이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#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
*/
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