Submission #1248309

#TimeUsernameProblemLanguageResultExecution timeMemory
1248309ll931110Skyscraper (JOI16_skyscraper)C++20
15 / 100
1 ms840 KiB
#ifdef ONLINE_JUDGE #include <bits/stdc++.h> #endif #include <algorithm> #include <cstdio> #include <cstring> #include <iostream> #include <set> #include <stack> #include <map> #include <queue> #include <vector> #define maxn 105 #define maxk 1005 using namespace std; template <unsigned M_> struct ModInt { static constexpr unsigned M = M_; unsigned x; constexpr ModInt() : x(0U) {} constexpr ModInt(unsigned x_) : x(x_ % M) {} constexpr ModInt(unsigned long long x_) : x(x_ % M) {} constexpr ModInt(int x_) : x(((x_ %= static_cast<int>(M)) < 0) ? (x_ + static_cast<int>(M)) : x_) {} constexpr ModInt(long long x_) : x(((x_ %= static_cast<long long>(M)) < 0) ? (x_ + static_cast<long long>(M)) : x_) {} ModInt &operator+=(const ModInt &a) { x = ((x += a.x) >= M) ? (x - M) : x; return *this; } ModInt &operator-=(const ModInt &a) { x = ((x -= a.x) >= M) ? (x + M) : x; return *this; } ModInt &operator*=(const ModInt &a) { x = (static_cast<unsigned long long>(x) * a.x) % M; return *this; } ModInt &operator/=(const ModInt &a) { return (*this *= a.inv()); } ModInt pow(long long e) const { if (e < 0) return inv().pow(-e); ModInt a = *this, b = 1U; for (; e; e >>= 1) { if (e & 1) b *= a; a *= a; } return b; } ModInt inv() const { unsigned a = M, b = x; int y = 0, z = 1; for (; b; ) { const unsigned q = a / b; const unsigned c = a - q * b; a = b; b = c; const int w = y - static_cast<int>(q) * z; y = z; z = w; } return ModInt(y); } ModInt operator+() const { return *this; } ModInt operator-() const { ModInt a; a.x = x ? (M - x) : 0U; return a; } ModInt operator+(const ModInt &a) const { return (ModInt(*this) += a); } ModInt operator-(const ModInt &a) const { return (ModInt(*this) -= a); } ModInt operator*(const ModInt &a) const { return (ModInt(*this) *= a); } ModInt operator/(const ModInt &a) const { return (ModInt(*this) /= a); } template <class T> friend ModInt operator+(T a, const ModInt &b) { return (ModInt(a) += b); } template <class T> friend ModInt operator-(T a, const ModInt &b) { return (ModInt(a) -= b); } template <class T> friend ModInt operator*(T a, const ModInt &b) { return (ModInt(a) *= b); } template <class T> friend ModInt operator/(T a, const ModInt &b) { return (ModInt(a) /= b); } explicit operator bool() const { return x; } bool operator==(const ModInt &a) const { return (x == a.x); } bool operator!=(const ModInt &a) const { return (x != a.x); } friend std::ostream &operator<<(std::ostream &os, const ModInt &a) { return os << a.x; } }; constexpr unsigned MOD = 1'000'000'007; using Mint = ModInt<MOD>; Mint dp[maxn][maxn][maxk][3]; int n, K; int a[maxn]; int inf = 1e6; Mint solve() { // dp[i][j][k][m] = number of ways to fill i numbers, there are j segments, diff of k and l ends have been filled so far. a[n + 1] = inf; dp[0][0][0][0] = 1; for (int i = 1; i <= n; i++) { for (int j = 1; j <= i; j++) { for (int k = 0; k <= K; k++) { for (int m = 0; m <= 2; m++) { int cost_diff = (2 * j - m) * (a[i + 1] - a[i]); if (cost_diff > k) { continue; } // Case 1: put a[i] in a new segment dp[i][j][k][m] += dp[i - 1][j - 1][k - cost_diff][m]; // Case 2: put a[i] in a new segment at the endpoint if (m > 0) { dp[i][j][k][m] += (3 - m) * dp[i - 1][j - 1][k - cost_diff][m - 1]; } // Case 3: extending a[i] from a segment, not at endpoint dp[i][j][k][m] += (2 * j - m) * dp[i - 1][j][k - cost_diff][m]; // Case 4: extending a[i] from a segment, using endpoint if (m == 1) { dp[i][j][k][m] += 2 * j * dp[i - 1][j][k - cost_diff][m - 1]; } else if (m == 2) { if (i == n) { dp[i][j][k][m] += dp[i - 1][j][k - cost_diff][m - 1]; } else if (j > 1) { dp[i][j][k][m] += (j - 1) * dp[i - 1][j][k - cost_diff][m - 1]; } } // Case 5: join two segments together if (m == 2) { if (i == n) { dp[i][j][k][m] += dp[i - 1][j + 1][k - cost_diff][m]; } else { dp[i][j][k][m] += j * (j - 1) * dp[i - 1][j + 1][k - cost_diff][m]; } } else if (m == 1) { dp[i][j][k][m] += j * j * dp[i - 1][j + 1][k - cost_diff][m]; } else { dp[i][j][k][m] += j * (j + 1) * dp[i - 1][j + 1][k - cost_diff][m]; } } } } } Mint ans = 0; for (int k = 0; k <= K; k++) { ans += dp[n][1][k][2]; } return ans; } int main() { ios_base::sync_with_stdio(false); cin.tie(0); cin >> n >> K; for (int i = 1; i <= n; i++) { cin >> a[i]; } sort(a + 1, a + n + 1); cout << solve() << endl; return 0; }
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