結果
問題 | No.2608 Divide into two |
ユーザー | ktr216 |
提出日時 | 2024-01-19 21:52:30 |
言語 | C++14 (gcc 12.3.0 + boost 1.83.0) |
結果 |
AC
|
実行時間 | 322 ms / 1,000 ms |
コード長 | 4,359 bytes |
コンパイル時間 | 2,046 ms |
コンパイル使用メモリ | 176,920 KB |
実行使用メモリ | 472,772 KB |
最終ジャッジ日時 | 2024-09-28 04:21:47 |
合計ジャッジ時間 | 3,763 ms |
ジャッジサーバーID (参考情報) |
judge4 / judge2 |
(要ログイン)
テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 314 ms
472,772 KB |
testcase_01 | AC | 322 ms
472,192 KB |
testcase_02 | AC | 298 ms
472,064 KB |
ソースコード
#include <bits/stdc++.h> using namespace std; #define double long double using ll = long long; using ull = unsigned long long; using VB = vector<bool>; using VVB = vector<VB>; using VVVB = vector<VVB>; using VC = vector<char>; using VVC = vector<VC>; using VI = vector<int>; using VVI = vector<VI>; using VVVI = vector<VVI>; using VVVVI = vector<VVVI>; using VL = vector<ll>; using VVL = vector<VL>; using VVVL = vector<VVL>; using VVVVL = vector<VVVL>; using VVVVVL = vector<VVVVL>; using VD = vector<double>; using VVD = vector<VD>; using VVVD = vector<VVD>; //using P = pair<int, int>; #define REP(i, n) for (ll i = 0; i < (int)(n); i++) #define FOR(i, a, b) for (ll i = a; i < (ll)(b); i++) #define ALL(a) (a).begin(),(a).end() constexpr int INF = 1001001001; constexpr ll LINF = 1001001001001001001ll; constexpr int DX[] = {1, 0, -1, 0}; constexpr int DY[] = {0, 1, 0, -1}; template< typename T1, typename T2> inline bool chmax(T1 &a, T2 b) {return a < b && (a = b, true); } template< typename T1, typename T2> inline bool chmin(T1 &a, T2 b) {return a > b && (a = b, true); } const ll MOD = 998244353; //const ll MOD = 1000000007; const int MAX_N = 1000000; int par[MAX_N]; int rnk[MAX_N]; int siz[MAX_N]; void init(int n) { REP(i,n) { par[i] = i; rnk[i] = 0; siz[i] = 1; } } int find(int x) { if (par[x] == x) { return x; } else { return par[x] = find(par[x]); } } void unite(int x, int y) { x = find(x); y = find(y); if (x == y) return; int s = siz[x] + siz[y]; if (rnk[x] < rnk[y]) { par[x] = y; } else { par[y] = x; if (rnk[x] == rnk[y]) rnk[x]++; } siz[find(x)] = s; } bool same(int x, int y) { return find(x) == find(y); } int size(int x) { return siz[find(x)]; } ll mod_pow(ll x, ll n, ll mod) { ll res = 1; x %= mod; while (n > 0) { if (n & 1) res = res * x % mod; x = x * x % mod; n >>= 1; } return res; } ll gcd(ll x, ll y) { if (y == 0) return x; return gcd(y, x % y); } typedef pair<ll, int> P0; struct edge { int to; ll cost; }; const int MAX_V = 20000000; //const ll LINF = 1LL<<60; int V; vector<edge> G[MAX_V]; ll D[MAX_V]; void dijkstra(ll s) { priority_queue<P0, vector<P0>, greater<P0> > que; fill(D, D + V, LINF); D[s] = 0; que.push(P0(0, s)); while (!que.empty()) { P0 p = que.top(); que.pop(); int v = p.second; if (D[v] < p.first) continue; for (edge e : G[v]) { if (D[e.to] > D[v] + e.cost) { D[e.to] = D[v] + e.cost; que.push(P0(D[e.to], e.to)); } } } } /* double EPS = 1e-10; double add(double a, double b) { if (abs(a + b) < EPS * (abs(a) + abs(b))) return 0; return a + b; } struct P { double x, y; P() {} P(double x, double y) : x(x), y(y) { } P operator + (P p) { return P(add(x, p.x), add(y, p.y)); } P operator - (P p) { return P(add(x, -p.x), add(y, -p.y)); } P operator * (double d) { return P(x * d, y * d); } double dot(P p) { return add(x * p.x, y * p.y); } double det(P p) { return add(x * p.y, -y * p.x); } }; bool on_seg(P p1, P p2, P q) { return () } P intersection(P p1, P p2, P q1, P q2) { return p1 + (p2 - p1) * ((q2 - q1).det(q1 - p1) / (q2 - q1).det(p2 - p1)); } */ /* VL f(400010, 1); ll C(ll n, ll k) { return f[n] * mod_pow(f[k], MOD - 2, MOD) % MOD * mod_pow(f[n - k], MOD - 2, MOD) % MOD; } */ bool fcomp(const VL& a, const VL& b) { if (a[0] * (b[0] + b[1]) > b[0] * (a[0] + a[1])) return true; if (a[0] * (b[0] + b[1]) < b[0] * (a[0] + a[1])) return false; return a[2] < b[2]; } int main() { ios::sync_with_stdio(false); std::cin.tie(nullptr); int t; cin >> t; while (t--) { int n; cin >> n; int s = n * (n + 1) / 2; if (s % 2) { cout << -1 << endl; continue; } VI ans(n, 0); int r = s / 2; for (int i = n; i > 0; i--) { if (r <= i) { ans[r - 1] = 1; break; } r -= i; ans[i - 1] = 1; } REP(i, n) cout << ans[i]; cout << endl; } }