Submission #39416

#	Time	Username	Problem	Language	Result	Execution time	Memory
39416		farmersrice	Ice Hockey World Championship (CEOI15_bobek)	C++14	60 / 100	1000 ms	395436 KiB

bobek

#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++) {
		if (values[i] % 2 == 0) {
			set1[size1++] = values[i];
		} else {
			set2[size2++] = values[i];
		}
	}	

	dfs1(0, 0);
	dfs2(0, 0);

	sort(one.begin(), one.end());
	sort(two.begin(), two.end());

	//assert(one.size() + two.size() < 5000000);

	ll answer = 0;
	int twopointer = two.size() - 1;
	for (int i = 0; i < (int) (one.size()); i++) {
		if (one[i] > m) break;
		while (twopointer >= 0 && two[twopointer] > m - one[i]) {
			twopointer--;
		}
		answer += twopointer + 1;
	}

	cout << answer << endl;
}

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 10 / 10
• Subtask 4: 10 / 10
• Subtask 5: 0 / 10
• Subtask 6: 10 / 10
• Subtask 7: 10 / 10
• Subtask 8: 0 / 10
• Subtask 9: 0 / 10
• Subtask 10: 0 / 10