第一场

1.容斥计数

给定 n,m,qn,m,q,求满足如下要求的序列个数模 qq 的结果:

  • 长度为 nn
  • 存在两个不同的子序列,满足子序列中的数区间与为1
  • 序列中的数的大小范围在 [0,2m)[0,2^m)

数据范围: n5000,m5000,q109n \leq 5000,m \leq 5000,q \leq 10^9

分析什么时候存在两个子序列满足上述条件。考虑该序列长度为 kk 的全部奇数组成的子序列 BB,容易证明若该序列存在区间与为 1 的子序列,则 BB 是最大的满足区间与是 1 的子序列。考虑容斥,即计算满足比 BB 小的子序列区间与都不为 1 的 bbBB 的个数,记这个条件为 PP 。也就是

xB,i__lg(x),xB中唯一满足2i&x=0的元素\forall x \in B,\exists i \leq \_\_\lg(x),x是B中唯一满足2^i \And x= 0的元素

考虑枚举式子中的 ii ,称这个 ii 为特殊位当且仅当存在 BB 中的元素满足上述条件。根据逻辑推理的知识,容易证明这两个条件是等价的。

于是我们记 dpi,jdp_{i,j} 表示满足长度为 ii,恰好有 jj 个特殊位的的奇数序列个数。考虑转移,假设有如下状态,数字纵向排列

101111101101\begin{array}{cccc} 1 & 0 & 1 & 1\\ 1 & 1 & 1 & 0\\ 1 &1 &0 &1\end{array}

对应状态 (3,4)(3,4)

它可以向如下矩阵转移

101111110011011\begin{array}{ccccc} 1 & 0 & 1 & 1 & 1\\ 1 & 1 & 1 & 0 & \textbf 0\\ 1 &1 &0 &1 & 1\end{array}

对应状态 (3,5)(3,5)

注意粗体的 00 可以有 ii 种选择,因此转移系数是 ii

还可以向如下矩阵转移

10111111011101111110\begin{array}{ccccc} 1 & 0 & 1 & 1 & 1\\ 1 & 1 & 1 & 0 & 1\\ 1 &1 &0 &1 & 1\\ \textbf 1 &\textbf 1 &\textbf 1 &\textbf 1 &\textbf 0\end{array}

对应状态 (4,5)(4,5)

注意,粗体的数字 1111011110 可以被插在任意行,它所对应的序列是不同的,因此转移系数也是 ii

综上可以有转移方程

dpi+1,j+1=i(dpi,j+dpi+1,j)dp_{i + 1,j + 1}=i(dp_{i,j}+dp_{i+1,j})

求出 dpdp 之后,我们固定 BB 的长度 k(k2)k(k\geq2) 以及特殊位的个数 tt ,就有满足 PP 的序列个数为

P(k,t)=dpk,t(2k1k)mt1P(k,t)=dp_{k,t}(2^k-1-k)^{m-t-1}

加上偶数,我们就有结果

B(k,t)=(nk)(2m1)nk(m1t)P(k,t)B=n(2m1)n1+k=2n(nk)(2m1)nkt=km1(m1t)P(k,t)B(k,t)=\dbinom{n}{k}(2^{m-1})^{n-k}\dbinom{m-1}{t}P(k,t)\\|B|=n(2^{m-1})^{n-1}+\sum_{k=2}^{n}\dbinom{n}{k}(2^{m-1})^{n-k} \sum_{t=k}^{m-1}\dbinom{m-1}{t}P(k,t)

于是引用上一题的答案,我们就有最终式子

(k=1n(nk)2(m1)(nk)(2k1)m)B\big(\sum_{k=1}^{n}\dbinom{n}{k}2^{(m-1)(n-k)}(2^k-1)^m \big)-|B|

时间复杂度 O((n+m)2)O((n + m)^2)

code:

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#pragma GCC optimize("O3")
#include <bits/stdc++.h>


#include <utility>

namespace atcoder {

namespace internal {

// @param m `1 <= m`
// @return x mod m
constexpr long long safe_mod(long long x, long long m) {
x %= m;
if (x < 0) x += m;
return x;
}

// Fast modular multiplication by barrett reduction
// Reference: https://en.wikipedia.org/wiki/Barrett_reduction
// NOTE: reconsider after Ice Lake
struct barrett {
unsigned int _m;
unsigned long long im;

// @param m `1 <= m < 2^31`
barrett(unsigned int m) : _m(m), im((unsigned long long)(-1) / m + 1) {}

// @return m
unsigned int umod() const { return _m; }

// @param a `0 <= a < m`
// @param b `0 <= b < m`
// @return `a * b % m`
unsigned int mul(unsigned int a, unsigned int b) const {
// [1] m = 1
// a = b = im = 0, so okay

// [2] m >= 2
// im = ceil(2^64 / m)
// -> im * m = 2^64 + r (0 <= r < m)
// let z = a*b = c*m + d (0 <= c, d < m)
// a*b * im = (c*m + d) * im = c*(im*m) + d*im = c*2^64 + c*r + d*im
// c*r + d*im < m * m + m * im < m * m + 2^64 + m <= 2^64 + m * (m + 1) < 2^64 * 2
// ((ab * im) >> 64) == c or c + 1
unsigned long long z = a;
z *= b;
#ifdef _MSC_VER
unsigned long long x;
_umul128(z, im, &x);
#else
unsigned long long x =
(unsigned long long)(((unsigned __int128)(z)*im) >> 64);
#endif
unsigned int v = (unsigned int)(z - x * _m);
if (_m <= v) v += _m;
return v;
}
};

// @param n `0 <= n`
// @param m `1 <= m`
// @return `(x ** n) % m`
constexpr long long pow_mod_constexpr(long long x, long long n, int m) {
if (m == 1) return 0;
unsigned int _m = (unsigned int)(m);
unsigned long long r = 1;
unsigned long long y = safe_mod(x, m);
while (n) {
if (n & 1) r = (r * y) % _m;
y = (y * y) % _m;
n >>= 1;
}
return r;
}

// Reference:
// M. Forisek and J. Jancina,
// Fast Primality Testing for Integers That Fit into a Machine Word
// @param n `0 <= n`
constexpr bool is_prime_constexpr(int n) {
if (n <= 1) return false;
if (n == 2 || n == 7 || n == 61) return true;
if (n % 2 == 0) return false;
long long d = n - 1;
while (d % 2 == 0) d /= 2;
constexpr long long bases[3] = {2, 7, 61};
for (long long a : bases) {
long long t = d;
long long y = pow_mod_constexpr(a, t, n);
while (t != n - 1 && y != 1 && y != n - 1) {
y = y * y % n;
t <<= 1;
}
if (y != n - 1 && t % 2 == 0) {
return false;
}
}
return true;
}
template <int n> constexpr bool is_prime = is_prime_constexpr(n);

// @param b `1 <= b`
// @return pair(g, x) s.t. g = gcd(a, b), xa = g (mod b), 0 <= x < b/g
constexpr std::pair<long long, long long> inv_gcd(long long a, long long b) {
a = safe_mod(a, b);
if (a == 0) return {b, 0};

// Contracts:
// [1] s - m0 * a = 0 (mod b)
// [2] t - m1 * a = 0 (mod b)
// [3] s * |m1| + t * |m0| <= b
long long s = b, t = a;
long long m0 = 0, m1 = 1;

while (t) {
long long u = s / t;
s -= t * u;
m0 -= m1 * u; // |m1 * u| <= |m1| * s <= b

// [3]:
// (s - t * u) * |m1| + t * |m0 - m1 * u|
// <= s * |m1| - t * u * |m1| + t * (|m0| + |m1| * u)
// = s * |m1| + t * |m0| <= b

auto tmp = s;
s = t;
t = tmp;
tmp = m0;
m0 = m1;
m1 = tmp;
}
// by [3]: |m0| <= b/g
// by g != b: |m0| < b/g
if (m0 < 0) m0 += b / s;
return {s, m0};
}

// Compile time primitive root
// @param m must be prime
// @return primitive root (and minimum in now)
constexpr int primitive_root_constexpr(int m) {
if (m == 2) return 1;
if (m == 167772161) return 3;
if (m == 469762049) return 3;
if (m == 754974721) return 11;
if (m == 998244353) return 3;
int divs[20] = {};
divs[0] = 2;
int cnt = 1;
int x = (m - 1) / 2;
while (x % 2 == 0) x /= 2;
for (int i = 3; (long long)(i)*i <= x; i += 2) {
if (x % i == 0) {
divs[cnt++] = i;
while (x % i == 0) {
x /= i;
}
}
}
if (x > 1) {
divs[cnt++] = x;
}
for (int g = 2;; g++) {
bool ok = true;
for (int i = 0; i < cnt; i++) {
if (pow_mod_constexpr(g, (m - 1) / divs[i], m) == 1) {
ok = false;
break;
}
}
if (ok) return g;
}
}
template <int m> constexpr int primitive_root = primitive_root_constexpr(m);

} // namespace internal

} // namespace atcoder


#include <cassert>
#include <numeric>
#include <type_traits>

namespace atcoder {

namespace internal {

#ifndef _MSC_VER
template <class T>
using is_signed_int128 =
typename std::conditional<std::is_same<T, __int128_t>::value ||
std::is_same<T, __int128>::value,
std::true_type,
std::false_type>::type;

template <class T>
using is_unsigned_int128 =
typename std::conditional<std::is_same<T, __uint128_t>::value ||
std::is_same<T, unsigned __int128>::value,
std::true_type,
std::false_type>::type;

template <class T>
using make_unsigned_int128 =
typename std::conditional<std::is_same<T, __int128_t>::value,
__uint128_t,
unsigned __int128>;

template <class T>
using is_integral = typename std::conditional<std::is_integral<T>::value ||
is_signed_int128<T>::value ||
is_unsigned_int128<T>::value,
std::true_type,
std::false_type>::type;

template <class T>
using is_signed_int = typename std::conditional<(is_integral<T>::value &&
std::is_signed<T>::value) ||
is_signed_int128<T>::value,
std::true_type,
std::false_type>::type;

template <class T>
using is_unsigned_int =
typename std::conditional<(is_integral<T>::value &&
std::is_unsigned<T>::value) ||
is_unsigned_int128<T>::value,
std::true_type,
std::false_type>::type;

template <class T>
using to_unsigned = typename std::conditional<
is_signed_int128<T>::value,
make_unsigned_int128<T>,
typename std::conditional<std::is_signed<T>::value,
std::make_unsigned<T>,
std::common_type<T>>::type>::type;

#else

template <class T> using is_integral = typename std::is_integral<T>;

template <class T>
using is_signed_int =
typename std::conditional<is_integral<T>::value && std::is_signed<T>::value,
std::true_type,
std::false_type>::type;

template <class T>
using is_unsigned_int =
typename std::conditional<is_integral<T>::value &&
std::is_unsigned<T>::value,
std::true_type,
std::false_type>::type;

template <class T>
using to_unsigned = typename std::conditional<is_signed_int<T>::value,
std::make_unsigned<T>,
std::common_type<T>>::type;

#endif

template <class T>
using is_signed_int_t = std::enable_if_t<is_signed_int<T>::value>;

template <class T>
using is_unsigned_int_t = std::enable_if_t<is_unsigned_int<T>::value>;

template <class T> using to_unsigned_t = typename to_unsigned<T>::type;

} // namespace internal

} // namespace atcoder

#include <cassert>
#include <numeric>
#include <type_traits>

#ifdef _MSC_VER
#include <intrin.h>
#endif

namespace atcoder {

namespace internal {

struct modint_base {};
struct static_modint_base : modint_base {};

template <class T> using is_modint = std::is_base_of<modint_base, T>;
template <class T> using is_modint_t = std::enable_if_t<is_modint<T>::value>;

} // namespace internal

template <int m, std::enable_if_t<(1 <= m)>* = nullptr>
struct static_modint : internal::static_modint_base {
using mint = static_modint;

public:
static constexpr int mod() { return m; }
static mint raw(int v) {
mint x;
x._v = v;
return x;
}

static_modint() : _v(0) {}
template <class T, internal::is_signed_int_t<T>* = nullptr>
static_modint(T v) {
long long x = (long long)(v % (long long)(umod()));
if (x < 0) x += umod();
_v = (unsigned int)(x);
}
template <class T, internal::is_unsigned_int_t<T>* = nullptr>
static_modint(T v) {
_v = (unsigned int)(v % umod());
}
static_modint(bool v) { _v = ((unsigned int)(v) % umod()); }

unsigned int val() const { return _v; }

mint& operator++() {
_v++;
if (_v == umod()) _v = 0;
return *this;
}
mint& operator--() {
if (_v == 0) _v = umod();
_v--;
return *this;
}
mint operator++(int) {
mint result = *this;
++*this;
return result;
}
mint operator--(int) {
mint result = *this;
--*this;
return result;
}

mint& operator+=(const mint& rhs) {
_v += rhs._v;
if (_v >= umod()) _v -= umod();
return *this;
}
mint& operator-=(const mint& rhs) {
_v -= rhs._v;
if (_v >= umod()) _v += umod();
return *this;
}
mint& operator*=(const mint& rhs) {
unsigned long long z = _v;
z *= rhs._v;
_v = (unsigned int)(z % umod());
return *this;
}
mint& operator/=(const mint& rhs) { return *this = *this * rhs.inv(); }

mint operator+() const { return *this; }
mint operator-() const { return mint() - *this; }

mint pow(long long n) const {
assert(0 <= n);
mint x = *this, r = 1;
while (n) {
if (n & 1) r *= x;
x *= x;
n >>= 1;
}
return r;
}
mint inv() const {
if (prime) {
assert(_v);
return pow(umod() - 2);
} else {
auto eg = internal::inv_gcd(_v, m);
assert(eg.first == 1);
return eg.second;
}
}

friend mint operator+(const mint& lhs, const mint& rhs) {
return mint(lhs) += rhs;
}
friend mint operator-(const mint& lhs, const mint& rhs) {
return mint(lhs) -= rhs;
}
friend mint operator*(const mint& lhs, const mint& rhs) {
return mint(lhs) *= rhs;
}
friend mint operator/(const mint& lhs, const mint& rhs) {
return mint(lhs) /= rhs;
}
friend bool operator==(const mint& lhs, const mint& rhs) {
return lhs._v == rhs._v;
}
friend bool operator!=(const mint& lhs, const mint& rhs) {
return lhs._v != rhs._v;
}

private:
unsigned int _v;
static constexpr unsigned int umod() { return m; }
static constexpr bool prime = internal::is_prime<m>;
};

template <int id> struct dynamic_modint : internal::modint_base {
using mint = dynamic_modint;

public:
static int mod() { return (int)(bt.umod()); }
static void set_mod(int m) {
assert(1 <= m);
bt = internal::barrett(m);
}
static mint raw(int v) {
mint x;
x._v = v;
return x;
}

dynamic_modint() : _v(0) {}
template <class T, internal::is_signed_int_t<T>* = nullptr>
dynamic_modint(T v) {
long long x = (long long)(v % (long long)(mod()));
if (x < 0) x += mod();
_v = (unsigned int)(x);
}
template <class T, internal::is_unsigned_int_t<T>* = nullptr>
dynamic_modint(T v) {
_v = (unsigned int)(v % mod());
}
dynamic_modint(bool v) { _v = ((unsigned int)(v) % mod()); }

unsigned int val() const { return _v; }

mint& operator++() {
_v++;
if (_v == umod()) _v = 0;
return *this;
}
mint& operator--() {
if (_v == 0) _v = umod();
_v--;
return *this;
}
mint operator++(int) {
mint result = *this;
++*this;
return result;
}
mint operator--(int) {
mint result = *this;
--*this;
return result;
}

mint& operator+=(const mint& rhs) {
_v += rhs._v;
if (_v >= umod()) _v -= umod();
return *this;
}
mint& operator-=(const mint& rhs) {
_v += mod() - rhs._v;
if (_v >= umod()) _v -= umod();
return *this;
}
mint& operator*=(const mint& rhs) {
_v = bt.mul(_v, rhs._v);
return *this;
}
mint& operator/=(const mint& rhs) { return *this = *this * rhs.inv(); }

mint operator+() const { return *this; }
mint operator-() const { return mint() - *this; }

mint pow(long long n) const {
assert(0 <= n);
mint x = *this, r = 1;
while (n) {
if (n & 1) r *= x;
x *= x;
n >>= 1;
}
return r;
}
mint inv() const {
auto eg = internal::inv_gcd(_v, mod());
assert(eg.first == 1);
return eg.second;
}

friend mint operator+(const mint& lhs, const mint& rhs) {
return mint(lhs) += rhs;
}
friend mint operator-(const mint& lhs, const mint& rhs) {
return mint(lhs) -= rhs;
}
friend mint operator*(const mint& lhs, const mint& rhs) {
return mint(lhs) *= rhs;
}
friend mint operator/(const mint& lhs, const mint& rhs) {
return mint(lhs) /= rhs;
}
friend bool operator==(const mint& lhs, const mint& rhs) {
return lhs._v == rhs._v;
}
friend bool operator!=(const mint& lhs, const mint& rhs) {
return lhs._v != rhs._v;
}

private:
unsigned int _v;
static internal::barrett bt;
static unsigned int umod() { return bt.umod(); }
};
template <int id> internal::barrett dynamic_modint<id>::bt = 998244353;

using modint998244353 = static_modint<998244353>;
using modint1000000007 = static_modint<1000000007>;
using modint = dynamic_modint<-1>;

namespace internal {

template <class T>
using is_static_modint = std::is_base_of<internal::static_modint_base, T>;

template <class T>
using is_static_modint_t = std::enable_if_t<is_static_modint<T>::value>;

template <class> struct is_dynamic_modint : public std::false_type {};
template <int id>
struct is_dynamic_modint<dynamic_modint<id>> : public std::true_type {};

template <class T>
using is_dynamic_modint_t = std::enable_if_t<is_dynamic_modint<T>::value>;

} // namespace internal

} // namespace atcoder


//#define int long long
using ll = long long;
const ll mod = 998244353;
const int N = 1e7 + 7;



typedef atcoder::modint MINT;

void solve() {

int n,m,q;std::cin>>n>>m>>q;
MINT::set_mod(q);

std::vector<std::vector<MINT>> dp(n + m + 2,std::vector<MINT>(m + n + 2,0));

dp[0][0] = 1;
for(int i = 1;i <= n;i++){
for(int j = 1;j < m;j++)
dp[i][j] = i * (dp[i][j - 1] + dp[i - 1][j - 1]);
}

MINT ans1 = 0;


std::vector<std::vector<MINT> > C(5050,std::vector<MINT>(5050,0));
std::vector<MINT> po2(5050,0);


po2[0] = 1;
for(int i = 1;i <= 5000;i++)po2[i] = po2[i - 1] * 2;
C[0][0] = 1;
for(int i = 1;i <= 5000;i++){
C[i][0] = 1;
for(int j = 1;j <= i;j++)
C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]);
}

for(int i = 2;i <= n;i++){
MINT cur = 1,cur1 = 1;
for(int j = 1;j < m;j++) cur *= (po2[i] - 1),cur1 *= po2[n - i];
ans1 += C[n][i] * cur1 * cur;
}
//std::cout<<ans1<<'\n';

MINT ans2 = 0;
std::vector<MINT> pw(m + 2);
for(int i = 2;i <= n;i++){
MINT tp = 0;
pw[0] = 1;
for(int j = 1;j <= m;j++)pw[j] = pw[j - 1] * (po2[i] - 1 - i);
for(int j = m - 1;j >= i;j--){
tp += C[m - 1][j] * dp[i][j] * pw[m - j - 1];
}
MINT cur = 1;
for(int j = 1;j < m;j++) cur *= po2[n - i];
tp *= C[n][i] * cur;
ans2 += tp;
}

std::cout<<(ans1 - ans2).val()<<'\n';
}
// 1 3 10
//(n + 1) ^2 - (n)^ = 2n + 1

signed main() {
std::ios::sync_with_stdio(false);
std::cin.tie(0);
int t = 1;
// int res = 0;
// for(int i = 1;i <= 1145;i++) res |= i;
// std::cerr<<res<<'\n';
//std::cin >> t;
while (t--) {
solve();
std::cout.flush();
}
return 0;
}

2.模意义下的不等式

给定一数组 aa ,初始为空。定义一次操作为

  • 将数组末尾 tt 个元素删去
  • 将元素 xx 添加至数组末尾

每次操作给定 t,xt,x,保证操作合法,求每次操作后的

i=1a(j=iaai)mod220\bigoplus_{i=1}^{|a|}(\sum_{j=i}^{|a|}a_i) \mod 2^{20}

数据范围:q5105,x109q\leq 5 \cdot 10^5, x \leq 10^9

首先考虑到模数,只需要考虑 20 位及以下的贡献。异或,拆位分析奇偶。

设当前位数为 dd ,我们记前缀和为 SiS_i,则对于某个位置 iij=iaai\sum_{j=i}^{|a|}a_i 的第 dd 位是 1 的充要条件是

(SaSi)mod2d+12d(1)(S_{|a|}-S_i) \mod 2^{d + 1} \geq 2^d \tag{1}

因为模一个整的 kk22 进制数相当于将原数 kk 位以上的二进制位消去。类比十进制即可证明。

SS 是好维护的,我们考虑 (1)(1) 如何维护。这是一个循环群上的不等式,先将 SiS_i 处理为逆元,即将上式变为

Sa+(Si)2d(1)S_{|a|}+(-S_i) \geq 2^d \tag{1}

上式所有量都是模意义下的量。我们分类讨论

  1. 0Sa<2d0\leq S_{|a|}<2^d,显然地就有 (Si)[2dSa,2d+1Sa](-S_i) \in [2^d-S_{|a|},2^{d+1}-S_{|a|}]
  2. Sa2dS_{|a|}\geq 2^d,这时候区间分成了两段,(0,2d+1Sa][Sa2d,2d+1)(0,2^{d+1}-S_{|a|}]\cup[S_{|a|} - 2^d,2^{d+1})

画一个圆圈即可理解。考虑到 $1e6 $ 的值域和 6s 的时限,我们可以开 20 个模意义下的权值树状数组,每次操作后按位求出区间内的 11 的奇偶性即可。

code

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#include <bits/stdc++.h>

using ll = long long;
const int mod[24] = {1,2,4,8,1<<4,1<<5,1<<6,1<<7,1<<8,1<<9,1<<10,1<<11,1<<12,1<<13,1<<14,1<<15,1<<16,1<<17,1<<18,1<<19,1<<20,1<<21};

//const ll mod = 1e9 + 7;
// #define int long long

class bit {
public:
int a[1<<22];
int n;
bit(int n) : n(n + 4) {
for(int i = 1;i <= n;i++)a[i] = 0;
}

bit() { n = 0; }

void add(int x, int y) {
x++;
for (int i = x; i <= n; i += i & -i) {
a[i] += y;
}
}
int get(int x) {
x++;
int res = 0;
if(x <= 0) return 0;
for (int i = x; i; i -= i & -i) {
res += a[i];
}
return res;
}

int get(int l, int r) {
// l++,r++;
l = std::max(l,0);
r = std::min(n,r);
if(l == 0) return get(r);
return get(r) - get(l - 1);
}
};

inline void read(int &x) {
x = 0;
char c = getchar();
while (!isdigit(c)) c = getchar();
while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
}

int sum[500050];
int idx;

void solve() {
std::vector<bit> a(21);

auto add = [&](int x) {
for (int i = 0; i < 21; i++){
int tp = x % mod[i + 1];
//std::cerr<<tp<<' '<<x<<'\n';
a[i].add(tp, 1);
}
};

auto rm = [&](int x) {
for (int i = 0; i < 21; i++){
int tp = x % mod[i + 1];
a[i].add(tp, -1);
}
};

for (int i = 0; i < 21; i++) {
a[i] = bit(1 << (i + 1));
}
add(0);

int q;
read(q);
//std::stack<int> sum;
while (q--) {
int t, x;
read(t), read(x);

for (int i = 1; i <= t; i++)
rm(sum[idx]), idx--;
if (idx == 0)
sum[idx + 1] = x;
else
sum[idx + 1] = x + sum[idx];
idx++;
// std::cerr<<sum.back()<<'\n';0
add(sum[idx]);

int res = 0;
for (int i = 0; i < 21; i++) {
int l = 0, r = 0;
int l1 = 0, r1 = 0;
int tp = (sum[idx] + 1) % mod[i + 1];
//if(tp == 0)tp = (1ll << (i + 1));
int tp1 = 0;
if (tp < (1ll << i)) {
l = tp, r = (1 << i) + tp - 1;
//std::cerr<<l<<' '<<r<<'\n';
tp1 = a[i].get(l, r);
} else {
l = 0, r = tp - (1<<i) - 1, l1 = tp, r1 = (1 << i + 1) - 1;
//std::cerr<<l<<' '<<r<<' '<<l1<<' '<<r1<<'\n';
tp1 = a[i].get(l, r) ^ a[i].get(l1, r1);
}
// r &= (1<<i);
res |= (tp1 & 1) * (1ll << i);
}

std::cout << res << '\n';
std::cout.flush();
}
}

signed main() {
//std::ios::sync_with_stdio(false);
//std::cin.tie(0);
int t = 1;
// std::cin >> t;
while (t--) {
solve();
// std::cout.flush();
}
// std::cout<<5 * inv(2ll) % mod<<'\n';
return 0;
}