Submission #232190

#TimeUsernameProblemLanguageResultExecution timeMemory
232190VEGAnnXOR Sum (info1cup17_xorsum)C++14
0 / 100
304 ms12152 KiB
#include <bits/stdc++.h> //#pragma GCC optimize("unroll-loops") //#pragma GCC optimize("-O3") //#pragma GCC optimize("Ofast") //#pragma GCC optimize("fast-math") //#pragma GCC optimize("no-stack-protector") #include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> #include <ext/rope> #define sz(x) int(x.size()) #define all(x) x.begin(),x.end() #define PB push_back #define MP make_pair #define pii pair<int, int> #define pll pair<ll, ll> #define pil pair<int, ll> #define pli pair<ll, int> #define pdd pair<ld, ld> #define ft first #define sd second using namespace std; using namespace __gnu_cxx; using namespace __gnu_pbds; typedef long long ll; typedef long double ld; template<class T> using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>; const int N = 1000100; const int M = 100100; const int BIG = int(5e6); const int oo = 2e9; const ll OO = 1e18; const int md = 998244353; const int PW = 30; ll sum; int a[N], ans = 0, n, mask, mn, mx, l1, r1, l2, r2; int vc[2][N], k1, k0, MASK; int main() { #ifdef _LOCAL freopen("in.txt","r",stdin); //freopen("output.txt","w",stdout); #else // freopen("mining.in","r",stdin); freopen("mining.out","w",stdout); ios_base::sync_with_stdio(0); cin.tie(0); #endif cin >> n; for (int i = 0; i < n; i++) { cin >> a[i]; vc[1][i] = a[i]; } int ite = 1; for (int bt = 0; bt < PW; bt++, ite ^= 1){ mask = (1 << bt); MASK = (1 << (bt + 1)) - 1; k1 = k0 = 0; for (int i = 0; i < n; i++) if (a[i] & mask) k1++; for (int i = 0; i < n; i++) if (vc[ite ^ 1][i] & (1 << i)) vc[ite][k1++] = vc[ite ^ 1][i]; else vc[ite][k0++] = vc[ite ^ 1][i]; sum = 0; l1 = n, r1 = n - 1, l2 = n, r2 = n - 1; for (int i = 0; i < n; i++){ int cr = (vc[ite][i] & MASK); mn = (1 << bt) - cr; mx = (1 << (bt + 1)) - 1 - cr; while (l1 > 0 && vc[ite][l1 - 1] >= mn) l1--; while (r1 >= 0 && vc[ite][r1] > mx) r1--; sum += r1 - l1 + 1; if (cr >= mn && cr <= mx) sum--; mn += (1 << (bt + 1)); mx = (1 << (bt + 2)) - 1 - cr; while (l2 > 0 && vc[ite][l2 - 1] >= mn) l2--; while (r2 >= 0 && vc[ite][r2] > mx) r2--; sum += r2 - l2 + 1; if (cr >= mn && cr <= mx) sum--; } sum /= 2; if (sum & 1) ans += (1 << bt); } for (int i = 0; i < n; i++) ans ^= (a[i] + a[i]); cout << ans; return 0; }
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