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;
}