結果
問題 | No.2878 𝕀𝔾ℕ𝕀𝕋𝕀𝕆ℕ |
ユーザー | ktr216 |
提出日時 | 2024-09-08 12:31:33 |
言語 | C++14 (gcc 12.3.0 + boost 1.83.0) |
結果 |
AC
|
実行時間 | 5 ms / 2,000 ms |
コード長 | 4,317 bytes |
コンパイル時間 | 1,708 ms |
コンパイル使用メモリ | 178,192 KB |
実行使用メモリ | 11,588 KB |
最終ジャッジ日時 | 2024-09-08 12:31:35 |
合計ジャッジ時間 | 2,221 ms |
ジャッジサーバーID (参考情報) |
judge4 / judge3 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 4 ms
9,596 KB |
testcase_01 | AC | 5 ms
11,588 KB |
testcase_02 | AC | 4 ms
9,732 KB |
testcase_03 | AC | 4 ms
11,588 KB |
testcase_04 | AC | 4 ms
11,588 KB |
testcase_05 | AC | 3 ms
9,544 KB |
testcase_06 | AC | 4 ms
11,588 KB |
testcase_07 | AC | 4 ms
11,584 KB |
ソースコード
#include <bits/stdc++.h> using namespace std; #define double long double using ll = long long; using ull = unsigned long long; using VB = vector<bool>; using VVB = vector<VB>; using VVVB = vector<VVB>; using VC = vector<char>; using VVC = vector<VC>; using VI = vector<int>; using VVI = vector<VI>; using VVVI = vector<VVI>; using VVVVI = vector<VVVI>; using VL = vector<ll>; using VVL = vector<VL>; using VVVL = vector<VVL>; using VVVVL = vector<VVVL>; using VVVVVL = vector<VVVVL>; using VD = vector<double>; using VVD = vector<VD>; using VVVD = vector<VVD>; using VT = vector<string>; using VVT = vector<VT>; //using P = pair<int, int>; #define REP(i, n) for (ll i = 0; i < (ll)(n); i++) #define FOR(i, a, b) for (ll i = a; i < (ll)(b); i++) #define ALL(a) (a).begin(),(a).end() constexpr int INF = 1001001001; constexpr ll LINF = 1001001001001001001ll; //constexpr ll LINF = 8e18; //constexpr int DX[] = {-1, -1, 0, 0, 1, 1}; //constexpr int DY[] = {-1, 0, -1, 1, 0, 1}; constexpr int DX[] = {1, 0, -1, 0, 0, 0}; constexpr int DY[] = {0, 1, 0, -1, 0, 0}; constexpr int DZ[] = {0, 0, 0, 0, 1, -1}; template< typename T1, typename T2> inline bool chmax(T1 &a, T2 b) {return a < b && (a = b, true); } template< typename T1, typename T2> inline bool chmin(T1 &a, T2 b) {return a > b && (a = b, true); } const ll MOD = 998244353; //const ll MOD = 1000000007; const int MAX_N = 1000000; int par[MAX_N]; int rnk[MAX_N]; int siz[MAX_N]; void init(int n) { REP(i,n) { par[i] = i; rnk[i] = 0; siz[i] = 1; } } int find(int x) { if (par[x] == x) { return x; } else { return par[x] = find(par[x]); } } void unite(int x, int y) { x = find(x); y = find(y); if (x == y) return; int s = siz[x] + siz[y]; if (rnk[x] < rnk[y]) { par[x] = y; } else { par[y] = x; if (rnk[x] == rnk[y]) rnk[x]++; } siz[find(x)] = s; } bool same(int x, int y) { return find(x) == find(y); } int size(int x) { return siz[find(x)]; } ll mod_pow(ll x, ll n, ll mod) { ll res = 1; x %= mod; while (n > 0) { if (n & 1) res = res * x % mod; x = x * x % mod; n >>= 1; } return res; } ll gcd(ll x, ll y) { if (y == 0) return x; return gcd(y, x % y); } typedef pair<ll, int> P0; struct edge { int to; ll cost; }; const int MAX_V = 250050; //const ll LINF = 1LL<<60; int V; vector<edge> G[MAX_V]; ll D[MAX_V]; void dijkstra(ll s) { priority_queue<P0, vector<P0>, greater<P0> > que; fill(D, D + V, LINF); D[s] = 0; que.push(P0(0, s)); while (!que.empty()) { P0 p = que.top(); que.pop(); int v = p.second; if (D[v] < p.first) continue; for (edge e : G[v]) { if (D[e.to] > D[v] + e.cost) { D[e.to] = D[v] + e.cost; que.push(P0(D[e.to], e.to)); } } } } /* double EPS = 1e-10; double add(double a, double b) { if (abs(a + b) < EPS * (abs(a) + abs(b))) return 0; return a + b; } struct P { double x, y; P() {} P(double x, double y) : x(x), y(y) { } P operator + (P p) { return P(add(x, p.x), add(y, p.y)); } P operator - (P p) { return P(add(x, -p.x), add(y, -p.y)); } P operator * (double d) { return P(x * d, y * d); } double dot(P p) { return add(x * p.x, y * p.y); } double det(P p) { return add(x * p.y, -y * p.x); } }; bool on_seg(P p1, P p2, P q) { return () } P intersection(P p1, P p2, P q1, P q2) { return p1 + (p2 - p1) * ((q2 - q1).det(q1 - p1) / (q2 - q1).det(p2 - p1)); } */ /* VL f(400010, 1); ll C(ll n, ll k) { return f[n] * mod_pow(f[k], MOD - 2, MOD) % MOD * mod_pow(f[n - k], MOD - 2, MOD) % MOD; } */ bool fcomp(VL x, VL y) { return x[1] * y[0] < x[0] * y[1]; } // 拡張ユークリッドの互除法の実装例 // ax + by = \pm gcd(a, b) となるx,yを返す pair<long long, long long> extgcd(long long a, long long b) { if (b == 0) return make_pair(1, 0); long long x, y; tie(y, x) = extgcd(b, a % b); y -= a / b * x; return make_pair(x, y); } int main() { ios::sync_with_stdio(false); std::cin.tie(nullptr); int i; cin >> i; cout << "IGNITION"[i - 1] << endl; }