{ "componentChunkName": "component---src-templates-post-js", "path": "/er-fen-tan-suo-noying-yong/", "result": {"data":{"ghostPost":{"id":"Ghost__Post__60fa72003986b000013a3951","title":"二分探索の応用","slug":"er-fen-tan-suo-noying-yong","featured":false,"feature_image":"https://ghost.tech.anti-pattern.co.jp/content/images/2022/04/1_XGofnd5bRfFGPVvXDCnuhg.png","excerpt":"皆さんご存知の二分探索。この応用について今回は書いていきたいと思う。\n\n二分探索の一般化\n二分探索とはつまり、範囲を特定することである。\n\n例億マス計算\nそこで今回は二分探索を使う下記問題を解いていく\n\nC - 億マス計算AtCoder is a programming contest site for anyone from beginners to\nexperts. We hold weekly programming contests online.AtCoderAtCoder Inc.\n[https://atcoder.jp/contests/arc037/tasks/arc037_c]\n\n愚直に2重ループで解くと\n\nO(N²)になるため、時間が足りなくなる\n\nblog-binary_search_2blog-binary_search_2. GitHub Gist: instantly share code,\nnotes, and snippets.Gist262588213843476\n[https://gist.github.com/kooooohe/8cc33e0f1e","custom_excerpt":null,"visibility":"public","created_at_pretty":"23 July, 2021","published_at_pretty":"15 June, 2021","updated_at_pretty":"21 April, 2022","created_at":"2021-07-23T16:38:40.000+09:00","published_at":"2021-06-16T00:00:00.000+09:00","updated_at":"2022-04-21T17:19:32.000+09:00","meta_title":null,"meta_description":null,"og_description":null,"og_image":null,"og_title":null,"twitter_description":null,"twitter_image":null,"twitter_title":null,"authors":[{"name":"Kohei Kondo","slug":"kooooohe","bio":null,"profile_image":"https://ghost.tech.anti-pattern.co.jp/content/images/2022/04/MVIMG_20180910_102813.png","twitter":"@kooooohe_","facebook":null,"website":null}],"primary_author":{"name":"Kohei Kondo","slug":"kooooohe","bio":null,"profile_image":"https://ghost.tech.anti-pattern.co.jp/content/images/2022/04/MVIMG_20180910_102813.png","twitter":"@kooooohe_","facebook":null,"website":null},"primary_tag":{"name":"C++","slug":"c","description":null,"feature_image":null,"meta_description":null,"meta_title":null,"visibility":"public"},"tags":[{"name":"C++","slug":"c","description":null,"feature_image":null,"meta_description":null,"meta_title":null,"visibility":"public"},{"name":"algorithm","slug":"algorithm","description":null,"feature_image":null,"meta_description":null,"meta_title":null,"visibility":"public"},{"name":"binary_search","slug":"binary_search","description":null,"feature_image":null,"meta_description":null,"meta_title":null,"visibility":"public"}],"plaintext":"皆さんご存知の二分探索。この応用について今回は書いていきたいと思う。\n\n二分探索の一般化\n二分探索とはつまり、範囲を特定することである。\n\n例億マス計算\nそこで今回は二分探索を使う下記問題を解いていく\n\nC - 億マス計算AtCoder is a programming contest site for anyone from beginners to\nexperts. We hold weekly programming contests online.AtCoderAtCoder Inc.\n[https://atcoder.jp/contests/arc037/tasks/arc037_c]\n\n愚直に2重ループで解くと\n\nO(N²)になるため、時間が足りなくなる\n\nblog-binary_search_2blog-binary_search_2. GitHub Gist: instantly share code,\nnotes, and snippets.Gist262588213843476\n[https://gist.github.com/kooooohe/8cc33e0f1eb035b878db003b0779a71b]#include \n#include \n#include \n\nusing namespace std;\n\nint main() {\n\n int N, M;\n cin >> N >> M;\n vector a(N);\n vector b(N);\n vector r(N * N);\n\n for (int i = 0; i < N; ++i) {\n cin >> a[i];\n }\n for (int i = 0; i < N; ++i) {\n cin >> b[i];\n }\n\n int ind = 0;\n // O(N)\n for (int i = 0; i < N; ++i) {\n // O(N)\n for (int j = 0; j < N; ++j) {\n r[ind] = a[i] * b[j];\n ind++;\n }\n }\n // O(log_N^2)\n sort(r.begin(), r.end());\n cout\n\n\n\nこれを 二分探索 in 二分探索でとくと\n\nO(log_(MAX(A)*MAX(B))*N*log_N)\n\nで解けるようになる。\n\n今回、 std:upper_bound を使う。\n\nこれは、指定した数よりも大きい最初のイテレータを返す。これを使用し、x以下の個数を求めることができる。\n\n※内部で二分探索を使用している\n\n> 詳細はこちら https://cpprefjp.github.io/reference/algorithm/upper_bound.html\nstd:upper_bound とstd:lower_boundの使い方は下記がわかりやすい\n\n\n\nlower_boundとupper_boundの使い方 - Qiita1.lower_boundとupper_bound\nlower_boundとupper_boundはC++のSTLライブラリの関数なのじゃ… 俗に言う二分探索に似たやつなのじゃ… 違いとしては\nlower_boundは...Qiitaganyariya\n[https://qiita.com/ganyariya/items/33f1326154b85db465c3]\n\n考え方ざっくり\n\n * 掛け算した結果をもとに二分探索する(外側)\n * ↑の結果をa[i]で割った数値以下の数値を二分探索で探し、カウントする\n\nblog-binary_searchblog-binary_search. GitHub Gist: instantly share code, notes,\nand snippets.Gist262588213843476\n[https://gist.github.com/kooooohe/01db8a05584db938eb8e2e9f52e4e743]\n\n#include \n#include \n#include \n \nusing namespace std;\n \nconst long long INF = 1LL << 60;\nint main() {\n \n int N, M;\n cin >> N >> M;\n vector a(N);\n vector b(N);\n \n for (int i = 0; i < N; ++i) {\n cin >> a[i];\n }\n for (int i = 0; i < N; ++i) {\n cin >> b[i];\n }\n // O(log_N)\n sort(b.begin(), b.end());\n \n long long left = 0;\n long long right = INF;\n \n // O(log_(MAX(A)*MAX(B)))\n while (right - left > 1) {\n long long mid = (right + left) / 2;\n long long cnt = 0;\n // O(N)\n for (int i = 0; i < N; ++i) {\n // O(log_N)\n cnt += upper_bound(b.begin(), b.end(), mid / a[i]) - b.begin();\n }\n if (cnt >= M) {\n right = mid;\n } else {\n left = mid;\n }\n }\n \n cout << right << endl;\n}","html":"

皆さんご存知の二分探索。この応用について今回は書いていきたいと思う。

二分探索の一般化

二分探索とはつまり、範囲を特定することである。

例

億マス計算

そこで今回は二分探索を使う下記問題を解いていく

C - 億マス計算
AtCoder is a programming contest site for anyone from beginners to experts. We hold weekly programming contests online.
AtCoderAtCoder Inc.

愚直に2重ループで解くと

O(N²)になるため、時間が足りなくなる

blog-binary_search_2
blog-binary_search_2. GitHub Gist: instantly share code, notes, and snippets.
Gist262588213843476
#include <algorithm>\n#include <iostream>\n#include <vector>\n\nusing namespace std;\n\nint main() {\n\n  int N, M;\n  cin >> N >> M;\n  vector<long long> a(N);\n  vector<long long> b(N);\n  vector<long long> r(N * N);\n\n  for (int i = 0; i < N; ++i) {\n    cin >> a[i];\n  }\n  for (int i = 0; i < N; ++i) {\n    cin >> b[i];\n  }\n\n  int ind = 0;\n  // O(N)\n  for (int i = 0; i < N; ++i) {\n    // O(N)\n    for (int j = 0; j < N; ++j) {\n      r[ind] = a[i] * b[j];\n      ind++;\n    }\n  }\n  // O(log_N^2)\n  sort(r.begin(), r.end());\n  cout\n\n
\n

これを 二分探索 in 二分探索でとくと

O(log_(MAX(A)*MAX(B))*N*log_N)

で解けるようになる。

今回、 std:upper_bound を使う。

これは、指定した数よりも大きい最初のイテレータを返す。これを使用し、x以下の個数を求めることができる。

※内部で二分探索を使用している

詳細はこちら https://cpprefjp.github.io/reference/algorithm/upper_bound.html

std:upper_bound とstd:lower_boundの使い方は下記がわかりやすい

lower_boundとupper_boundの使い方 - Qiita
1.lower_boundとupper_bound lower_boundとupper_boundはC++のSTLライブラリの関数なのじゃ… 俗に言う二分探索に似たやつなのじゃ… 違いとしては lower_boundは...
Qiitaganyariya

考え方ざっくり

  • 掛け算した結果をもとに二分探索する(外側)
  • ↑の結果をa[i]で割った数値以下の数値を二分探索で探し、カウントする
blog-binary_search
blog-binary_search. GitHub Gist: instantly share code, notes, and snippets.
Gist262588213843476

#include <algorithm>\n#include <iostream>\n#include <vector>\n \nusing namespace std;\n \nconst long long INF = 1LL << 60;\nint main() {\n \n  int N, M;\n  cin >> N >> M;\n  vector<long long> a(N);\n  vector<long long> b(N);\n \n  for (int i = 0; i < N; ++i) {\n    cin >> a[i];\n  }\n  for (int i = 0; i < N; ++i) {\n    cin >> b[i];\n  }\n  // O(log_N)\n  sort(b.begin(), b.end());\n \n  long long left = 0;\n  long long right = INF;\n \n  // O(log_(MAX(A)*MAX(B)))\n  while (right - left > 1) {\n    long long mid = (right + left) / 2;\n    long long cnt = 0;\n    // O(N)\n    for (int i = 0; i < N; ++i) {\n      // O(log_N)\n      cnt += upper_bound(b.begin(), b.end(), mid / a[i]) - b.begin();\n    }\n    if (cnt >= M) {\n      right = mid;\n    } else {\n      left = mid;\n    }\n  }\n \n  cout << right << endl;\n}\n
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