2; }
template <typename T_char>T_char TL(T_char cX) { return tolower(cX); };
template <typename T_char>T_char TU(T_char cX) { return toupper(cX); };
typedef pair<LL, LL> pii;
const int vy[] = { -1, -1, -1, 0, 1, 1, 1, 0 }, vx[] = { -1, 0, 1, 1, 1, 0, -1, -1 };
const int dx[4] = { 0,1,0,-1 }, dy[4] = { 1,0,-1,0 };
int popcnt(unsigned long long n) { int cnt = 0; for (int i = 0; i < 64; i++)if ((n >> i) & 1)cnt++; return cnt; }
int d_sum(LL n) { int ret = 0; while (n > 0) { ret += n % 10; n /= 10; }return ret; }
int d_cnt(LL n) { int ret = 0; while (n > 0) { ret++; n /= 10; }return ret; }
LL gcd(LL a, LL b) { if (b == 0)return a; return gcd(b, a%b); };
LL lcm(LL a, LL b) { LL g = gcd(a, b); return a / g*b; };
# define ALL(qpqpq) (qpqpq).begin(),(qpqpq).end()
# define UNIQUE(wpwpw) sort(ALL((wpwpw)));(wpwpw).erase(unique(ALL((wpwpw))),(wpwpw).end())
# define LOWER(epepe) transform(ALL((epepe)),(epepe).begin(),TL<char>)
# define UPPER(rprpr) transform(ALL((rprpr)),(rprpr).begin(),TU<char>)
# define FOR(i,tptpt,ypypy) for(LL i=(tptpt);i<(ypypy);i++)
# define REP(i,upupu) FOR(i,0,upupu)
# define INIT std::ios::sync_with_stdio(false);std::cin.tie(0)
int n;
int ans[33][33];
void f(int num){
if(num < 10)cout << "00" << num;
else if(num < 100)cout << "0" << num;
else cout << num;
return;
}
int main(){
cin >> n;
int x = 1, y = 1;
int bx = 1, by = 0;
int idx = 1;
while(idx <= n * n){
ans[y][x] = idx;
idx++;
int nx = x + bx, ny = y + by;
if(nx < 1 || nx > n || ny < 1 || ny > n || ans[ny][nx] != 0){
if(bx == 1 && by == 0){
bx = 0;
by = 1;
}else if(bx == 0 && by == 1){
bx = -1;
by = 0;
}else if(bx == -1 && by == 0){
bx = 0;
by = -1;
}else{
bx = 1;
by = 0;
}
}
x += bx;
y += by;
}
for(int i = 1;i <= n;i++){
for(int j = 1;j <= n;j++){
if(j != 1)cout << " ";
f(ans[i][j]);
}
cout << endl;
}
}
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