이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h> // Tomasz Nowak
using namespace std; // XIII LO Szczecin
using LL = long long; // Poland
#define FOR(i, l, r) for(int i = (l); i <= (r); ++i)
#define REP(i, n) FOR(i, 0, (n) - 1)
template<class T> int size(T &&x) {
return int(x.size());
}
template<class A, class B> ostream& operator<<(ostream &out, const pair<A, B> &p) {
return out << '(' << p.first << ", " << p.second << ')';
}
template<class T> auto operator<<(ostream &out, T &&x) -> decltype(x.begin(), out) {
out << '{';
for(auto it = x.begin(); it != x.end(); ++it)
out << *it << (it == prev(x.end()) ? "" : ", ");
return out << '}';
}
void dump() {}
template<class T, class... Args> void dump(T &&x, Args... args) {
cerr << x << "; ";
dump(args...);
}
#ifdef DEBUG
struct Nl{~Nl(){cerr << '\n';}};
# define debug(x...) cerr << (strcmp(#x, "") ? #x ": " : ""), dump(x), Nl(), cerr << ""
#else
# define debug(...) 0 && cerr
#endif
mt19937_64 rng(0);
int rd(int l, int r) {
return uniform_int_distribution<int>(l, r)(rng);
}
// end of templates
int n;
vector<int> curr_vec;
int curr_mul = 1;
set<int> answers;
bool can_mul(int a, int b) {
return a <= n / b and a * b <= n;
}
void backtrack() {
debug(curr_mul, curr_vec);
if(curr_mul == n) {
answers.emplace(accumulate(curr_vec.begin(), curr_vec.end(), 0));
return;
}
// add new
if(can_mul(curr_mul, 2)) {
curr_mul *= 2;
curr_vec.emplace_back(1);
backtrack();
curr_mul /= 2;
curr_vec.pop_back();
}
// increase last number
if(size(curr_vec) == 1 or (size(curr_vec) >= 2 and curr_vec.back() + 1 <= curr_vec.end()[-2]))
if(can_mul(curr_mul / (curr_vec.back() + 1), curr_vec.back() + 2)) {
curr_mul /= curr_vec.back() + 1;
++curr_vec.back();
curr_mul *= curr_vec.back() + 1;
backtrack();
curr_mul /= curr_vec.back() + 1;
--curr_vec.back();
curr_mul *= curr_vec.back() + 1;
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cin >> n;
backtrack();
cout << size(answers) << '\n';
for(int a : answers)
cout << a << ' ';
cout << '\n';
}
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