Submission #440971

#	Time	Username	Problem	Language	Result	Execution time	Memory
440971		zxcvbnm	XOR Sum (info1cup17_xorsum)	C++14	100 / 100	802 ms	31508 KiB

xorsum

#include<bits/stdc++.h>
using namespace std;
using ll = long long;
const int L = 29;
vector<int> p = {1};

vector<int> merge(vector<int> a, int pw) {
    vector<int> b, c;
    int n = a.size();
    int id = n;
    for(int i = 0; i < n; i++) {
        if (a[i] >= p[pw+1]) {
            id = i;
            break;
        }
    }
    for(int i = id; i < n; i++) {
        a[i] -= p[pw+1];
        b.push_back(a[i]);
    }
    for(int i = id; i < n; i++) {
        a.pop_back();
    }

    // sudedame du surikiuotus masyvus i viena, islaikydami nemazejancia tvarka
    int id1 = 0, id2 = 0;
    n = a.size();
    int m = b.size();
    while(id1 < n && id2 < m) {
        // visada galime deti mazesni elementa nesugadindami tvarkos
        if (a[id1] < b[id2]) {
            c.push_back(a[id1++]);
        } else {
            c.push_back(b[id2++]);
        }
    }

    // jei dar liko elementu is a
    while(id1 < n) {
        c.push_back(a[id1++]);
    }

    // jei dar liko elementu is b
    while(id2 < m) {
        c.push_back(b[id2++]);
    }
    return c;
}
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    cin >> n;
    vector<int> a(n);
    for(int i = 0; i < n; i++) {
        cin >> a[i];
    }

    for(int i = 1; i <= L+1; i++) {
        p.push_back(p[i-1]*2);
    }

    vector<int> cnt(L+1, 0);
    vector<int> x(n);

    for(int j = L; j >= 0; j--) {
        if (j == L) {
            for(int i = 0; i < n; i++) {
                x[i] = (a[i] % p[j+1]);
            }
            sort(x.begin(), x.end());
        } else {
            x = merge(x, j);
        }

        int pos1 = n, pos2 = n, pos3 = n;
        for(int i = 0; i < n; i++) {
            pos1 = max(pos1, i+1);
            pos2 = max(pos2, i+1);
            pos3 = max(pos3, i+1);

            while(pos1 >= i+1 && 1*p[j]-x[i] <= x[pos1-1]) pos1--;
            while(pos2 >= i+1 && 2*p[j]-x[i] <= x[pos2-1]) pos2--;
            while(pos3 >= i+1 && 3*p[j]-x[i] <= x[pos3-1]) pos3--;

            int curr = (pos2 - pos1) + (n - pos3);
            cnt[j] += curr;

        }
    }

    ll ans = 0;
    for(int i = 0; i <= L; i++) {
//        cout << cnt[i] << " ";
        if (cnt[i] % 2) {
            ans += p[i];
        }
    }
    cout << ans << "\n";
    return 0;
}

• Subtask 1: 7 / 7
• Subtask 2: 11 / 11
• Subtask 3: 21 / 21
• Subtask 4: 38 / 38
• Subtask 5: 23 / 23