Submission #39414

#TimeUsernameProblemLanguageResultExecution timeMemory
39414farmersriceIce Hockey World Championship (CEOI15_bobek)C++14
50 / 100
1000 ms395436 KiB
#include <bits/stdc++.h> //#pragma GCC optimize ("O3") //#pragma GCC target ("sse4") //#pragma GCC target ("avx,tune=native") //Use above if bruteforcing with lots of small operations. Or just use it anytime, there's no downside. AVX is better slightly /* TASK: hidden LANG: C++11 */ using namespace std; typedef long long ll; typedef pair<int, int> pair2; typedef pair<int, pair<int, int> > pair3; typedef pair<int, pair<int, pair<int, int> > > pair4; #define MAXN 43 #define INF 1000000000000000000LL #define mp make_pair #define add push_back #define remove pop int n; ll values[MAXN]; ll set1[MAXN]; ll set2[MAXN]; int size1, size2; ll m; vector<ll> one; vector<ll> two; void dfs1(int depth, ll sum) { if (depth == size1) { one.add(sum); return; } dfs1(depth + 1, sum + set1[depth]); dfs1(depth + 1, sum); } void dfs2(int depth, ll sum) { if (depth == size2) { two.add(sum); return; } dfs2(depth + 1, sum + set2[depth]); dfs2(depth + 1, sum); } int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); cin >> n >> m; for (int i = 0; i < n; i++) { cin >> values[i]; } sort(values, values + n); for (int i = 0; i < n; i++) { //cin >> values[i]; if (values[i] % 2 == 0) { set1[size1++] = values[i]; } else { set2[size2++] = values[i]; } } dfs1(0, 0); dfs2(0, 0); sort(one.rbegin(), one.rend()); sort(two.begin(), two.end()); assert(one.size() + two.size() < 5000000); ll answer = 0; int twopointer = 0; for (int i = 0; i < (int) (one.size()); i++) { while (twopointer < (int) (two.size()) && one[i] + two[twopointer] <= m) { twopointer++; } answer += twopointer; } cout << answer << endl; }
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