제출 #401310

#제출 시각아이디문제언어결과실행 시간메모리
401310arwaeystoamnegTents (JOI18_tents)C++17
100 / 100
1794 ms35760 KiB
// EXPLOSION! #define _CRT_SECURE_NO_WARNINGS #include<bits/stdc++.h> #include<unordered_set> #include<unordered_map> #include<chrono> using namespace std; typedef pair<int, int> pii; typedef long long ll; typedef pair<ll, ll> pll; typedef long double ld; typedef vector<int> vi; typedef vector<ll> vll; typedef vector<pair<int, int>> vpi; typedef vector<pair<ll, ll>> vpll; #define FOR(i,a,b) for (int i = (a); i < (b); ++i) #define F0R(i,a) FOR(i,0,a) #define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i) #define R0F(i,a) ROF(i,0,a) #define trav(a,x) for (auto& a: x) #define pb push_back #define mp make_pair #define rsz resize #define sz(x) int(x.size()) #define all(x) x.begin(),x.end() #define f first #define s second #define cont continue #define endl '\n' //#define ednl '\n' #define test int testc;cin>>testc;while(testc--) #define pr(a, b) trav(x,a)cerr << x << b; cerr << endl; #define message cout << "Hello World" << endl; const int dx[4] = { 1,0,-1,0 }, dy[4] = { 0,1,0,-1 }; // for every grid problem!! const ll linf = 4000000000000000000LL; const ll inf = 1000000007;//998244353 void pv(vi a) { trav(x, a)cout << x << " "; cout << endl; }void pv(vll a) { trav(x, a)cout << x << " "; cout << endl; }void pv(vector<vi>a) { F0R(i, sz(a)) { cout << i << endl; pv(a[i]); cout << endl; } }void pv(vector<vll>a) { F0R(i, sz(a)) { cout << i << endl; pv(a[i]); }cout << endl; }void pv(vector<string>a) { trav(x, a)cout << x << endl; cout << endl; } void setIO(string s) { ios_base::sync_with_stdio(0); cin.tie(0); #ifdef arwaeystoamneg if (sz(s)) { freopen((s + ".in").c_str(), "r", stdin); if (s != "test1") freopen((s + ".out").c_str(), "w", stdout); } #endif } template<int MOD, int RT> struct mint { static const int mod = MOD; static constexpr mint rt() { return RT; } // primitive root for FFT int v; explicit operator int() const { return v; } // explicit -> don't silently convert to int mint() { v = 0; } mint(ll _v) { v = int((-MOD < _v&& _v < MOD) ? _v : _v % MOD); if (v < 0) v += MOD; } friend bool operator==(const mint& a, const mint& b) { return a.v == b.v; } friend bool operator!=(const mint& a, const mint& b) { return !(a == b); } friend bool operator<(const mint& a, const mint& b) { return a.v < b.v; } mint& operator+=(const mint& m) { if ((v += m.v) >= MOD) v -= MOD; return *this; } mint& operator-=(const mint& m) { if ((v -= m.v) < 0) v += MOD; return *this; } mint& operator*=(const mint& m) { v = int((ll)v * m.v % MOD); return *this; } mint& operator/=(const mint& m) { return (*this) *= inv(m); } friend mint power(mint a, ll p) {// MAKE SURE YOU ARE USING THE CORRECT VERSION OF POW!!!!!!!! mint ans = 1; assert(p >= 0); for (; p; p /= 2, a *= a) if (p & 1) ans *= a; return ans; } friend mint inv(const mint& a) { assert(a.v != 0); return power(a, MOD - 2); } mint operator-() const { return mint(-v); } mint& operator++() { return *this += 1; } mint& operator--() { return *this -= 1; } friend mint operator+(mint a, const mint& b) { return a += b; } friend mint operator-(mint a, const mint& b) { return a -= b; } friend mint operator*(mint a, const mint& b) { return a *= b; } friend mint operator/(mint a, const mint& b) { return a /= b; } }; typedef mint<inf, 5> mi; // 5 is primitive root for both common mods const int MAX = 3005;//3005; mi dp[MAX][MAX]; int main() { setIO(""); int n, m; cin >> n >> m; FOR(i, 1, MAX)dp[1][i] = dp[i][1] = 4 * i + (i * (i - 1)) / 2; FOR(i, 2, MAX) { FOR(j, 2, MAX) { /* mi ans = dp[i - 1][j - 1]; ans += 4 * (dp[i - 1][j - 1] + 1); ans += (j - 1) * (1 + dp[i - 1][j - 2]); ans += (i - 1) * (1 + dp[i - 2][j - 1]); // only y ans += 4 * (j - 1) * (1 + dp[i - 1][j - 2]); ans += (i - 1) * (j - 1) * (1 + dp[i - 2][j - 2]); // only x ans += 4 * (i - 1) * (1 + dp[i - 2][j - 1]); ans += (i - 1) * (j - 1) * (1 + dp[i - 2][j - 2]); // both ans += 16 * (i - 1) * (j - 1) * (1 + dp[i - 2][j - 2]); if (i >= 3) ans += 4 * (i - 1) * (j - 1) * (i - 2) * (1 + dp[i - 3][j - 2]); if (j >= 3) ans += 4 * (i - 1) * (j - 1) * (j - 2) * (1 + dp[i - 2][j - 3]); if (i >= 3 && j >= 3) ans += (i - 1) * (j - 1) * (i - 2) * (j - 2) * (1 + dp[i - 3][j - 3]); dp[i][j] = ans; */ mi ans = dp[i][j - 1]; ans += 4 * i * (dp[i - 1][j - 1] + 1); ans += i * (j - 1) * (dp[i - 1][j - 2] + 1); ans += i * (i - 1) * (dp[i - 2][j - 1] + 1) * inv((mi)2); dp[i][j] = ans; } } cout << (int)dp[n][m] << endl; }
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