結果
問題 | No.526 フィボナッチ数列の第N項をMで割った余りを求める |
ユーザー | Takuto-Onikubo |
提出日時 | 2022-12-30 23:04:22 |
言語 | C++14 (gcc 12.3.0 + boost 1.83.0) |
結果 |
AC
|
実行時間 | 73 ms / 2,000 ms |
コード長 | 449 bytes |
コンパイル時間 | 1,310 ms |
コンパイル使用メモリ | 165,460 KB |
実行使用メモリ | 5,376 KB |
最終ジャッジ日時 | 2024-05-04 11:04:38 |
合計ジャッジ時間 | 2,236 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge2 |
(要ログイン)
テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 1 ms
5,248 KB |
testcase_01 | AC | 2 ms
5,376 KB |
testcase_02 | AC | 2 ms
5,376 KB |
testcase_03 | AC | 2 ms
5,376 KB |
testcase_04 | AC | 2 ms
5,376 KB |
testcase_05 | AC | 2 ms
5,376 KB |
testcase_06 | AC | 1 ms
5,376 KB |
testcase_07 | AC | 2 ms
5,376 KB |
testcase_08 | AC | 2 ms
5,376 KB |
testcase_09 | AC | 3 ms
5,376 KB |
testcase_10 | AC | 17 ms
5,376 KB |
testcase_11 | AC | 73 ms
5,376 KB |
testcase_12 | AC | 73 ms
5,376 KB |
testcase_13 | AC | 71 ms
5,376 KB |
testcase_14 | AC | 71 ms
5,376 KB |
コンパイルメッセージ
In file included from /home/linuxbrew/.linuxbrew/Cellar/gcc@12/12.3.0/include/c++/12/istream:39, from /home/linuxbrew/.linuxbrew/Cellar/gcc@12/12.3.0/include/c++/12/sstream:38, from /home/linuxbrew/.linuxbrew/Cellar/gcc@12/12.3.0/include/c++/12/complex:45, from /home/linuxbrew/.linuxbrew/Cellar/gcc@12/12.3.0/include/c++/12/ccomplex:39, from /home/linuxbrew/.linuxbrew/Cellar/gcc@12/12.3.0/include/c++/12/x86_64-pc-linux-gnu/bits/stdc++.h:54, from main.cpp:1: In member function 'std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(long long int) [with _CharT = char; _Traits = std::char_traits<char>]', inlined from 'int main()' at main.cpp:22:11: /home/linuxbrew/.linuxbrew/Cellar/gcc@12/12.3.0/include/c++/12/ostream:202:25: warning: 'pre' may be used uninitialized [-Wmaybe-uninitialized] 202 | { return _M_insert(__n); } | ~~~~~~~~~^~~~~ main.cpp: In function 'int main()': main.cpp:12:6: note: 'pre' was declared here 12 | ll pre, prepre; | ^~~
ソースコード
#include <bits/stdc++.h> #define rep(i,n) for(int i=0; i<n; ++i) using namespace std; using std::abs; using P = pair<long long, long long>; typedef long long ll; const ll INF = 100000000000000000LL; int main(void) { ll N, M; cin >> N >> M; ll pre, prepre; rep(i, N) { if(i == 0) prepre = 0; else if(i == 1) pre = 1; else { ll tmp = (prepre + pre) % M; prepre = pre; pre = tmp; } } cout << pre << endl; }