結果
問題 | No.731 等差数列がだいすき |
ユーザー |
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提出日時 | 2018-09-07 21:42:00 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
WA
|
実行時間 | - |
コード長 | 2,125 bytes |
コンパイル時間 | 2,334 ms |
コンパイル使用メモリ | 199,068 KB |
最終ジャッジ日時 | 2025-01-06 12:56:06 |
ジャッジサーバーID (参考情報) |
judge3 / judge1 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 16 WA * 2 |
ソースコード
#include <bits/stdc++.h> using namespace std; using int64 = long long; int main() { int N, A[1000]; cin >> N; for(int i = 0; i < N; i++) cin >> A[i]; tuple< double, double, double > ret(1e30, 1e30, 1e30); { double left = -1e10, right = 1e10; auto check2 = [&](double real_a) { auto check = [&](double real_d) { double a = real_a; double cost = 0.0; for(int j = 0; j < N; j++) { cost += (A[j] - a) * (A[j] - a); a += real_d; } ret = min(ret, make_tuple(cost, real_a, real_d)); return cost; }; double low = -1e10, high = 1e10; for(int j = 0; j < 300; j++) { double latte = (low * 2 + high) / 3; double malta = (low + high * 2) / 3; if(check(latte) < check(malta)) high = malta; else low = latte; } return check(low); }; for(int i = 0; i < 300; i++) { double aa = (left * 2 + right) / 3; double bb = (left + right * 2) / 3; if(check2(aa) < check2(bb)) right = bb; else left = aa; } } { tuple< double, double, double > ret(1e30, 1e30, 1e30); double left = -40000000, right = 40000000; auto check2 = [&](double real_d) { auto check = [&](double a) { double b = a; double cost = 0.0; for(int j = 0; j < N; j++) { cost += (A[j] - a) * (A[j] - a); a += real_d; } ret = min(ret, make_tuple(cost, b, real_d)); return cost; }; double low = -40000000, high = 40000000; for(int j = 0; j < 300; j++) { double latte = (low * 2 + high) / 3; double malta = (low + high * 2) / 3; if(check(latte) < check(malta)) high = malta; else low = latte; } return check(low); }; for(int i = 0; i < 300; i++) { double aa = (left * 2 + right) / 3; double bb = (left + right * 2) / 3; if(check2(aa) < check2(bb)) right = bb; else left = aa; } } cout << fixed << setprecision(10) << get< 1 >(ret) << " " << get< 2 >(ret) << endl << get< 0 >(ret) << endl; }