# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
125477 | eriksuenderhauf | Ice Hockey World Championship (CEOI15_bobek) | C++11 | 365 ms | 16880 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
//#pragma GCC optimize("O3")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>
#define mem(a,v) memset((a), (v), sizeof (a))
#define enl printf("\n")
#define case(t) printf("Case #%d: ", (t))
#define ni(n) scanf("%d", &(n))
#define nl(n) scanf("%lld", &(n))
#define nai(a, n) for (int i = 0; i < (n); i++) ni(a[i])
#define nal(a, n) for (int i = 0; i < (n); i++) nl(a[i])
#define pri(n) printf("%d\n", (n))
#define prl(n) printf("%lld\n", (n))
#define pii pair<int, int>
#define pil pair<int, long long>
#define pll pair<long long, long long>
#define vii vector<pii>
#define vil vector<pil>
#define vll vector<pll>
#define vi vector<int>
#define vl vector<long long>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef cc_hash_table<int,int,hash<int>> ht;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> oset;
const double pi = acos(-1);
const int MOD = 1e9 + 7;
const int INF = 1e9 + 7;
const int MAXN = 42;
const double eps = 1e-9;
ll a[MAXN];
ll l[1 << 21], r[1 << 21];
int main() {
srand(time(0));
int n; ll m;
scanf("%d %lld", &n, &m);
nal(a, n);
random_shuffle(a, a + n);
int x = n / 2, y = (n+1) / 2;
for (int i = 0; i < (1 << x); i++) {
ll cur = 0;
for (int j = 0; j < x; j++)
if ((i >> j) & 1)
cur += a[j];
l[i] = cur;
}
sort(l, l+(1<<x));
for (int i = 0; i < (1 << y); i++) {
ll cur = 0;
for (int j = 0; j < y; j++)
if ((i >> j) & 1)
cur += a[j+x];
r[i] = cur;
}
sort(r, r+(1<<y));
ll j = (1<<y)-1, ans = 0;
for (int i = 0; i < (1<<x); i++) {
while (j > 0 && l[i] + r[j] > m) j--;
if (j >= 0 && l[i] + r[j] <= m) ans += j + (i>0 ? 1 : 0);
//for (j = 0; j < (1 << y); j++)
// if (l[i] + r[j] <= m)
// ans++;
}
prl(ans);
return 0;
}
Compilation message (stderr)
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