結果
| 問題 |
No.2766 Delicious Multiply Spice
|
| コンテスト | |
| ユーザー |
👑 |
| 提出日時 | 2024-05-31 21:39:51 |
| 言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
AC
|
| 実行時間 | 2 ms / 2,000 ms |
| コード長 | 2,638 bytes |
| コンパイル時間 | 4,280 ms |
| コンパイル使用メモリ | 260,344 KB |
| 最終ジャッジ日時 | 2025-02-21 17:37:53 |
|
ジャッジサーバーID (参考情報) |
judge2 / judge3 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 8 |
| other | AC * 31 |
ソースコード
#include<bits/stdc++.h>
#include<atcoder/all>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
using namespace atcoder;
typedef long long ll;
typedef vector<int> vi;
typedef vector<long long> vl;
typedef vector<vector<int>> vvi;
typedef vector<vector<long long>> vvl;
typedef long double ld;
typedef pair<int, int> P;
ostream& operator<<(ostream& os, const modint& a) {os << a.val(); return os;}
template <int m> ostream& operator<<(ostream& os, const static_modint<m>& a) {os << a.val(); return os;}
template <int m> ostream& operator<<(ostream& os, const dynamic_modint<m>& a) {os << a.val(); return os;}
template<typename T> istream& operator>>(istream& is, vector<T>& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;}
template<typename U, typename T> ostream& operator<<(ostream& os, const pair<U, T>& p){os << p.first << ' ' << p.second; return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<T>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : ""); return os;}
template<typename T> ostream& operator<<(ostream& os, const set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename T> ostream& operator<<(ostream& os, const unordered_set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename S, auto op, auto e> ostream& operator<<(ostream& os, const atcoder::segtree<S, op, e>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}
template<typename S, auto op, auto e, typename F, auto mapping, auto composition, auto id> ostream& operator<<(ostream& os, const atcoder::lazy_segtree<S, op, e, F, mapping, composition, id>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}
template<typename T> void chmin(T& a, T b){a = min(a, b);}
template<typename T> void chmax(T& a, T b){a = max(a, b);}
const long long e18 = 10000;
int main(){
long long n;
cin >> n;
map<long long, string> ma;
auto dfs = [&](auto dfs, long long x) -> void{
if((x - 1) % 2 == 0){
if(ma.find((x - 1) / 2) == ma.end()){
ma[(x - 1) / 2] = ma[x] + 'A';
dfs(dfs, (x - 1) / 2);
}
}
if((x - 1) % 3 == 0){
if(ma.find((x - 1) / 3) == ma.end()){
ma[(x - 1) / 3] = ma[x] + 'B';
dfs(dfs, (x - 1) / 3);
}
}
};
dfs(dfs, n);
string s = ma[1];
reverse(s.begin(), s.end());
cout << s << "\n";
return 0;
}