Submission #58457

#	Time	Username	Problem	Language	Result	Execution time	Memory
58457		Benq	Bootfall (IZhO17_bootfall)	C++14	65 / 100	1020 ms	6516 KiB

bootfall

#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 = 500*500+1;

typedef array<int,MX> B;

int N, a[500];
bitset<MX> all;
B b;

int add(int a, int b) { return (a+b)%MOD; }

int sub(int a, int b) { return (a-b+MOD)%MOD; }

B ad(B b, int x) {
    B b2;
    F0R(i,MX) {
        b2[i] = b[i];
        if (i >= x) b2[i] = add(b2[i],b[i-x]);
    }
    return b2;
}

B del(B b2, int x) { // b[i]+b[i-x] = b2[i], b[i] = b2[i]-b[i-x]
    B b;
    F0R(i,MX) {
        b[i] = b2[i];
        if (i >= x) b[i] = sub(b[i],b[i-x]);
    }
    return b;
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    all.flip();
    cin >> N;
    b[0] = 1;
    int sum = 0;
    F0R(i,N) {
        cin >> a[i];
        sum += a[i];
        b = ad(b,a[i]);
    }
    if (sum % 2 != 0 || !b[sum/2]) {
        cout << 0;
        exit(0);
    }
    F0R(i,N) {
        auto x = del(b,a[i]);
        /*F0R(i,11) cout << x[i] << " ";
        cout << "\n";*/
        int sum2 = sum-a[i];
        bitset<MX> cur;
        F0R(j,MX) if (2*j < sum2 && x[j]) {
            cur[sum2-2*j] = 1;
        }
        all &= cur;
    }
    vi v;
    FOR(i,1,MX) if (all[i]) v.pb(i);
    cout << sz(v) << "\n";
    for (int i: v) cout << i << " ";
}

/* 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: 6 / 6
• Subtask 2: 7 / 7
• Subtask 3: 15 / 15
• Subtask 4: 16 / 16
• Subtask 5: 21 / 21
• Subtask 6: 0 / 35