結果
問題 | No.1089 三変数方程式 |
ユーザー | stoq |
提出日時 | 2021-09-03 19:01:30 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
AC
|
実行時間 | 17 ms / 2,000 ms |
コード長 | 4,355 bytes |
コンパイル時間 | 4,481 ms |
コンパイル使用メモリ | 267,516 KB |
実行使用メモリ | 9,216 KB |
最終ジャッジ日時 | 2024-05-08 22:20:05 |
合計ジャッジ時間 | 5,438 ms |
ジャッジサーバーID (参考情報) |
judge4 / judge5 |
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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 | 16 ms
9,088 KB |
testcase_01 | AC | 16 ms
9,088 KB |
testcase_02 | AC | 16 ms
9,088 KB |
testcase_03 | AC | 16 ms
9,216 KB |
testcase_04 | AC | 17 ms
9,088 KB |
testcase_05 | AC | 17 ms
9,216 KB |
testcase_06 | AC | 16 ms
9,088 KB |
testcase_07 | AC | 16 ms
9,088 KB |
testcase_08 | AC | 16 ms
9,216 KB |
testcase_09 | AC | 16 ms
9,216 KB |
testcase_10 | AC | 16 ms
9,088 KB |
testcase_11 | AC | 17 ms
9,088 KB |
testcase_12 | AC | 16 ms
9,216 KB |
testcase_13 | AC | 16 ms
9,088 KB |
testcase_14 | AC | 17 ms
9,216 KB |
testcase_15 | AC | 17 ms
9,088 KB |
testcase_16 | AC | 16 ms
9,088 KB |
ソースコード
#define MOD_TYPE 1 #pragma region Macros #include <bits/stdc++.h> using namespace std; #include <atcoder/all> using namespace atcoder; #if 0 #include <boost/multiprecision/cpp_dec_float.hpp> #include <boost/multiprecision/cpp_int.hpp> using Int = boost::multiprecision::cpp_int; using lld = boost::multiprecision::cpp_dec_float_100; #endif #if 1 #pragma GCC target("avx2") #pragma GCC optimize("O3") #pragma GCC optimize("unroll-loops") #endif using ll = long long int; using ld = long double; using pii = pair<int, int>; using pll = pair<ll, ll>; using pld = pair<ld, ld>; template <typename Q_type> using smaller_queue = priority_queue<Q_type, vector<Q_type>, greater<Q_type>>; #if MOD_TYPE == 1 constexpr ll MOD = ll(1e9 + 7); #else #if MOD_TYPE == 2 constexpr ll MOD = 998244353; #else constexpr ll MOD = 1000003; #endif #endif using mint = static_modint<MOD>; constexpr int INF = (int)1e9 + 10; constexpr ll LINF = (ll)4e18; constexpr double PI = acos(-1.0); constexpr double EPS = 1e-11; constexpr int Dx[] = {0, 0, -1, 1, -1, 1, -1, 1, 0}; constexpr int Dy[] = {1, -1, 0, 0, -1, -1, 1, 1, 0}; #define REP(i, m, n) for (ll i = m; i < (ll)(n); ++i) #define rep(i, n) REP(i, 0, n) #define REPI(i, m, n) for (int i = m; i < (int)(n); ++i) #define repi(i, n) REPI(i, 0, n) #define MP make_pair #define MT make_tuple #define YES(n) cout << ((n) ? "YES" : "NO") << "\n" #define Yes(n) cout << ((n) ? "Yes" : "No") << "\n" #define possible(n) cout << ((n) ? "possible" : "impossible") << "\n" #define Possible(n) cout << ((n) ? "Possible" : "Impossible") << "\n" #define Yay(n) cout << ((n) ? "Yay!" : ":(") << "\n" #define all(v) v.begin(), v.end() #define NP(v) next_permutation(all(v)) #define dbg(x) cerr << #x << ":" << x << "\n"; #define UNIQUE(v) v.erase(unique(all(v)), v.end()) struct io_init { io_init() { cin.tie(nullptr); ios::sync_with_stdio(false); cout << setprecision(30) << setiosflags(ios::fixed); }; } io_init; template <typename T> inline bool chmin(T &a, T b) { if (a > b) { a = b; return true; } return false; } template <typename T> inline bool chmax(T &a, T b) { if (a < b) { a = b; return true; } return false; } inline ll CEIL(ll a, ll b) { return (a + b - 1) / b; } template <typename A, size_t N, typename T> inline void Fill(A (&array)[N], const T &val) { fill((T *)array, (T *)(array + N), val); } template <typename T> vector<T> compress(vector<T> &v) { vector<T> val = v; sort(all(val)), val.erase(unique(all(val)), val.end()); for (auto &&vi : v) vi = lower_bound(all(val), vi) - val.begin(); return val; } template <typename T, typename U> constexpr istream &operator>>(istream &is, pair<T, U> &p) noexcept { is >> p.first >> p.second; return is; } template <typename T, typename U> constexpr ostream &operator<<(ostream &os, pair<T, U> p) noexcept { os << p.first << " " << p.second; return os; } ostream &operator<<(ostream &os, mint m) { os << m.val(); return os; } random_device seed_gen; mt19937_64 engine(seed_gen()); struct BiCoef { vector<mint> fact_, inv_, finv_; BiCoef(int n) noexcept : fact_(n, 1), inv_(n, 1), finv_(n, 1) { fact_.assign(n, 1), inv_.assign(n, 1), finv_.assign(n, 1); for (int i = 2; i < n; i++) { fact_[i] = fact_[i - 1] * i; inv_[i] = -inv_[MOD % i] * (MOD / i); finv_[i] = finv_[i - 1] * inv_[i]; } } mint C(ll n, ll k) const noexcept { if (n < k || n < 0 || k < 0) return 0; return fact_[n] * finv_[k] * finv_[n - k]; } mint P(ll n, ll k) const noexcept { return C(n, k) * fact_[k]; } mint H(ll n, ll k) const noexcept { return C(n + k - 1, k); } mint Ch1(ll n, ll k) const noexcept { if (n < 0 || k < 0) return 0; mint res = 0; for (int i = 0; i < n; i++) res += C(n, i) * mint(n - i).pow(k) * (i & 1 ? -1 : 1); return res; } mint fact(ll n) const noexcept { if (n < 0) return 0; return fact_[n]; } mint inv(ll n) const noexcept { if (n < 0) return 0; return inv_[n]; } mint finv(ll n) const noexcept { if (n < 0) return 0; return finv_[n]; } }; BiCoef bc(500010); #pragma endregion void solve() { int n; cin >> n; int cnt = 0; for (int x = 0; x <= n; x++) { for (int y = 0; y <= n; y++) { if (n - x - y >= 0) cnt++; } } cout << cnt << "\n"; } int main() { solve(); }