| # |
Submission time |
Handle |
Problem |
Language |
Result |
Execution time |
Memory |
| 1046168 |
2024-08-06T10:54:29 Z |
pan |
cmp (balkan11_cmp) |
C++17 |
|
692 ms |
107096 KB |
#include "cmp.h"
#include <bits/stdc++.h>
//#include "bits_stdc++.h"
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define endl '\n'
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define input(x) scanf("%lld", &x);
#define input2(x, y) scanf("%lld%lld", &x, &y);
#define input3(x, y, z) scanf("%lld%lld%lld", &x, &y, &z);
#define input4(x, y, z, a) scanf("%lld%lld%lld%lld", &x, &y, &z, &a);
#define print(x, y) printf("%lld%c", x, y);
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define all(x) x.begin(), x.end()
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
#define FOR(i, x, n) for (ll i =x; i<=n; ++i)
#define RFOR(i, x, n) for (ll i =x; i>=n; --i)
using namespace std;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
//using namespace __gnu_pbds;
//#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
//#define ordered_multiset tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>
typedef long long ll;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<pi, ll> pii;
typedef pair<pi, pi> piii;
ll buffer[6] = {1, 4097, 5121, 5377, 5441, 5457};
void remember(int n)
{
for (ll i=0; i<6; ++i) bit_set(n/(1LL << 2*i) + buffer[i]);
}
int compare(int b)
{
ll lo = 0, hi = 6;
while (lo!=hi)
{
ll mid = (lo+hi)/2;
if (bit_get(b/(1LL << 2*mid) + buffer[mid])) hi = mid;
else lo = mid+1;
}
if (lo==0) return 0; // same
ll nxt = (b/(1LL << 2*(lo-1)))&3;
if (nxt <= 1LL)
{
if (bit_get((b/(1LL << 2*(lo))*4) + buffer[lo-1])) return 1;
return -1;
}
else
{
if (bit_get((b/(1LL << 2*(lo))*4) + 3 + buffer[lo-1])) return -1;
return 1;
}
}
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Correct |
692 ms |
107096 KB |
Output is correct - maxAccess = 10, score = 100 |
