結果

問題 No.1179 Quadratic Equation
コンテスト
ユーザー kaikey
提出日時 2020-08-21 23:15:45
言語 C++14
(gcc 13.3.0 + boost 1.89.0)
結果
WA  
実行時間 -
コード長 1,830 bytes
記録
記録タグの例:
初AC ショートコード 純ショートコード 純主流ショートコード 最速実行時間
コンパイル時間 1,486 ms
コンパイル使用メモリ 167,608 KB
実行使用メモリ 6,824 KB
最終ジャッジ日時 2024-10-15 06:19:47
合計ジャッジ時間 2,067 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge2
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 * 7 WA * 4
権限があれば一括ダウンロードができます
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 double) [with _CharT = char; _Traits = std::char_traits<char>]',
    inlined from 'int main()' at main.cpp:52:31:
/home/linuxbrew/.linuxbrew/Cellar/gcc@12/12.3.0/include/c++/12/ostream:233:25: warning: 'ans1' may be used uninitialized [-Wmaybe-uninitialized]
  233 |       { return _M_insert(__f); }
      |                ~~~~~~~~~^~~~~
main.cpp: In function 'int main()':
main.cpp:21:58: note: 'ans1' was declared here
   21 |                 long double lb = -1000, ub = -B / 2 * A, ans1, ans2;
      |                                                          ^~~~

ソースコード

diff #
raw source code 

#include "bits/stdc++.h"
#include <random>
using namespace std; typedef unsigned long long _ulong; typedef long long int lint; typedef pair<lint, lint> plint; typedef pair<double long, double long> pld;
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((lint)(x).size())
#define FOR(i, begin, end) for(lint i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(lint 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)
#define endk '\n'
template<class T>bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; }
const lint MOD = 1e9 + 7, INF = 1e18;
lint dx[8] = { 0, -1, 1, 0, 1, -1, 1, -1 }, dy[8] = { 1, 0, 0, -1, -1, -1, 1, 1 };

double A, B, C;
int main() {
	cin >> A >> B >> C;
	if (B * B < 4 * A * C) cout << "imaginary" << endk;
	else if(B *B != 4 * A * C){
		long double lb = -1000, ub = -B / 2 * A, ans1, ans2;
		REP(i, 1000) {
			long double mid = (lb + ub) / 2;
			if (A * mid * mid + B * mid + C <= 0) lb = mid;
			else ub = mid;
		}
		if (lb == -1000) {
			lb = -1000, ub = -B / 2 * A;
			REP(i, 1000) {
				long double mid = (lb + ub) / 2;
				if (A * mid * mid + B * mid + C >= 0) lb = mid;
				else ub = mid;
			}
			ans1 = lb;
		}
		lb = -B / 2 * A, ub = 1000;
		REP(i, 1000) {
			long double mid = (lb + ub) / 2;
			if (A * mid * mid + B * mid + C <= 0) lb = mid;
			else ub = mid;
		}
		ans2 = lb;
		if (lb == 1000) {
			lb = -B / 2 * A, ub = 1000;
			REP(i, 1000) {
				long double mid = (lb + ub) / 2;
				if (A * mid * mid + B * mid + C >= 0) lb = mid;
				else ub = mid;
			}
			ans2 = ub;
		}
		cout << setprecision(30) << ans1 << " " << ans2 << endk;
	}
	else {
		cout << setprecision(30) << -B  / 2 * A << endk;
	}
}
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