結果
問題 | No.2483 Yet Another Increasing XOR Problem |
ユーザー |
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提出日時 | 2023-09-22 23:56:16 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 23 ms / 2,000 ms |
コード長 | 2,744 bytes |
コンパイル時間 | 4,151 ms |
コンパイル使用メモリ | 260,520 KB |
最終ジャッジ日時 | 2025-02-17 01:11:49 |
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 * 2 |
other | AC * 23 |
コンパイルメッセージ
main.cpp: In function ‘std::pair<atcoder::static_modint<998244353>, atcoder::static_modint<998244353> > solve(ll, int)’: main.cpp:103:1: warning: control reaches end of non-void function [-Wreturn-type] 103 | } | ^
ソースコード
#include<bits/stdc++.h>using namespace std;//* ATCODER#include<atcoder/all>using namespace atcoder;typedef modint998244353 mint;//*//* BOOST MULTIPRECISION#include<boost/multiprecision/cpp_int.hpp>using namespace boost::multiprecision;//*/typedef long long ll;#define rep(i, s, n) for (int i = (int)(s); i < (int)(n); i++)#define rrep(i, s, n) for (int i = (int)(n)-1; i >= (int)(s); i--)template <typename T> bool chmin(T &a, const T &b) {if (a <= b) return false;a = b;return true;}template <typename T> bool chmax(T &a, const T &b) {if (a >= b) return false;a = b;return true;}template <typename T> T max(vector<T> &a){assert(!a.empty());T ret = a[0];for (int i=0; i<(int)a.size(); i++) chmax(ret, a[i]);return ret;}template <typename T> T min(vector<T> &a){assert(!a.empty());T ret = a[0];for (int i=0; i<(int)a.size(); i++) chmin(ret, a[i]);return ret;}template <typename T> T sum(vector<T> &a){T ret = 0;for (int i=0; i<(int)a.size(); i++) ret += a[i];return ret;}//defmodfactconst int COMinitMAX = 998244;mint fact[COMinitMAX+1], factinv[COMinitMAX+1];void modfact(){fact[0] = 1;for (int i=1; i<=COMinitMAX; i++){fact[i] = fact[i-1] * i;}factinv[COMinitMAX] = fact[COMinitMAX].inv();for (int i=COMinitMAX-1; i>=0; i--){factinv[i] = factinv[i+1] * (i+1);}}mint cmb(int a, int b){if (a<b || b<0) return mint(0);return fact[a]*factinv[b]*factinv[a-b];}//--------const mint inv2 = mint(2).inv();map<pair<ll,int>, pair<mint, mint>> mp;// kosu, anspair<mint,mint> solve(ll m, int k){if (m < 0) return pair(0, 0);chmin(m, (1LL << k+1));if (mp.find(pair(m, k)) != mp.end()){return mp[pair(m, k)];}//cout << m << ' ' << k << endl;if (k == 0){if (m == 0){return pair(0, 0);}else if(m == 1){return pair(2, 2);}else if(m >= 2){return pair(4, 4 + inv2);}}else{pair<mint, mint> ret1 = solve(m, k-1);pair<mint, mint> ret2 = solve(m - (1LL << k), k-1);pair<mint, mint> ret =\pair(ret1.first * 2 + ret2.first * 2,\ret1.second * 2\+ ret2.second * inv2.pow(1LL << k)\+ ret2.second * mint(2).pow(1LL << k));mp[pair(m, k)] = ret;return ret;}}int main(){ll k = 0;ll m; cin >> m;while((1LL << k) < m){k++;}if (k == 0){cout << 1 << '\n';return 0;}mint teisu1 = mint(2).pow((1LL << k) - 2) * inv2.pow((1LL << k-1));mint ans = mint(2).pow(1LL << k-1) - 1;//cout << ans.val() << '\n';ans += mint(2).pow((1LL << k-1) - (m - (1LL << k-1))) * (mint(2).pow(m - (1LL << k-1)) - 1);//cout << ans.val() << '\n';//cout << solve(m - (1LL << k-1), k-1).second.val() << '\n';ans += solve(m - (1LL << k-1), k-1).second * teisu1;cout << ans.val() << '\n';}