제출 #58632

#제출 시각아이디문제언어결과실행 시간메모리
58632BenqNice sequence (IZhO18_sequence)C++14
15 / 100
19 ms1392 KiB
#include <bits/stdc++.h> #include <ext/pb_ds/tree_policy.hpp> #include <ext/pb_ds/assoc_container.hpp> using namespace std; using namespace __gnu_pbds; typedef long long ll; typedef long double ld; typedef complex<ld> cd; typedef pair<int, int> pi; typedef pair<ll,ll> pl; typedef pair<ld,ld> pd; typedef vector<int> vi; typedef vector<ld> vd; typedef vector<ll> vl; typedef vector<pi> vpi; typedef vector<pl> vpl; typedef vector<cd> vcd; template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>; #define FOR(i, a, b) for (int i=a; i<(b); i++) #define F0R(i, a) for (int i=0; i<(a); i++) #define FORd(i,a,b) for (int i = (b)-1; i >= a; i--) #define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--) #define sz(x) (int)(x).size() #define mp make_pair #define pb push_back #define f first #define s second #define lb lower_bound #define ub upper_bound #define all(x) x.begin(), x.end() const int MOD = 1000000007; const ll INF = 1e18; const int MX = 100001; int T; vi tri(int N, int M) { // N > M if (N % M == 0) { vi z; F0R(i,N-1) z.pb(1); return z; } array<int,3> ans = {-1,-1,-1}; for (int k = 0; k < N; k += M) { int l = max(k-(N-1),-(M-1)); int r = min(N-1,k+M-1); ans = max(ans,{r-l+1,l,r}); // k-(N-1), -(M-1) // N-1, k+M-1 } vi res; FOR(i,ans[1],ans[2]+1) { if (i % M == 0) res.pb(1); else res.pb(-1); } /*cout << N << " " << M << " " << ans.f << " " << ans.s << "\n"; exit(0);*/ pi x = {0,0}; F0R(i,M) if (res[i] == 1) x.f ++; else x.s ++; pi y = {0,0}; F0R(i,N) if (res[i] == 1) y.f ++; else y.s ++; // cout << x.f << pi z = {y.s+x.s,y.f+x.f}; F0R(i,sz(res)) { if (res[i] == 1) res[i] *= z.f; else res[i] *= z.s; } return res; } vi flip(vi z) { F0R(i,sz(z)) z[i] *= -1; return z; } vi triN(int N, vi z) { if (sz(z) < N) return z; int tmp = 0; F0R(i,N) tmp += z[i]; if (tmp > 0) z = flip(z); return z; } vi triM(int N, vi z) { if (sz(z) < N) return z; int tmp = 0; F0R(i,N) tmp += z[i]; if (tmp < 0) z = flip(z); return z; } bool testN(int M, vi z) { F0R(i,sz(z)-M+1) { int ret = 0; F0R(j,M) ret += z[i+j]; if (ret >= 0) return 0; } return 1; } bool testM(int M, vi z) { F0R(i,sz(z)-M+1) { int ret = 0; F0R(j,M) ret += z[i+j]; if (ret <= 0) return 0; } return 1; } void solve() { int N,M; cin >> N >> M; // int N = rand() % 10+1,M = rand() % 10+1; // cin >> N >> M; vi z = tri(max(N,M),min(M,N)); z = triN(N,z); z = triM(M,z); if (!testN(N,z) || !testM(M,z)) { cout << N << " " << M << "\n"; cout << sz(z) << "\n"; cout << testN(N,z) << " " << testM(M,z) << "\n"; for (int i: z) cout << i << " "; cout << "\n"; exit(0); } cout << sz(z) << "\n"; for (int i: z) cout << i << " "; cout << "\n"; } int main() { ios_base::sync_with_stdio(0); cin.tie(0); cin >> T; F0R(i,T) solve(); } /* Look for: * the exact constraints (multiple sets are too slow for n=10^6 :( ) * special cases (n=1?) * overflow (ll vs int?) * array bounds */
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...