P6824 「EZEC-4」可乐 题解

题目链接：可乐

一开始想着 0-1 Trie，枚举 \(x\) 去写，然后判断就行了。然后想起南京区域赛的 C 题，其实和这个也有点大同小异的感觉，可以用更朴素的办法，找到对于一个 \(a_i\) 而言，满足题意的所有 \(x\) 去 \(+1\)。这玩意很容易办到的，稍微讨论下：

类似 0-1 Trie 的按位讨论，从高位开始，我们知道如果 \(x\) 与 \(a_i\) 的某一位相同，就取 \(0\) 否则取 \(1\)。然后注意到 \(k\) 的某一位只有 \(0\) 或者 \(1\) 情况，所以当为 \(0\) 时，显然 \(x==a_i\) 这一位情况，并且情况唯一。当为 \(1\) 时，如果让异或结果为 \(0\) 则左半 \(01\) 异或树上的是所有数都能取到，很显然这个范围为 \([\ curr,curr+(1<<i)\ )\)。其实和数位 dp 或者说试填法的思想是一样的，不懂的画出 \(01\) Trie 自行理解就行。其中 \(curr\) 为使 \(a_i \oplus curr=0\)。当然另一边我们就直接让 \(x\) 带上可以使 \(a_i \oplus x=1\) 的那个数，比如 \(a_i\) 这一位为 \(1\)，那么 \(x\) 这一位就该为 \(0\)，反过来就该为 \(1\)。由于贡献的数是连续的很明显，所以我们可以直接差分区间，然后最后前缀和恢复，统计最大的覆盖数量即可，这里注意到为 \(\le k\)，所以最后那个 \(x \oplus a_i=k\) 的 \(x\) 也是要计入贡献的，因此 \(wa\) 了好几发。

参照代码

#include <bits/stdc++.h>

// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")

#define isPbdsFile

#ifdef isPbdsFile

#include <bits/extc++.h>

#else

#include <ext/pb_ds/priority_queue.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/trie_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/list_update_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/exception.hpp>
#include <ext/rope>

#endif

using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> tii;
typedef tuple<ll, ll, ll> tll;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef __int128 i128;
#define hash1 unordered_map
#define hash2 gp_hash_table
#define hash3 cc_hash_table
#define stdHeap std::priority_queue
#define pbdsHeap __gnu_pbds::priority_queue
#define sortArr(a, n) sort(a+1,a+n+1)
#define all(v) v.begin(),v.end()
#define yes cout<<"YES"
#define no cout<<"NO"
#define Spider ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define MyFile freopen("..\\input.txt", "r", stdin),freopen("..\\output.txt", "w", stdout);
#define forn(i, a, b) for(int i = a; i <= b; i++)
#define forv(i, a, b) for(int i=a;i>=b;i--)
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
#define endl '\n'
//用于Miller-Rabin
[[maybe_unused]] static int Prime_Number[13] = {0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};

template <typename T>
int disc(T* a, int n)
{
    return unique(a + 1, a + n + 1) - (a + 1);
}

template <typename T>
T lowBit(T x)
{
    return x & -x;
}

template <typename T>
T Rand(T l, T r)
{
    static mt19937 Rand(time(nullptr));
    uniform_int_distribution<T> dis(l, r);
    return dis(Rand);
}

template <typename T1, typename T2>
T1 modt(T1 a, T2 b)
{
    return (a % b + b) % b;
}

template <typename T1, typename T2, typename T3>
T1 qPow(T1 a, T2 b, T3 c)
{
    a %= c;
    T1 ans = 1;
    for (; b; b >>= 1, (a *= a) %= c)if (b & 1)(ans *= a) %= c;
    return modt(ans, c);
}

template <typename T>
void read(T& x)
{
    x = 0;
    T sign = 1;
    char ch = getchar();
    while (!isdigit(ch))
    {
        if (ch == '-')sign = -1;
        ch = getchar();
    }
    while (isdigit(ch))
    {
        x = (x << 3) + (x << 1) + (ch ^ 48);
        ch = getchar();
    }
    x *= sign;
}

template <typename T, typename... U>
void read(T& x, U&... y)
{
    read(x);
    read(y...);
}

template <typename T>
void write(T x)
{
    if (typeid(x) == typeid(char))return;
    if (x < 0)x = -x, putchar('-');
    if (x > 9)write(x / 10);
    putchar(x % 10 ^ 48);
}

template <typename C, typename T, typename... U>
void write(C c, T x, U... y)
{
    write(x), putchar(c);
    write(c, y...);
}


template <typename T11, typename T22, typename T33>
struct T3
{
    T11 one;
    T22 tow;
    T33 three;

    bool operator<(const T3 other) const
    {
        if (one == other.one)
        {
            if (tow == other.tow)return three < other.three;
            return tow < other.tow;
        }
        return one < other.one;
    }

    T3() { one = tow = three = 0; }

    T3(T11 one, T22 tow, T33 three) : one(one), tow(tow), three(three)
    {
    }
};

template <typename T1, typename T2>
void uMax(T1& x, T2 y)
{
    if (x < y)x = y;
}

template <typename T1, typename T2>
void uMin(T1& x, T2 y)
{
    if (x > y)x = y;
}

constexpr int MX = 1e6;
constexpr int N = 1 << static_cast<int>(ceil(log2(MX))) + 1;
int n, k;
int cnt[N];
constexpr int T = ceil(log2(N));

inline void solve()
{
    cin >> n >> k;
    while (n--)
    {
        int val;
        cin >> val;
        int x = 0;
        forv(i, T, 0)
        {
            const int idx = val >> i & 1; //当前val有无数,0/1
            if (k >> i & 1)
            {
                int curr = x | idx << i; //必须相同
                cnt[curr]++, cnt[curr + (1 << i)]--; //差分[curr,curr+(1<<i)-1] +1
                x |= (idx ^ 1) << i; //使其往右子树走算右子树情况
            }
            else x |= idx << i; //只能往左子树走
        }
        cnt[x]++, cnt[x + 1]--; //注意x^val==k的x也要计算贡献
    }
    forn(i, 1, N-1)cnt[i] += cnt[i - 1]; //前缀和恢复
    cout << *max_element(cnt, cnt + N); //最大答案
}

signed int main()
{
    // MyFile
    Spider
    //------------------------------------------------------
    // clock_t start = clock();
    int test = 1;
    //    read(test);
    // cin >> test;
    forn(i, 1, test)solve();
    //    while (cin >> n, n)solve();
    //    while (cin >> test)solve();
    // clock_t end = clock();
    // cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
}

\[时间复杂度为:\ O(n\log{V_{max}}) \]