제출 #62246

#제출 시각아이디문제언어결과실행 시간메모리
62246BenqXOR Sum (info1cup17_xorsum)C++11
100 / 100
1138 ms54276 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 = 1000001; int N, x; array<int,MX> A, A2; int get(int ind) { int cur = 0; if (ind < 0) cur -= 2*x, ind += N; if (ind >= N) cur += 2*x, ind -= N; return (A[ind]&(2*x-1))+cur; } void sor() { int co = 0; F0R(i,N) if (!(A[i]&x)) co ++; int co0 = 0; F0R(i,N) { if (!(A[i]&x)) A2[co0++] = A[i]; else A2[co++] = A[i]; } swap(A,A2); } ll tri() { sor(); ll tot = 0; F0R(i,N) if ((2*A[i])&x) tot ++; int i1 = -N, i2 = -N; // first one >= x-i, first one >= 2*x-i F0Rd(i,N) { int v = A[i]&(2*x-1); while (get(i1) < x-v) i1 ++; while (get(i2) < 2*x-v) i2 ++; tot += i2-i1; } return tot/2; } int main() { ios_base::sync_with_stdio(0); cin.tie(0); cin >> N; F0R(i,N) { // A[i] = rand() % (1<<29); cin >> A[i]; } int ans = 0; F0R(i,30) { x = 1<<i; if (tri()&1) ans ^= x; } cout << ans; } /* 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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