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답안 #319112

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
319112 2020-11-04T01:08:47 Z mohamedsobhi777 XOR Sum (info1cup17_xorsum) C++14
0 / 100
 1600 ms 23900 KB 
#include <bits/stdc++.h>

#pragma GCC optimize("-Ofast")
//#pragma GCC optimize("trapv")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,sse4.2,popcnt,abm,mmx,avx2,tune=native")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-funroll-loops")

#define I inline void
#define S struct
#define vi vector<int>
#define vii vector<pair<int, int>>
#define pii pair<int, int>
#define pll pair<ll, ll>

using namespace std;
using ll = long long;
using ld = long double;

const int N = 2e6 + 7, mod = 1e9 + 7;
const ll inf = 2e18;

// How interesting!

int n;
int a[N];
ll ans;
int cnts[N];

int bit[N];

void add(int x)
{
        ++x;
        for (; x < N; x += x & -x)
                bit[x]++;
}

int get(int x)
{
        if (x < 0)
                return 0;
        ++x;
        int ret = 0;
        for (; x; x -= x & -x)
                ret += bit[x];
        return ret;
}

int main()
{
        ios_base::sync_with_stdio(0);
        cin.tie(0);
        //freopen("in.in", "r", stdin);

        cin >> n;
        for (int i = 0; i < n; ++i)
                cin >> a[i];

        for (int r = 0; r < 20; ++r)
        {
                memset(bit, 0, sizeof bit);
                int R = (1 << r) * 2;
                vector<int> vec;

                for (int j = 0; j < n; ++j)
                {
                        add(a[j] % R);
                        vec.push_back(a[j] % R);
                        int le = (1 << r), ri = R - 1;

                        int z = a[j] % R;

                        if (z <= (1 << r))
                        {
                                le -= z;
                                ri -= z;
                                int m = get(ri) - get(le - 1);
                                if (m%2)
                                        ans ^= (1 << r);
                        }
                        else
                        {
                                int xx = le - z;
                                int yy = ri - z;

                                xx = (xx % R + R) % R;
                                yy = (yy % R + R) % R;
                                if (xx > yy)
                                        swap(xx, yy);
                                int m = j + 1 - (get(yy - 1) - get(xx));
                                if (m%2)
                                        ans ^= (1 << r);
                        }
                }
        }

        cout << ans;
        return 0;
}

/*
        - bounds sir (segtree = 4N, eulerTour = 2N, ...)
        - a variable defined twice?
        - will overflow?
        - is it a good complexity?
        - don't mess up indices (0-indexed vs 1-indexed)
        - reset everything between testcases. 
*/

Subtask #1
0 / 7.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 29 ms 8172 KB Output isn't correct
2 Halted 0 ms 0 KB -

Subtask #2
0 / 11.0

# 결과 실행 시간 메모리 Grader output
1 Execution timed out 1678 ms 23900 KB Time limit exceeded
2 Halted 0 ms 0 KB -

Subtask #3
0 / 21.0

# 결과 실행 시간 메모리 Grader output
1 Execution timed out 1678 ms 23900 KB Time limit exceeded
2 Halted 0 ms 0 KB -

Subtask #4
0 / 38.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 29 ms 8172 KB Output isn't correct
2 Halted 0 ms 0 KB -

Subtask #5
0 / 23.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 29 ms 8172 KB Output isn't correct
2 Halted 0 ms 0 KB -