Submission #39408

#	Time	Username	Problem	Language	Result	Execution time	Memory
39408		farmersrice	Ice Hockey World Championship (CEOI15_bobek)	C++14	50 / 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;

int main() {
	ios_base::sync_with_stdio(false); 
	cin.tie(NULL);

	cin >> n >> m;

	for (int i = 0; i < n; i++) {
		cin >> values[i];

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

	sort(set1, set1 + size1);
	sort(set2, set2 + size2);

	for (int i = 0; i < (1 << size1); i++) {
		ll sum = 0;
		for (int j = 0; j < size1; j++) {
			if (i & (1 << j)) {
				sum += set1[j];
			}
		}
		one.add(sum);
	}

	for (int i = 0; i < (1 << size2); i++) {
		ll sum = 0;
		for (int j = 0; j < size2; j++) {
			if (i & (1 << j)) {
				sum += set2[j];
			}
		}
		two.add(sum);
	}

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

	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;
}

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