結果

問題 No.887 Collatz
ユーザー hitonanodehitonanode
提出日時 2019-09-20 21:24:46
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 2 ms / 2,000 ms
コード長 4,241 bytes
コンパイル時間 1,583 ms
コンパイル使用メモリ 166,500 KB
実行使用メモリ 5,376 KB
最終ジャッジ日時 2024-09-14 16:11:14
合計ジャッジ時間 2,420 ms
ジャッジサーバーID
(参考情報)
judge3 / judge4
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 28
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include <bits/stdc++.h>
using namespace std;
using lint = long long int;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
struct fast_ios
{
fast_ios()
{
cin.tie(0);
ios::sync_with_stdio(false);
cout << fixed << setprecision(20);
};
} fast_ios_;
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((lint)(x).size())
#define POW2(n) (1LL << (n))
#define FOR(i, begin, end) for (int i = (begin), i##_end_ = (end); i < i##_end_; i++)
#define IFOR(i, begin, end) for (int i = (end)-1, i##_begin_ = (begin); i >= i##_begin_; i--)
#define REP(i, n) FOR(i, 0, n)
#define IREP(i, n) IFOR(i, 0, n)
template <typename T>
void ndarray(vector<T> &vec, int len)
{
vec.resize(len);
}
template <typename T, typename... Args>
void ndarray(vector<T> &vec, int len, Args... args)
{
vec.resize(len);
for (auto &v : vec)
ndarray(v, args...);
}
template <typename T>
bool mmax(T &m, const T q)
{
if (m < q)
{
m = q;
return true;
}
else
return false;
}
template <typename T>
bool mmin(T &m, const T q)
{
if (m > q)
{
m = q;
return true;
}
else
return false;
}
template <typename T1, typename T2>
pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first + r.first, l.second + r.second); }
template <typename T1, typename T2>
pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first - r.first, l.second - r.second); }
template <typename T>
istream &operator>>(istream &is, vector<T> &vec)
{
for (auto &v : vec)
is >> v;
return is;
}
///// This part below is only for debug, not used /////
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &vec)
{
os << "[";
for (auto v : vec)
os << v << ",";
os << "]";
return os;
}
template <typename T>
ostream &operator<<(ostream &os, const deque<T> &vec)
{
os << "deq[";
for (auto v : vec)
os << v << ",";
os << "]";
return os;
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &vec)
{
os << "{";
for (auto v : vec)
os << v << ",";
os << "}";
return os;
}
template <typename T>
ostream &operator<<(ostream &os, const unordered_set<T> &vec)
{
os << "{";
for (auto v : vec)
os << v << ",";
os << "}";
return os;
}
template <typename T>
ostream &operator<<(ostream &os, const multiset<T> &vec)
{
os << "{";
for (auto v : vec)
os << v << ",";
os << "}";
return os;
}
template <typename T>
ostream &operator<<(ostream &os, const unordered_multiset<T> &vec)
{
os << "{";
for (auto v : vec)
os << v << ",";
os << "}";
return os;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &pa)
{
os << "(" << pa.first << "," << pa.second << ")";
return os;
}
template <typename TK, typename TV>
ostream &operator<<(ostream &os, const map<TK, TV> &mp)
{
os << "{";
for (auto v : mp)
os << v.first << "=>" << v.second << ",";
os << "}";
return os;
}
template <typename TK, typename TV>
ostream &operator<<(ostream &os, const unordered_map<TK, TV> &mp)
{
os << "{";
for (auto v : mp)
os << v.first << "=>" << v.second << ",";
os << "}";
return os;
}
#define dbg(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << endl;
///// END /////
/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
using namespace __gnu_pbds; // find_by_order(), order_of_key()
template<typename TK> using pbds_set = tree<TK, null_type, less<TK>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename TK, typename TV> using pbds_map = tree<TK, TV, less<TK>, rb_tree_tag, tree_order_statistics_node_update>;
*/
int main()
{
lint n, i = 0, maxi;
cin >> n;
lint m = n;
maxi = n;
while (m != 1)
{
i++;
if (m % 2) m = m * 3 + 1;
else m /= 2;
}
cout << i << endl;
REP(_, i)
{
if (n % 2) n = n * 3 + 1;
else n /= 2;
mmax(maxi, n);
}
cout << maxi << endl;
}
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