Submission #57735

#	Time	Username	Problem	Language	Result	Execution time	Memory
57735		Benq	San (COCI17_san)	C++14	120 / 120	130 ms	11292 KiB

san

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;

int N;
ll K,ans;
pi p[40];
vl tmp[40];

void zz(vl a, vl b) {
    int ind = sz(b);
    F0R(i,sz(a)) {
        while (ind > 0 && a[i]+b[ind-1] >= K) ind --;
        ans += sz(b)-ind;
    }
}

void comb(int a, int b) {
    if (a == -1) {
        if (b == N) {
            zz({0},{0});
        } else {
            zz({0},tmp[b]);
        }
    } else {
        if (b == N) {
            zz(tmp[a],{0});
        } else {
            if (p[a].f > p[b].f) return;
            zz(tmp[a],tmp[b]);
        }
    }
}
int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> N >> K;
    F0R(i,N) cin >> p[i].f >> p[i].s;
    
    F0R(i,N/2+1) {
        tmp[i].pb(p[i].s);
        F0Rd(j,i) if (p[j].f <= p[i].f) 
            for (auto a: tmp[j]) tmp[i].pb(a+p[i].s);
    }
    FORd(i,N/2+1,N) {
        tmp[i].pb(p[i].s);
        FOR(j,i+1,N) if (p[i].f <= p[j].f)
            for (auto a: tmp[j]) tmp[i].pb(a+p[i].s);
    }
    F0R(i,N) sort(all(tmp[i]));
    FOR(i,-1,N/2+1) FOR(j,N/2+1,N+1) comb(i,j);
    cout << ans;
}

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/

• Subtask 1: 24 / 24
• Subtask 2: 24 / 24
• Subtask 3: 24 / 24
• Subtask 4: 24 / 24
• Subtask 5: 24 / 24