結果

問題 No.731 等差数列がだいすき
ユーザー ei1333333ei1333333
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
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
sample AC * 3
other AC * 16 WA * 2
権限があれば一括ダウンロードができます

ソースコード

diff #

#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;
}
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