PROBLEM LINK:
Setter: Mohammed Ehab
Tester: Ramazan Rakhmatullin
Editorialist: Ishmeet Singh Saggu
DIFFICULTY:
Easy
PREREQUISITES:
Bitwise Operation and Maths
PROBLEM:
Given a sequence of positive integers a_1,a_2,…,a_N, if you take each of its 2^N subsequences and write down the sum of elements of this subsequence, Output the bitwise OR of the written integers.
EXPLANATION:
Let the sequence formed by writing down the sum of elements of each 2^N subsequences of the sequence a is : b_1, b_2, ... b_{2^N}. Note the maximum element in the sequence b is the sum of all elements in the sequence a, so the final answer which we have to compute by doing bitwise OR of all the elements of the sequence b will have the same number of bits in its binary representation as of the maximum number in sequence b.
So this observation gives us a hint that if for each bit position, if we can find out that is it possible to make it 1 by summing some or all elements of sequence a then we can compute the answer. Now let’s see how to check it.
- We will start from the lowest bit i.e. the bit-0 which represents 2^0 and go till the maximum bit.
- We will count the number of elements in the sequence a which have bit-0 equals to 1 in its binary representation and it will be the final count of bit-0 let us represent it as dp[0]. If dp[0] \geq 1 then we can make the bit-0 equals to 1 in the answer else bit-0 will be 0, then we will go to the next bit i.e. the bit-1 then again we will count as we did for the bit-0 but to the count of bit-1 we will add the floor(dp[0]/2), hence for bit-1, dp[1] = count in sequence + floor(dp[0]/2).
- So for bit-i, dp[i] = count in sequence + floor(dp[i-1]/2), where i > 0
- To construct the answer, if dp[i]>0 then the bit-i in the answer will be 1 else it will be 0 because if dp[i] > 0 then it is always possible to select a subsequence such that bit-i is 1 when all the numbers in the subsequence are added.
To give you an intuition about why the above process works. Suppose we consider a bit-i, and if there is at least one element in the sequence a which have 1 at bit-i in its binary representation then, as it is the subsequence(where we will select only this element) then we will have 1 at bit-i in the binary representation of the final answer. Suppose we don’t have elements with bit-i equals 1, but we have 2 numbers whose bit-(i-1) is 1 or 4 numbers whose bit-(i-2) is 1 , then we can make 1 at bit-i by selecting these numbers as subsequence. Similarly, you can think of many other possible constructions. One more example is if you have 2 numbers with bit-(i-2) equals 1 and 1 number with bit-(i-1) equals 1 (and for simplicity consider rest all bits are 0 in these 3 numbers) then if we add these 3 numbers together we will get a number with bit-i equals 1.
TIME COMPLEXITY:
- Time complexity per test case is O(N * log(max(a_i))).
SOLUTIONS:
Setter's Solution
Tester's Solution
#include <bits/stdc++.h>
#pragma clang diagnostic push
#pragma clang diagnostic ignored "-Wunused-const-variable"
#define popcnt(x) __builtin_popcount(x)
#define fr first
#define sc second
#define m_p make_pair
#define low_bo(a, x) lower_bound(a.begin(), a.end(), x) - a.begin()
#define up_bo(a, x) upper_bound(a.begin(), a.end(), x) - a.begin()
#define unique(a) a.resize(unique(a.begin(), a.end()) - a.begin())
#define popcnt(x) __builtin_popcount(x)
//#include <ext/pb_ds/assoc_container.hpp>
//using namespace __gnu_pbds;
//gp_hash_table<int, int> table;
//#pragma GCC optimize("O3")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
//#pragma GCC target("avx,tune=native")
//float __attribute__((aligned(32)))
/*char memory[(int)1e8];
char memorypos;
inline void * operator new(size_t n){
char * ret = memory + memorypos;
memorypos += n;
return (void *)ret;
}
inline void operator delete(void *){}
*/
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef unsigned int uint;
template<typename T>
class Modular {
public:
using Type = typename decay<decltype(T::value)>::type;
constexpr Modular() : value() {}
template<typename U>
Modular(const U &x) {
value = normalize(x);
}
static Type inverse(Type a, Type mod) {
Type b = mod, x = 0, y = 1;
while (a != 0) {
Type t = b / a;
b -= a * t;
x -= t * y;
swap(a, b);
swap(x, y);
}
if (x < 0)
x += mod;
return x;
}
template<typename U>
static Type normalize(const U &x) {
Type v;
if (-mod() <= x && x < mod()) v = static_cast<Type>(x);
else v = static_cast<Type>(x % mod());
if (v < 0) v += mod();
return v;
}
const Type &operator()() const { return value; }
template<typename U>
explicit operator U() const { return static_cast<U>(value); }
constexpr static Type mod() { return T::value; }
Modular &operator+=(const Modular &other) {
if ((value += other.value) >= mod()) value -= mod();
return *this;
}
Modular &operator-=(const Modular &other) {
if ((value -= other.value) < 0) value += mod();
return *this;
}
template<typename U>
Modular &operator+=(const U &other) { return *this += Modular(other); }
template<typename U>
Modular &operator-=(const U &other) { return *this -= Modular(other); }
Modular &operator++() { return *this += 1; }
Modular &operator--() { return *this -= 1; }
Modular operator++(int) {
Modular result(*this);
*this += 1;
return result;
}
Modular operator--(int) {
Modular result(*this);
*this -= 1;
return result;
}
Modular operator-() const { return Modular(-value); }
template<typename U = T>
typename enable_if<is_same<typename Modular<U>::Type, int>::value, Modular>::type &operator*=(const Modular &rhs) {
#ifdef _WIN32
uint64_t x = static_cast<int64_t>(value) * static_cast<int64_t>(rhs.value);
uint32_t xh = static_cast<uint32_t>(x >> 32), xl = static_cast<uint32_t>(x), d, m;
asm(
"divl %4; \n\t"
: "=a" (d), "=d" (m)
: "d" (xh), "a" (xl), "r" (mod())
);
value = m;
#else
value = normalize(static_cast<int64_t>(value) * static_cast<int64_t>(rhs.value));
#endif
return *this;
}
template<typename U = T>
typename enable_if<is_same<typename Modular<U>::Type, int64_t>::value, Modular>::type &
operator*=(const Modular &rhs) {
int64_t q = static_cast<int64_t>(static_cast<long double>(value) * rhs.value / mod());
value = normalize(value * rhs.value - q * mod());
return *this;
}
template<typename U = T>
typename enable_if<!is_integral<typename Modular<U>::Type>::value, Modular>::type &operator*=(const Modular &rhs) {
value = normalize(value * rhs.value);
return *this;
}
Modular &operator/=(const Modular &other) { return *this *= Modular(inverse(other.value, mod())); }
template<typename U>
friend const Modular<U> &abs(const Modular<U> &v) { return v; }
template<typename U>
friend bool operator==(const Modular<U> &lhs, const Modular<U> &rhs);
template<typename U>
friend bool operator<(const Modular<U> &lhs, const Modular<U> &rhs);
template<typename U>
friend std::istream &operator>>(std::istream &stream, Modular<U> &number);
private:
Type value;
};
template<typename T>
bool operator==(const Modular<T> &lhs, const Modular<T> &rhs) { return lhs.value == rhs.value; }
template<typename T, typename U>
bool operator==(const Modular<T> &lhs, U rhs) { return lhs == Modular<T>(rhs); }
template<typename T, typename U>
bool operator==(U lhs, const Modular<T> &rhs) { return Modular<T>(lhs) == rhs; }
template<typename T>
bool operator!=(const Modular<T> &lhs, const Modular<T> &rhs) { return !(lhs == rhs); }
template<typename T, typename U>
bool operator!=(const Modular<T> &lhs, U rhs) { return !(lhs == rhs); }
template<typename T, typename U>
bool operator!=(U lhs, const Modular<T> &rhs) { return !(lhs == rhs); }
template<typename T>
bool operator<(const Modular<T> &lhs, const Modular<T> &rhs) { return lhs.value < rhs.value; }
template<typename T>
Modular<T> operator+(const Modular<T> &lhs, const Modular<T> &rhs) { return Modular<T>(lhs) += rhs; }
template<typename T, typename U>
Modular<T> operator+(const Modular<T> &lhs, U rhs) { return Modular<T>(lhs) += rhs; }
template<typename T, typename U>
Modular<T> operator+(U lhs, const Modular<T> &rhs) { return Modular<T>(lhs) += rhs; }
template<typename T>
Modular<T> operator-(const Modular<T> &lhs, const Modular<T> &rhs) { return Modular<T>(lhs) -= rhs; }
template<typename T, typename U>
Modular<T> operator-(const Modular<T> &lhs, U rhs) { return Modular<T>(lhs) -= rhs; }
template<typename T, typename U>
Modular<T> operator-(U lhs, const Modular<T> &rhs) { return Modular<T>(lhs) -= rhs; }
template<typename T>
Modular<T> operator*(const Modular<T> &lhs, const Modular<T> &rhs) { return Modular<T>(lhs) *= rhs; }
template<typename T, typename U>
Modular<T> operator*(const Modular<T> &lhs, U rhs) { return Modular<T>(lhs) *= rhs; }
template<typename T, typename U>
Modular<T> operator*(U lhs, const Modular<T> &rhs) { return Modular<T>(lhs) *= rhs; }
template<typename T>
Modular<T> operator/(const Modular<T> &lhs, const Modular<T> &rhs) { return Modular<T>(lhs) /= rhs; }
template<typename T, typename U>
Modular<T> operator/(const Modular<T> &lhs, U rhs) { return Modular<T>(lhs) /= rhs; }
template<typename T, typename U>
Modular<T> operator/(U lhs, const Modular<T> &rhs) { return Modular<T>(lhs) /= rhs; }
template<typename T, typename U>
Modular<T> power(const Modular<T> &a, const U &b) {
assert(b >= 0);
Modular<T> x = a, res = 1;
U p = b;
while (p > 0) {
if (p & 1) res *= x;
x *= x;
p >>= 1;
}
return res;
}
template<typename T>
bool IsZero(const Modular<T> &number) {
return number() == 0;
}
template<typename T>
string to_string(const Modular<T> &number) {
return to_string(number());
}
template<typename T>
std::ostream &operator<<(std::ostream &stream, const Modular<T> &number) {
return stream << number();
}
template<typename T>
std::istream &operator>>(std::istream &stream, Modular<T> &number) {
typename common_type<typename Modular<T>::Type, int64_t>::type x;
stream >> x;
number.value = Modular<T>::normalize(x);
return stream;
}
const int md = 1e9 + 7;
using Mint = Modular<std::integral_constant<decay<decltype(md)>::type, md>>;
ll sqr(ll x) {
return x * x;
}
int mysqrt(ll x) {
int l = 0, r = 1e9 + 1;
while (r - l > 1) {
int m = (l + r) / 2;
if (m * (ll) m <= x)
l = m;
else
r = m;
}
return l;
}
#ifdef ONPC
mt19937 rnd(513);
mt19937_64 rndll(231);
#else
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
mt19937_64 rndll(chrono::high_resolution_clock::now().time_since_epoch().count());
#endif
template<typename T>
T gcd(T a, T b) {
return a ? gcd(b % a, a) : b;
}
int gcdex(int a, int b, int &x, int &y) {
if (a == 0) {
x = 0;
y = 1;
return b;
}
int x1, y1;
int ret = gcdex(b % a, a, x1, y1);
x = y1 - (b / a) * x1;
y = x1;
return ret;
}
void setmin(int &x, int y) {
x = min(x, y);
}
void setmax(int &x, int y) {
x = max(x, y);
}
void setmin(ll &x, ll y) {
x = min(x, y);
}
void setmax(ll &x, ll y) {
x = max(x, y);
}
const ll llinf = 4e18 + 100;
const ld eps = 1e-9, PI = atan2(0, -1);
const int maxn = 1e5 + 100, maxw = 2e6 + 1111, inf = 1e9 + 100, sq = 450, LG = 18, mod = 1e9 + 933, mod1 = 1e9 + 993;
int main() {
#ifdef ONPC
freopen("../a.in", "r", stdin);
freopen("../a.out", "w", stdout);
#else
//freopen("a.in", "r", stdin);
//freopen("a.out", "w", stdout);
#endif // ONPC
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
ll ans = 0;
ll sum = 0;
for (int i = 0; i < n; i++) {
ll w;
cin >> w;
ans |= w;
sum += w;
ans |= sum;
}
cout << ans << '\n';
}
}
Editorialist's Solution
#include <bits/stdc++.h>
using namespace std;
void Solve() {
int n;
cin >> n;
vector<int> a(n);
for(int i = 0; i < n; i ++) {
cin >> a[i];
}
vector<int> dp(50, 0);
for(int i = 0; i < n; i ++) {
for(int j = 0; j < 30; j ++) {
dp[j] += a[i] % 2;
a[i] /= 2;
}
}
long long num = 0;
long long mul = 1;
for(int i = 0; i < 50; i ++) {
if(i) {
dp[i] += (dp[i-1]/2);
}
if(dp[i]) num += mul;
mul *= 2LL;
}
cout << num << "\n";
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int test_case = 1;
cin >> test_case;
for(int i = 1; i <= test_case; i ++) {
Solve();
}
return 0;
}
VIDEO EDITORIAL:
Feel free to share your approach. In case of any doubt or anything is unclear please ask it in the comment section. Any suggestions are welcomed. 
