제출 #373514

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
373514		qpwoeirut	XOR Sum (info1cup17_xorsum)	C++17	77 / 100	1690 ms	73980 KiB

xorsum

//xorsum.cpp created at 03/04/21 10:30:22

#include <bits/stdc++.h>

using namespace std;

#define LB lower_bound

const int MN = 1000000;
const int LG = 24;

int N;
int A[MN];
int val[MN];
int psum[(1 << LG) + 1];

int main() {
    cin.tie(0)->sync_with_stdio(0);

    cin >> N;
    for (int i=0; i<N; ++i) {
        cin >> A[i];
    }
    sort(A, A+N);

    int ans = 0;
    for (int i=0; i<LG; ++i) {
        fill(psum, psum + (1 << (i+1)) + 1, 0);
        for (int j=0; j<N; ++j) {
            ++psum[(A[j] & ((1 << (i + 1)) - 1)) + 1]; // add 1 for easy indexing stuff
        }
        for (int j=1; j<=(1 << (i + 1)); ++j) {
            psum[j] += psum[j - 1];
        }

        // range [2^i, 2^(i+1)) and [2^(i+1) + 2^i, 2^(i+2))
        int ct = 0;
        for (int j=0; j<N; ++j) {
            const int cur = A[j] & ((1 << (i + 1)) - 1);
            const int lo1 = (1 << i) - cur;
            ct += psum[(1 << (i + 1)) - cur] - psum[lo1 < 0 ? 0 : lo1] +
                  psum[min(1 << (i + 1), (1 << (i + 2)) - cur)] - psum[min(1 << (i + 1), (1 << (i + 1)) + lo1)] +
                  (((A[j] << 1) >> i) & 1);
        }

        ans |= (i == 0) ? ((ct >> 1) & 1) : ((ct & 2) << (i-1));
    }

    for (int i=LG; i<30; ++i) {
        for (int j=0; j<N; ++j) {
            val[j] = A[j] & ((1 << (i + 1)) - 1);
        }
        sort(val, val+N);

        int ct = 0;
        for (int j=0; j<N; ++j) {
            const int cur = A[j] & ((1 << (i + 1)) - 1);
            const int range1 = LB(val, val+N, (1 << (i + 1)) - cur) - LB(val, val+N, (1 << i) - cur);
            const int range2 = LB(val, val+N, (1LL << (i + 2)) - cur) - LB(val, val+N, (1 << (i + 1)) + (1 << i) - cur);

            ct += range1 + range2 + ((A[j] >> (i-1)) & 1);
        }

        ans |= ((ct >> 1) & 1) << i;
    }

    cout << ans << '\n';
}

/*
4  
3 9 6 6 

00110   1 1
01100   1 2
01001   1 3
01001   1 4
10010   2 2
01111   2 3
01111   2 4
01100   3 3
01100   3 4
01100   4 4
*/

• Subtask 1: 7 / 7
• Subtask 2: 11 / 11
• Subtask 3: 21 / 21
• Subtask 4: 38 / 38
• Subtask 5: 0 / 23