Submission #46730

#TimeUsernameProblemLanguageResultExecution timeMemory
46730TransBoat (APIO16_boat)C++14
58 / 100
2023 ms12620 KiB
#include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for (int i = (a); i < (b); i++) #define repd(i, a, b) for (int i = (a) - 1; i >= b; i--) #define pb push_back #define fi first #define se second #define sz(a) ((int)a.size()) typedef long long ll; const int mod = 1e9 + 7; const int maxn = 1e3 + 5; int n, a[maxn], b[maxn]; ll precal[maxn][maxn], inv[maxn]; map<int, int> mm; ll dp[maxn][maxn], f[maxn]; vector<ll> vi; ll power_mod(ll a, ll b) { ll res = 1; while (b) { if (b % 2) res = res * a % mod; a = a * a % mod; b >>= 1; } return res; } int main() { rep(i, 1, maxn) { inv[i] = power_mod(i, mod - 2); } cin >> n; // mm[0]; rep(i, 0, n) { cin >> a[i] >> b[i]; mm[a[i] - 1]++; mm[b[i]]++; } int cnt = 0; for (auto it = mm.begin(); it != mm.end(); it++) { it->se = cnt++; vi.pb(it->fi); } rep(i, 1, sz(vi)) { int range = vi[i] - vi[i - 1]; if (range < 2) continue; rep(j, 1, n + 1) { // if (j > range) { // precal[i][j] = precal[i][j - 1]; // continue; // } ll cur1 = 1; // Cij ll cur2 = (ll) range * (range - 1) / 2; // Cirange cur2 %= mod; ll res = cur1 * cur2 % mod; int tar = min(range - 2, j - 2); rep(k, 1, tar + 1) { cur1 = cur1 * (ll) (j - k - 1) % mod * inv[k] % mod; cur2 = cur2 * (ll) (range - k - 1) % mod * inv[k + 2] % mod; res = (res + cur1 * cur2 % mod) % mod; } precal[i][j] = res; } } // cout << precal[3][3] << ' ' << precal[3][2] << endl; ll ans = 0; rep(i, 0, n) { int hi = mm[b[i]] + 1; int lo = mm[a[i] - 1] + 1; dp[i][0] = 1; // ntc repd(k, hi, lo) { ll ways = (vi[k] - vi[k - 1]) * (i ? dp[i - 1][k - 1] : 1) % mod; int tot = 1; repd(j, i, 0) { if (b[j] < vi[k - 1] + 1 || a[j] > vi[k]) continue; tot++; // cout << i << ' ' << k << ' ' << vi[k] - vi[k - 1] << ' ' << tot << endl; ways += (((j ? dp[j - 1][k - 1] : 1) * precal[k][tot]) % mod); if (precal[k][tot] == 0) break; ways %= mod; } // cout << i << ' ' << k << ' ' << ways << endl; dp[i][k] = ways; } rep(k, 1, cnt) { dp[i][k] = (dp[i][k] + f[k]) % mod; // if (i == 2) cout << f[k] << endl; f[k] = dp[i][k]; dp[i][k] = (dp[i][k] + dp[i][k - 1]) % mod; ans = dp[i][k]; } } // cout << ans << endl; cout << ((ans - 1 + mod) % mod) << '\n'; return 0; }
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