제출 #467122

#제출 시각아이디문제언어결과실행 시간메모리
467122wind_reaperMoney (IZhO17_money)C++17
0 / 100
0 ms204 KiB
#include <bits/stdc++.h> #include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> using namespace std; using namespace __gnu_pbds; using namespace chrono; #define dbg(x) cout << "[" << #x << ' ' << x << "] "; // mt19937 rng((int) std::chrono::steady_clock::now().time_since_epoch().count()); template <class T> using ordered_set = tree <T, null_type, less <T>, rb_tree_tag, tree_order_statistics_node_update>; //***************** CONSTANTS ***************** const int MXN = 1'00; const int MXK = 20; //***************** GLOBAL VARIABLES ***************** int A[MXN], diff[MXN], N; //***************** AUXILIARY STRUCTS ***************** template<typename T> struct SparseTable{ int st[MXK][MXN]; int log[MXN]; function<T(const T&,const T&)> merge; SparseTable(const function<T(const T&, const T&)>& _merge){ merge = _merge; log[0] = log[1] = 0; for(int i = 2; i <= N; i++) log[i] = log[i>>1] + 1; for(int i = 1; i < N; i++) st[0][i] = diff[i]; for(int j = 1; j < MXK; j++){ for(int i = 1; i + (1 << j) <= N; i++) st[j][i] = merge(st[j-1][i], st[j-1][i + (1 << (j-1))]); } } int query(int L, int R){ int j = log[R - L + 1]; return merge(st[j][L], st[j][R - (1 << j) + 1]); } }; //***************** MAIN BODY ***************** void solve(){ cin >> N; for(int i = 1; i <= N; i++){ cin >> A[i]; diff[i-1] = A[i] - A[i-1]; } SparseTable<int> st([](int a, int b){return min(a, b); }); function<int(const int&)> find = [&](const int& i) -> int { int lo = i, hi = N - 1, ans = i; while(lo <= hi){ int mid = (lo + hi) >> 1; if(st.query(i, mid) >= 0) ans = mid, lo = mid + 1; else hi = mid - 1; } return ans + 1; }; // find the min j such that A[j...i] is non decreasing, use a sparse table on a diff array to find minimum j such that min of \ diff is >= 0 // then use pbds to find the number of segments this needs to be broken into ordered_set<pair<int, int>> s; // overall approach is nlogn int ans = 0; for(int i = 1; i <= N;){ int j = find(i); int inc = s.order_of_key({A[j], j}) - s.order_of_key({A[i], -1}) + 1; ans += inc; // dbg(j) dbg(i) dbg(ans) dbg(inc) dbg(s.order_of_key({A[j], j})) dbg(s.order_of_key({A[i], i})) cout << '\n'; for(; i <= j; i++) s.insert({A[i], i}); } cout << ans << '\n'; } //***************** ***************** int32_t main(){ ios_base::sync_with_stdio(NULL); cin.tie(NULL); #ifdef LOCAL auto begin = high_resolution_clock::now(); #endif int tc = 1; // cin >> tc; for (int t = 0; t < tc; t++) solve(); #ifdef LOCAL auto end = high_resolution_clock::now(); cout << fixed << setprecision(4); cout << "Execution Time: " << duration_cast<duration<double>>(end - begin).count() << "seconds" << endl; #endif return 0; } /* If code gives a WA, check for the following : 1. I/O format 2. Are you clearing all global variables in between tests if multitests are a thing 3. Can you definitively prove the logic 4. If the code gives an inexplicable TLE, and you are sure you have the best possible complexity, use faster io */ /* dp[i] -> minimum no. of subarrays to break into to finish this job if a subsegment is purely increasing in nature, the number of 3 6 -> 3 3 7 use an ordered set */

컴파일 시 표준 에러 (stderr) 메시지

money.cpp:81:2: warning: multi-line comment [-Wcomment]
   81 |  // find the min j such that A[j...i] is non decreasing, use a sparse table on a diff array to find minimum j such that min of \
      |  ^
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