結果
問題 |
No.1186 長方形の敷き詰め
|
ユーザー |
|
提出日時 | 2020-08-22 13:26:19 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 10 ms / 2,000 ms |
コード長 | 2,064 bytes |
コンパイル時間 | 1,890 ms |
コンパイル使用メモリ | 98,496 KB |
最終ジャッジ日時 | 2025-01-13 07:34:54 |
ジャッジサーバーID (参考情報) |
judge1 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 24 |
ソースコード
#include <iostream> #include <algorithm> #include <map> #include <set> #include <queue> #include <stack> #include <numeric> #include <bitset> #include <cmath> static const int MOD = 998244353; using ll = long long; using u32 = unsigned; using u64 = unsigned long long; using namespace std; template<class T> constexpr T INF = ::numeric_limits<T>::max()/32*15+208; template<u32 M = 1000000007> struct modint{ u32 val; modint(): val(0){} template<typename T> modint(T t){t %= (T)M; if(t < 0) t += (T)M; val = t;} modint pow(ll k) const { modint res(1), x(val); while(k){ if(k&1) res *= x; x *= x; k >>= 1; } return res; } template<typename T> modint& operator=(T t){t %= (T)M; if(t < 0) t += (T)M; val = t; return *this;} modint inv() const {return pow(M-2);} modint& operator+=(modint a){val += a.val; if(val >= M) val -= M; return *this;} modint& operator-=(modint a){if(val < a.val) val += M-a.val; else val -= a.val; return *this;} modint& operator*=(modint a){val = (u64)val*a.val%M; return *this;} modint& operator/=(modint a){return (*this) *= a.inv();} modint operator+(modint a) const {return modint(val) +=a;} modint operator-(modint a) const {return modint(val) -=a;} modint operator*(modint a) const {return modint(val) *=a;} modint operator/(modint a) const {return modint(val) /=a;} modint operator-(){return modint(M-val);} bool operator==(const modint a) const {return val == a.val;} bool operator!=(const modint a) const {return val != a.val;} bool operator<(const modint a) const {return val < a.val;} }; using mint = modint<MOD>; int main() { int n, m; cin >> n >> m; if(n > m) { puts("1"); return 0; }else if(n == 1){ puts("1"); return 0; } vector<mint> dp(m+1); dp[0] = 1; for (int i = 0; i < m; ++i) { dp[i+1] += dp[i]; if(i+n <= m) dp[i+n] += dp[i]; } cout << dp.back().val << "\n"; return 0; }