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problems/PA-05/Atcoder/abc184_f/editorial.md

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+# F - Programming Contest (editorial)
+
+Source: https://atcoder.jp/contests/abc184/editorial/376 by evima
+
+We can see that this problem is NP-hard, since it is a difficult version of the subset sum problem.
+
+Let us focus on the fairly small \(N\) (\(N \leq 40\)).
+
+If you search all the combinations of tasks to solve, it takes \(O(2^N)\) and it will TLE. Instead, we can use a technique called “meet-in-the-middle” searching.
+
+1. Split a sequence \(A\) into two so that the number of elements are halved. Let us call them \(B\) and \(C\).
+2. Enumerate all the possible sums of combinations from \(B\) and sort them, which we will call \(D\). It takes \(O(N 2^{N/2})\) time for this.
+3. Brute force all the possible combinations from \(C\) and obtain the maximum value, with \(B\) combined. Here, the optimal choice of \(B\) can be found by performing binary searching on \(D\). The time complexity for these operations are \(O(N 2^{N/2})\).
+
+Now, the problem can be solved in a total of \(O(N 2^{N/2})\) time.
+
+We can improve each step so that the problem can be solved in a total of \(O(2^{N/2})\) time.
+
+Let \(D_i\) be a sorted sequence, each of whose element is a possible sum of choices from \(B_1, B_2, \ldots, B_i\). Initially, \(D_0 = [0]\). \(D_x\) can be obtained by merging \(D_{x-1}\) and “\(D_{x-1}\), each of whose element is added by \(B_x\)”. Since \(|D_0| + |D_1| + |D_2| + \cdots + |D_x| \leq 1 + 2 + 4 + \cdots + 2^x = O(2^x)\), we can obtain \(D_{N/2}\) in a \(O(2^{N/2})\) time.
+
+Enumerate all the possible sums of choices from \(B\), enumerate all the possible sums of choices from \(C\) too, and combine them with sliding windows, so that the problem can be solved in a total of \(O(2^{N/2})\) time.

problems/PA-05/Atcoder/abc184_f/judge.json

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+{
+  "time_limit_s": 2.0,
+  "time_limit_source": "task page",
+  "memory_mb": 1024,
+  "memory_source": "task page",
+  "mode": "exact",
+  "validator_flags": "case_sensitive",
+  "comparison": "token-wise, whitespace-insensitive, case-sensitive (AtCoder's checker)",
+  "validator": null,
+  "tests_dir": "data",
+  "answer_suffix": ".out",
+  "n_sample": null,
+  "n_secret": 35,
+  "note": "no sample/secret split: the statement's samples are among the numbered tests"
+}

problems/PA-05/Atcoder/abc184_f/meta.json

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+{
+  "task_id": "abc184_f",
+  "contest": "abc184",
+  "type": "ABC",
+  "letter": "F",
+  "title": "Programming Contest",
+  "url": "https://atcoder.jp/contests/abc184/tasks/abc184_f",
+  "difficulty": 1423,
+  "score": 600,
+  "time_limit_sec": 2.0,
+  "memory_limit_mb": 1024,
+  "n_tests": 35,
+  "editorial_url": "https://atcoder.jp/contests/abc184/editorial/376",
+  "code_source": "model",
+  "pid": "A1482"
+}

problems/PA-05/Atcoder/abc184_f/solution.cpp

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+// abc184_f: F - Programming Contest
+// Written by a model (gpt-6-astra) from the official editorial, https://atcoder.jp/contests/abc184/editorial/376
+// Judged AC on all 35 tests, max 0.04 s of the 2 s limit (submission 1).
+
+#include <bits/stdc++.h>
+using namespace std;
+
+using ll = long long;
+
+vector<ll> subset_sums(const vector<ll>& a, int l, int r) {
+    vector<ll> sums{0};
+    for (int k = l; k < r; ++k) {
+        size_t n = sums.size();
+        vector<ll> next;
+        next.reserve(2 * n);
+        size_t i = 0, j = 0;
+        while (i < n || j < n) {
+            if (j == n || (i < n && sums[i] <= sums[j] + a[k])) {
+                next.push_back(sums[i++]);
+            } else {
+                next.push_back(sums[j++] + a[k]);
+            }
+        }
+        sums.swap(next);
+    }
+    return sums;
+}
+
+int main() {
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int N;
+    ll T;
+    cin >> N >> T;
+    vector<ll> A(N);
+    for (ll& x : A) cin >> x;
+
+    vector<ll> left = subset_sums(A, 0, N / 2);
+    vector<ll> right = subset_sums(A, N / 2, N);
+
+    ll answer = 0;
+    int j = static_cast<int>(right.size()) - 1;
+    for (ll x : left) {
+        while (j >= 0 && x + right[j] > T) --j;
+        if (j < 0) break;
+        answer = max(answer, x + right[j]);
+        if (answer == T) break;
+    }
+    cout << answer << '\n';
+    return 0;
+}

problems/PA-05/Atcoder/abc184_f/statement.md

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+# F - Programming Contest
+
+**Score:** 600 points
+
+## Problem Statement
+Takahashi will participate in a programming contest, which lasts for \(T\) minutes and presents \(N\) problems.<br />
+With his extrasensory perception, he already knows that it will take \(A_i\) minutes to solve the \(i\)-th problem.<br />
+He will choose zero or more problems to solve from the \(N\) problems so that it takes him no longer than \(T\) minutes in total to solve them.<br />
+Find the longest possible time it takes him to solve his choice of problems.
+
+## Constraints
+- All values in input are integers.
+- \(1 \le N \le 40\)
+- \(1 \le T \le 10^9\)
+- \(1 \le A_i \le 10^9\)
+
+## Input
+Input is given from Standard Input in the following format:
+```
+N T
+A_1 ... A_N
+```
+
+## Output
+Print the answer as an integer.
+
+## Sample Input 1
+```
+5 17
+2 3 5 7 11
+```
+
+### Sample Output 1
+```
+17
+```
+>If he chooses the 1-st, 2-nd, 3-rd, and 4-th problems, it takes him \(2+3+5+7=17\) minutes in total to solve them, which is the longest possible time not exceeding \(T=17\) minutes.
+
+---
+
+## Sample Input 2
+```
+6 100
+1 2 7 5 8 10
+```
+
+### Sample Output 2
+```
+33
+```
+>It is optimal to solve all the problems.
+
+---
+
+## Sample Input 3
+```
+6 100
+101 102 103 104 105 106
+```
+
+### Sample Output 3
+```
+0
+```
+>He cannot solve any of the problems.
+
+---
+
+## Sample Input 4
+```
+7 273599681
+6706927 91566569 89131517 71069699 75200339 98298649 92857057
+```
+
+### Sample Output 4
+```
+273555143
+```
+>If he chooses the 2-nd, 3-rd, and 7-th problems, it takes him \(273555143\) minutes in total to solve them.

problems/PA-05/Atcoder/abc196_d/data/1.in

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problems/PA-05/Atcoder/abc196_d/data/1.out

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problems/PA-05/Atcoder/abc196_d/data/12.in

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problems/PA-05/Atcoder/abc196_d/data/21.in

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problems/PA-05/Atcoder/abc196_d/data/22.out

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