結果
問題 | No.2241 Reach 1 |
ユーザー |
|
提出日時 | 2023-03-10 22:02:15 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 20 ms / 2,000 ms |
コード長 | 1,460 bytes |
コンパイル時間 | 1,761 ms |
コンパイル使用メモリ | 192,432 KB |
最終ジャッジ日時 | 2025-02-11 08:19:42 |
ジャッジサーバーID (参考情報) |
judge5 / judge3 |
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ファイルパターン | 結果 |
---|---|
other | AC * 35 |
ソースコード
#include "bits/stdc++.h"using namespace std;#define all(x) begin(x),end(x)template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << '(' << p.first << ", " << p.second << ')'; }template<typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type> ostream&operator<<(ostream &os, const T_container &v) { string sep; for (const T &x : v) os << sep << x, sep = " "; return os; }#define debug(a) cerr << "(" << #a << ": " << a << ")\n";typedef long long ll;typedef vector<int> vi;typedef vector<vi> vvi;typedef pair<int,int> pi;const int mxN = 1e5+1, oo = 1e9;int main() {int N; cin >> N;// remove trailing zeroes// do times a number + 1// need to make number of bits less.// it's possible in log(n) moves, have only few choices?// always do towards that.// multiplication by 3 could cancel a lot...// 6 do towards that 3 and then 2 left.// multiplication is so chaotic!// want to get (2^k-1)/something = my number// (2^k - 1)(2^k+1)// can do +1 anyway, gives a new bound.// want to get 2^b - 2^a -1. A lot of freedom...// how to get there? 100000000// how to get there?if(1<<__lg(N)==N) {cout << "1\n";} else {int ans = 3; // shift then multiply to get 2-power - 1 + 1if(N%2==1) ans--;cout << ans << '\n';}}