Submission #1013717

#TimeUsernameProblemLanguageResultExecution timeMemory
1013717caterpillowparentrises (BOI18_parentrises)C++17
100 / 100
27 ms24064 KiB
#include <bits/stdc++.h> using namespace std; #pragma GCC optimize("O3,unroll-loops") #pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt") using ll = long long; using pl = pair<ll, ll>; using pi = pair<int, int>; #define vt vector #define f first #define s second #define pb push_back #define all(x) x.begin(), x.end() #define size(x) ((int) (x).size()) #define FOR(i, a, b) for (int i = (a); i < (b); i++) #define ROF(i, a, b) for (int i = (b) - 1; i >= (a); i--) #define F0R(i, b) FOR (i, 0, b) #define endl '\n' const ll INF = 1e18; const int inf = 1e9; /* P = 1: "greedily open" some prefix and then "greedily close" some suffix if at some point there is a valid "turning point", then we can colour assign a first P = 2: some sequence is rgb-readable if and only if there exists a greedy prefix + suffix of assignments there cannot be two greedy prefix/suffix assignments since it would throw off the balance compute # of greedy assignments for each prefix and each suffix dp[i][j] = # of ways to greedy assign first i parenthesis such that there are j open parenthesis rdp[i][j] # of ways to greedy assign the 1-indexed suffix such that there are j closed parenthesis which is the same thing but just backwards */ const ll mod = 1e9 + 7; void add(ll& a, ll b) { a += b; a -= (a >= mod) * mod; } const ll bruh[2] = {1, -1}; void solve2() { const ll n = 300; vt<vt<ll>> dp(n + 1, vt<ll>(n + 1)); dp[0][0] = 1; F0R (i, n) { F0R (j, n) { add(dp[i + 1][j + 1], dp[i][j]); if (3 * j - i - 1 >= 0) add(dp[i + 1][j], dp[i][j]); } } int t; cin >> t; F0R (_, t) { int m; cin >> m; ll ans = 0; F0R (i, m + 1) { F0R (j, i + 1) { ll req_closed = 3 * j - 2 * i + m; if (req_closed % 3 != 0 || req_closed < 0 || req_closed > 3 * m) continue; add(ans, (dp[i][j] * dp[m - i][req_closed / 3]) % mod); } } cout << ans << endl; } } void solve1() { string st; cin >> st; int n = size(st); vt<int> seq(n); F0R (i, n) seq[i] = st[i] == '(' ? 1 : -1; vt<pi> pfx(n + 1), sfx(n + 1); F0R (i, n) { pi prv = pfx[i]; if (seq[i] == 1) prv.f++, prv.s++; else { if (prv.f == prv.s) prv.f--; else prv.s--; } if (prv.f < 0 || prv.s < 0) prv = {-inf, -inf}; pfx[i + 1] = prv; } ROF (i, 0, n) { pi prv = sfx[i + 1]; if (seq[i] == -1) prv.f++, prv.s++; else { if (prv.f == prv.s) prv.f--; else prv.s--; } if (prv.f < 0 || prv.s < 0) prv = {inf, inf}; sfx[i] = prv; } F0R (i, n + 1) { if (pfx[i].f == sfx[i].f && pfx[i].s == sfx[i].s) { // answer string out = ""; string rout = ""; int x = 0, y = 0; F0R (j, i) { if (seq[j] == 1) { x++, y++; out += 'G'; } else { if (x == y) out += 'R', x--; else out += 'B', y--; } } x = y = 0; ROF (j, i, n) { if (seq[j] == -1) { x--, y--; rout += 'G'; } else { if (x == y) rout += 'R', x++; else rout += 'B', y++; } } reverse(all(rout)); cout << out << rout << endl; return; } } cout << "impossible" << endl; } main() { cin.tie(0)->sync_with_stdio(0); int p; cin >> p; if (p == 1) { int t; cin >> t; F0R (i, t) solve1(); } else { solve2(); } }

Compilation message (stderr)

parentrises.cpp:138:1: warning: ISO C++ forbids declaration of 'main' with no type [-Wreturn-type]
  138 | main() {
      | ^~~~
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