CW 11.02 模拟赛 FSYo T1

题面

自出题
挂个 pdf
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算法

暴力

可能的答案只有 \(O(n^2)\) 个, 考虑每个答案 \(\rm{check}\) 是 \(O(n \log n)\) 的
总时间复杂度 \(O(n ^ 3 \log n)\)

/*O(answer * n * logn), 即 O(n^3logn) 的算法, 预期 60pts*/

/*对于每一种可能的答案, 首先对于每一个点, 计算其配对点是否存在, 若存在, 则继续, 否则退出*/
#include <bits/stdc++.h>
#define int long long
const int MAXN = 2e3 + 24;
const int MAXCNT = 4520641;

int n;
int a[MAXN], b[MAXN];
int b_copy[MAXN];
int Appear_Time[MAXN];
int num = 0;

int Possible_X[MAXCNT];
int cnt = 0;

int Ans[MAXCNT];
int lsy = 0;

class LSH
{
private:

public:
    void solve()
    {
        num = std::unique(b_copy + 1, b_copy + n + 1) - (b_copy + 1);
        for (int i = 1; i <= n; i++)
        {
            b[i] = std::lower_bound(b_copy + 1, b_copy + n + 1, b[i]) - b_copy;
            Appear_Time[b[i]]++;
        }
    }
} lsh;

class Solution
{
private:
    int Appear_Time_Copy[MAXN];
    void init()
    {
        for (int i = 1; i <= n; i++)
        {
            Appear_Time_Copy[i] = Appear_Time[i];
        }
    }
    bool check(int x)
    {
        init();
        for (int i = 1; i <= n; i++)
        {
            int Need = x ^ a[i];
            int pos = std::lower_bound(b_copy + 1, b_copy + n + 1, Need) - b_copy;
            if (b_copy[pos] != Need || Appear_Time_Copy[pos] == 0)
                return false;

            Appear_Time_Copy[pos]--;
        }
        return true;
    }

public:
    void solve()
    {
        for (int i = 1; i <= cnt; i++)
        {
            int NowX = Possible_X[i];
            if(check(NowX))
            {
                Ans[++lsy] = NowX;
            }
        }

        std::sort(Ans + 1, Ans + lsy + 1);
        lsy = std::unique(Ans + 1, Ans + lsy + 1) - (Ans + 1);
        printf("%lld\n", lsy); // 520
        for (int i = 1; i <= lsy; i++)
        {
            printf("%lld\n", Ans[i]);
        }
    }
} Sol;

signed main()
{
    //freopen("f.in", "r", stdin);
    //freopen("f.out", "w", stdout);

    scanf("%lld", &n);
    for (int i = 1; i <= n; i++)
        scanf("%lld", &a[i]);
    for (int i = 1; i <= n; i++)
        scanf("%lld", &b[i]), b_copy[i] = b[i];

    std::sort(b_copy + 1, b_copy + n + 1);

    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            Possible_X[++cnt] = a[i] ^ b[j];
        }
    }

    lsh.solve();
    Sol.solve();

    return 0;
}

/*
3
1 2 3
6 4 7

1
5
*/

/*
2
0 1
0 2

0
*/

优化

观察到符合条件的 \(x\) , 当且仅当 \(x\) 出现了 \(\geq n\) 次, 判断后复杂度变为 \(O(n ^ 2)\)
具体的, 使用 hash 存储