# ORSUM - Editorial

Setter: Satyam
Tester: Abhinav Sharma, Aryan
Editorialist: Lavish Gupta

Easy-Medium

# PROBLEM:

Let’s define Score of an array B of length M as \sum_{i=1}^{M} \sum_{j=i}^{M} f(i,j), where f(i,j) = B_i | B_{i+1} | \dots | B_j (Here | denotes the bitwise OR operation)

You are given an array A, consisting of N distinct elements.

Find the sum of Score over all N! possible permutations of A.

Since the answer can be large, output it modulo 998244353.

# QUICK EXPLANATION:

• Contribution of i^{th} bit in the Score is independent of the contribution of j^{th} bit, for i \neq j.
• Sum of Scores over all permutations can be written as - sum of contribution of i^{th} bit (for 1 \leq i \leq 30) over all permutations.
• Find total number of subarray (over all permutations) such that the i^{th} bit is not set in the OR of these subarrays. Using this, calculate number of subarrays num_i such that the i^{th} bit is set in the OR of these subarrays.
• Contribution of the i^{th} bit = 2^{i-1} \cdot num_i, where num_i = (N+1)! \cdot \bigg(\frac{N}{2} - \frac{K}{N-K+2} \bigg) and K represents the number of numbers in which the i^{th} is not set.

# EXPLANATION:

Score of an array is defined as the sum of the OR of all the possible subarrays. Now, an important observation regarding OR (which holds for other bitwise operators too) is that the result of the i^{th} bit is independent of the result of the j^{th} bit. This observation suggests us that if we can focus on a fixed bit, and can calculate the contribution of that bit in the Score, then we can solve our problem.

Let us focus on the i^{th} bit. Let num_i denotes the number of subarrays such that in the OR of these subarrays, the i^{th} bit is set. This is equivalent to saying that at least one element in each of these subarrays have the i^{th} bit set. Now, it is easier to calculate the number of subarrays in which none of the elements have the i^{th} bit set. Hence, we can write num_i as follows:

num_i = Total number of subarrays - number of subarrays in which none of the elements have the i^{th} bit set

Total number of subarrays

If we consider a single permutation, then the number of subarrays in that permutation is \frac{(N)(N+1)}{2}.
Total number of permutations = N!, where ! represents factorial.
Hence, total number of subarrays = N! \cdot \frac{(N)(N+1)}{2}

number of subarrays in which none of the elements have the i-th bit set

Let K represents the number of numbers in which the i^{th} bit is not set.
We will calculate the required number of subarrays by calculating the subarrays of length len (1 \leq len \leq K) for each len individually. Note that all the subarrays of length len \geq K+1 will have at least one element with the i^{th} bit set.

Now, the number of subarrays of length len satisfying the required property
= \binom{K}{len} \cdot (len)! \cdot (N-len)! \cdot (N-len+1)

First term denotes the number of ways to choose len elements out of the K elements which have the i^{th} bit as not set. The second term denotes the number of ways to arrange these len elements and third term denotes the number of ways to arrange the remaining elements. The fourth term denotes the number of positions where this subarray of length len can be placed in the entire array.

Summing this expression for all the values of len

After simplifying, the above expression can be written as

(K)! \cdot (N-K+1)! \cdot \binom{N-len+1}{N-K+1}

Summing over all values of len, we get

\sum_{len = 1}^{K} (K)! \cdot (N-K+1)! \cdot \binom{N-len+1}{N-K+1}

= (K)! \cdot (N-K+1)! \cdot \sum_{len = 1}^{K} \binom{N-len+1}{N-K+1})

The summation term can be represented as the sum of coefficients of x^{N-K+1} in (1+x)^{N-len+1}.
Coefficient of x^{N-K+1} : (1+x)^{N-K+1} + (1+x)^{N-K+2} + \cdots + (1+x)^{N}.

We can first simplify the Geometric Progression at the Right Hand Side, and say that the above value is equivalent to coefficient of x^{N-K+2} in (1+x)^{N+1}.

Substituting this value back in the original expression, we get the final value as \frac{(N+1)! \cdot K}{N-K+2}

Hence, num_i = (N+1)! \cdot \bigg(\frac{N}{2} - \frac{K}{N-K+2} \bigg)

Contribution of the i^{th} bit = 2^{i-1} \cdot num_i. So finally, we need to sum up this contribution over all values of i to get the final answer.

# TIME COMPLEXITY:

We can pre-compute all the factorials beforehand : O(N)
Then for each test case, calculate K for each i - O(N \cdot \log{A_i}), and use these values to get answer for each i.
Overall Time Complexity: O(N \cdot \log{A_i}) for each test case.

# SOLUTION:

Setter's Solution

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
using namespace std;
#define ll long long
const ll INF_MUL=1e13;
#define pb push_back
#define mp make_pair
#define nline "\n"
#define f first
#define s second
#define pll pair<ll,ll>
#define all(x) x.begin(),x.end()
#define vl vector<ll>
#define vvl vector<vector<ll>>
#define vvvl vector<vector<vector<ll>>>
#ifndef ONLINE_JUDGE
#define debug(x) cerr<<#x<<" "; _print(x); cerr<<nline;
#else
#define debug(x);
#endif
void _print(ll x){cerr<<x;}
void _print(char x){cerr<<x;}
void _print(string x){cerr<<x;}
template<class T,class V> void _print(pair<T,V> p) {cerr<<"{"; _print(p.first);cerr<<","; _print(p.second);cerr<<"}";}
template<class T>void _print(vector<T> v) {cerr<<" [ "; for (T i:v){_print(i);cerr<<" ";}cerr<<"]";}
template<class T>void _print(set<T> v) {cerr<<" [ "; for (T i:v){_print(i); cerr<<" ";}cerr<<"]";}
template<class T>void _print(multiset<T> v) {cerr<< " [ "; for (T i:v){_print(i);cerr<<" ";}cerr<<"]";}
template<class T,class V>void _print(map<T, V> v) {cerr<<" [ "; for(auto i:v) {_print(i);cerr<<" ";} cerr<<"]";}
typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
typedef tree<ll, null_type, less_equal<ll>, rb_tree_tag, tree_order_statistics_node_update> ordered_multiset;
typedef tree<pair<ll,ll>, null_type, less<pair<ll,ll>>, rb_tree_tag, tree_order_statistics_node_update> ordered_pset;
//--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
const ll MOD=998244353;
const ll MAX=500500;
vector<ll> fact(MAX+2,1),inv_fact(MAX+2,1);
ll binpow(ll a,ll b,ll MOD){
ll ans=1;
a%=MOD;
while(b){
if(b&1)
ans=(ans*a)%MOD;
b/=2;
a=(a*a)%MOD;
}
return ans;
}
ll inverse(ll a,ll MOD){
return binpow(a,MOD-2,MOD);
}
void precompute(ll MOD){
for(ll i=2;i<MAX;i++){
fact[i]=(fact[i-1]*i)%MOD;
}
inv_fact[MAX-1]=inverse(fact[MAX-1],MOD);
for(ll i=MAX-2;i>=0;i--){
inv_fact[i]=(inv_fact[i+1]*(i+1))%MOD;
}
}
ll nCr(ll a,ll b,ll MOD){
if((a<0)||(a<b)||(b<0))
return 0;
ll denom=(inv_fact[b]*inv_fact[a-b])%MOD;
return (denom*fact[a])%MOD;
}
void solve(){
ll n; cin>>n;
ll a[n+5],ans=0;
for(ll i=1;i<=n;i++){
cin>>a[i];
}
for(ll i=0;i<30;i++){
ll off=0;
for(ll j=1;j<=n;j++){
if(!(a[j]&(1<<i))){
off++;
}
}
ll now=(n*(n+1))/2;
now%=MOD;
now=(now*fact[n])%MOD;
for(ll len=1;len<=n;len++){
ll z=nCr(off,len,MOD)*(n-len+1);
z%=MOD;
z=(z*fact[len])%MOD;
now-=z*fact[n-len];
now%=MOD;
}
now+=MOD;
ans=(ans+now*(1<<i))%MOD;
}
cout<<ans<<nline;
return;
}
int main()
{
ll test_cases=1;
cin>>test_cases;
precompute(MOD);
while(test_cases--){
solve();
}
cout<<fixed<<setprecision(10);
cerr<<"Time:"<<1000*((double)clock())/(double)CLOCKS_PER_SEC<<"ms\n";
}

Tester's Solution
/* in the name of Anton */

/*
Compete against Yourself.
Author - Aryan (@aryanc403)
Atcoder library - https://atcoder.github.io/ac-library/production/document_en/
*/

#ifdef ARYANC403
#else
#pragma GCC optimize ("Ofast")
#pragma GCC target ("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
//#pragma GCC optimize ("-ffloat-store")
#include<bits/stdc++.h>
#define dbg(args...) 42;
#endif

// y_combinator from @neal template https://codeforces.com/contest/1553/submission/123849801
// http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2016/p0200r0.html
template<class Fun> class y_combinator_result {
Fun fun_;
public:
template<class T> explicit y_combinator_result(T &&fun): fun_(std::forward<T>(fun)) {}
template<class ...Args> decltype(auto) operator()(Args &&...args) { return fun_(std::ref(*this), std::forward<Args>(args)...); }
};
template<class Fun> decltype(auto) y_combinator(Fun &&fun) { return y_combinator_result<std::decay_t<Fun>>(std::forward<Fun>(fun)); }

using namespace std;
#define fo(i,n)   for(i=0;i<(n);++i)
#define repA(i,j,n)   for(i=(j);i<=(n);++i)
#define repD(i,j,n)   for(i=(j);i>=(n);--i)
#define all(x) begin(x), end(x)
#define sz(x) ((lli)(x).size())
#define pb push_back
#define mp make_pair
#define X first
#define Y second
#define endl "\n"

typedef long long int lli;
typedef long double mytype;
typedef pair<lli,lli> ii;
typedef vector<ii> vii;
typedef vector<lli> vi;

const auto start_time = std::chrono::high_resolution_clock::now();
void aryanc403()
{
#ifdef ARYANC403
auto end_time = std::chrono::high_resolution_clock::now();
std::chrono::duration<double> diff = end_time-start_time;
cerr<<"Time Taken : "<<diff.count()<<"\n";
#endif
}

long long readInt(long long l, long long r, char endd) {
long long x=0;
int cnt=0;
int fi=-1;
bool is_neg=false;
while(true) {
char g=getchar();
if(g=='-') {
assert(fi==-1);
is_neg=true;
continue;
}
if('0'<=g&&g<='9') {
x*=10;
x+=g-'0';
if(cnt==0) {
fi=g-'0';
}
cnt++;
assert(fi!=0 || cnt==1);
assert(fi!=0 || is_neg==false);

assert(!(cnt>19 || ( cnt==19 && fi>1) ));
} else if(g==endd) {
if(is_neg) {
x=-x;
}
assert(l<=x&&x<=r);
return x;
} else {
assert(false);
}
}
}
string readString(int l, int r, char endd) {
string ret="";
int cnt=0;
while(true) {
char g=getchar();
assert(g!=-1);
if(g==endd) {
break;
}
cnt++;
ret+=g;
}
assert(l<=cnt&&cnt<=r);
return ret;
}
long long readIntSp(long long l, long long r) {
}
long long readIntLn(long long l, long long r) {
}
string readStringLn(int l, int r) {
}
string readStringSp(int l, int r) {
}

assert(getchar()==EOF);
}

vi a(n);
for(int i=0;i<n-1;++i)
return a;
}

// #include<atcoder/dsu>
//     vector<vi> e(n);
//     atcoder::dsu d(n);
//     for(lli i=1;i<n;++i){
//         e[u].pb(v);
//         e[v].pb(u);
//         d.merge(u,v);
//     }
//     assert(d.size(0)==n);
//     return e;
// }

const lli INF = 0xFFFFFFFFFFFFFFFL;

lli seed;
inline lli rnd(lli l=0,lli r=INF)
{return uniform_int_distribution<lli>(l,r)(rng);}

class CMP
{public:
bool operator()(ii a , ii b) //For min priority_queue .
{    return ! ( a.X < b.X || ( a.X==b.X && a.Y <= b.Y ));   }};

void add( map<lli,lli> &m, lli x,lli cnt=1)
{
auto jt=m.find(x);
if(jt==m.end())         m.insert({x,cnt});
else                    jt->Y+=cnt;
}

void del( map<lli,lli> &m, lli x,lli cnt=1)
{
auto jt=m.find(x);
if(jt->Y<=cnt)            m.erase(jt);
else                      jt->Y-=cnt;
}

bool cmp(const ii &a,const ii &b)
{
return a.X<b.X||(a.X==b.X&&a.Y<b.Y);
}

const lli mod = 1000000007L;
// const lli maxN = 1000000007L;

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

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

#include <utility>

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

namespace atcoder {

namespace internal {

constexpr long long safe_mod(long long x, long long m) {
x %= m;
if (x < 0) x += m;
return x;
}

struct barrett {
unsigned int _m;
unsigned long long im;

explicit barrett(unsigned int m) : _m(m), im((unsigned long long)(-1) / m + 1) {}

unsigned int umod() const { return _m; }

unsigned int mul(unsigned int a, unsigned int b) const {

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;
}
};

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;
}

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);

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};

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

auto tmp = s;
s = t;
t = tmp;
tmp = m0;
m0 = m1;
m1 = tmp;
}
if (m0 < 0) m0 += b / s;
return {s, m0};
}

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);

unsigned long long floor_sum_unsigned(unsigned long long n,
unsigned long long m,
unsigned long long a,
unsigned long long b) {
unsigned long long ans = 0;
while (true) {
if (a >= m) {
ans += n * (n - 1) / 2 * (a / m);
a %= m;
}
if (b >= m) {
ans += n * (b / m);
b %= m;
}

unsigned long long y_max = a * n + b;
if (y_max < m) break;
n = (unsigned long long)(y_max / m);
b = (unsigned long long)(y_max % m);
std::swap(m, a);
}
return ans;
}

}  // 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

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());
}

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());
}

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

using namespace atcoder;
using mint = modint998244353;
//using mint = modint1000000007;
using vm = vector<mint>;
std::ostream& operator << (std::ostream& out, const mint& rhs) {
return out<<rhs.val();
}

// Ref - https://www.codechef.com/viewsolution/41909444 Line 827 - 850.
const int maxnCr=2e6+5;
array<mint,maxnCr+1> fac,inv;

mint nCr(lli n,lli r)
{
if(n<0||r<0||r>n)
return 0;
return fac[n]*inv[r]*inv[n-r];
}

void pre(lli n)
{
fac[0]=1;
for(int i=1;i<=n;++i)
fac[i]=i*fac[i-1];
inv[n]=fac[n].pow(mint(-2).val());
for(int i=n;i>0;--i)
inv[i-1]=i*inv[i];
assert(inv[0]==mint(1));
}

lli T,n,i,j,k,in,cnt,l,r,u,v,x,y;
lli m;
string s;
vi a;
//priority_queue < ii , vector < ii > , CMP > pq;// min priority_queue .

int main(void) {
ios_base::sync_with_stdio(false);cin.tie(NULL);
// freopen("txt.in", "r", stdin);
// freopen("txt.out", "w", stdout);
// cout<<std::fixed<<std::setprecision(35);
lli sumN = 5e5;
const lli B = 30;
pre(maxnCr);
while(T--)
{

sumN-=n;
sort(all(a));
assert(sz(a)==n);
mint ans=0;
for(lli b=0;b<B;++b){
lli k=0;
for(auto x:a)
if(x&(1LL<<b))
k++;
mint cur=nCr(n+1,2)*fac[n];
for(lli d=1;d<=n;++d)
cur-=nCr(n-d,k)*fac[k]*fac[n-k]*(n-d+1);
dbg(b,k,n,cur);
ans+=cur*(1LL<<b);
}
cout<<ans<<endl;
}   aryanc403();
return 0;
}

Editorialist's Solution
#include<bits/stdc++.h>
#define ll long long
using namespace std ;
const ll z = 998244353 ;

ll fact[100005] ;

ll power(ll a ,ll b , ll p){if(b == 0) return 1 ; ll c = power(a , b/2 , p) ; if(b%2 == 0) return ((c*c)%p) ; else return ((((c*c)%p)*a)%p) ;}
ll inverse(ll a ,ll n){return power(a , n-2 , n) ;}
ll ncr(ll n , ll r){if(n < r|| (n < 0) || (r < 0)) return 0 ; return ((((fact[n] * inverse(fact[r] , z)) % z) * inverse(fact[n-r] , z)) % z);}

void initialize()
{
fact[0]= 1 ;
for(ll i = 1 ; i < 100005 ; i++)
fact[i] = (fact[i-1]*i)%z ;
return ;
}

ll get_contri(ll n , ll k)
{
ll ans = fact[n+1] ;
ll mult = (n * inverse(2 , z))%z ;
mult -= (k * inverse(n-k+2 , z))%z ;

mult = ((mult%z)+z)%z ;

ans *= mult ;
return ans%z ;
}

void solve()
{
ll n;
cin >> n ;
vector<ll> cnt(30 , n) ;

for(int i = 0 ; i < n ; i++)
{
ll u ;
cin >> u ;

for(ll j = 0 ; j < 30 ; j++)
{
if(u & (1LL << j))
cnt[j]-- ;
}
}

ll ans = 0 ;

for(ll j = 0 ; j < 30 ; j++)
{
ll contri = get_contri(n , cnt[j]) ;
ans += ((1LL << j) * contri)%z ;
}
ans %= z ;
cout << ans << '\n' ;

return ;
}

int main()
{
ios_base::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
#ifndef ONLINE_JUDGE
freopen("inputf.txt" , "r" , stdin) ;
freopen("outputf.txt" , "w" , stdout) ;
freopen("error.txt" , "w" , stderr) ;
#endif

initialize() ;
ll t;
cin >> t ;
while(t--)
{
solve() ;
}

return 0;
}

5 Likes

Kudos to the author, nice problem.

4 Likes

Indeed a nice problem!

when you calculate for len, will not it include solutions for len +1 ? will not be there any inclusion - exclusion principle here ?

No, it won’t, as we’re counting the contribution of subarrays of length j for all j from 1 to S_i where S_i is the number of elements which have their i^{\text{th}} bit unset and all of these j computations are independent of each other.

One a side note, final answer is just

\displaystyle\sum_{i=0}^{30} \left(\frac{N(N+1)}{2} \times N! -\sum_{j=1}^{S_i}\binom{S_i}{j}\times (N-j)! \times j! \ \times (N-j+1) \right)

And this can be naively calculated in \mathcal{O}(N\log A_i) without doing the math to simplify the inner summation. CODE

1 Like

Very nice problem. I personally did not get it during the contest, but thought about it for about a week here and there, not as a programming problem but as a math problem. I arrived at the same conclusion using a different combinatorial argument.

Given n+2 tokens in a row, choose k+2 of them as blue and n-k as red. The red tokens represent the bits not set. Out of the blue tokens, choose two as “start” and “end” with the condition that start must come before end, and there is at least one blue token between start and end. I was able to prove that this token arrangement corresponds to a score-producing bit arrangement, with the subarray denoted by the tokens between start and end. You do have to multiply by k! (n-k)! because we’re talking about permutation and not just 0-1 arrangements.