제출 #536144

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
536144		akhan42	Bootfall (IZhO17_bootfall)	C++17	13 / 100	5 ms	1112 KiB

bootfall

#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")

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

using namespace std;
using namespace __gnu_pbds;

#define F first
#define S second
#define forn(i, n) for(int i = 0; i < n; ++i)
#define forbn(i, b, n) for(int i = b; i < n; ++i)
#define sz(v) (int)v.size()
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define all(v) v.begin(), v.end()
#define min3(a, b, c) min(a, min(b, c))
#define lc v << 1
#define rc (v << 1) + 1

typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<ii> vii;
typedef long long ll;
typedef complex<double> cd;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;

template <class T> inline void mineq(T &a, T b) { a = min(a, b); }
template <class T> inline void maxeq(T &a, T b) { a = max(a, b); }


const double PI = acos(-1);


void fft(vector<cd> & a, bool invert) {
    int n = a.size();

    for (int i = 1, j = 0; i < n; i++) {
        int bit = n >> 1;
        for (; j & bit; bit >>= 1)
            j ^= bit;
        j ^= bit;

        if (i < j)
            swap(a[i], a[j]);
    }

    for (int len = 2; len <= n; len <<= 1) {
    	double ang = 2 * PI / len * (invert ? -1 : 1);
        cd wlen(cos(ang), sin(ang));
        for (int i = 0; i < n; i += len) {
            cd w(1, 0);
            for (int j = 0; j < len / 2; j++) {
                cd u = a[i+j], v = a[i+j+len/2] * w;
                a[i+j] = u + v;
                a[i+j+len/2] = u - v;
                w *= wlen;
            }
        }
    }

    if (invert) {
        for (cd & x : a)
            x /= n;
    }
}


vector<int> multiply(vector<int> const& a, vector<int> const& b) {
    vector<cd> fa(a.begin(), a.end()), fb(b.begin(), b.end());
    int n = 1;
    while (n < sz(a) + sz(b))
        n <<= 1;

    fa.resize(n);
    fb.resize(n);

    fft(fa, false);
    fft(fb, false);
    for (int i = 0; i < n; i++)
        fa[i] *= fb[i];
    fft(fa, true);

    vector<int> result(n);
    for (int i = 0; i < n; i++)
        result[i] = round(fa[i].real());
    return result;
}


const int BASE = 350;
const int MX = BASE * BASE + 1;
using bts = bitset<MX>;


int last_bit(bts& a) {
	int i = a._Find_first();
	while(a._Find_next(i) != a.size())
		i = a._Find_next(i);
	return i;
}


bts sum(bts& a, bts& b) {
	int al = last_bit(a), bl = last_bit(b);

	vi v1, v2;
	forn(i, al + 1)
	v1.pb(a[i]);

	forn(i, bl + 1)
	v2.pb(b[i]);

	vi c = multiply(v1, v2);
	bts res;
	forn(i, min(sz(c), MX))
		if(c[i] > 0)
			res[i] = 1;
	return res;
}


vi pols[500];


vi mult(int l, int r) {
	if(l == r)
		return pols[l];

	int m = (l + r) / 2;
	vi r1 = mult(l, m);
	vi r2 = mult(m + 1, r);

	return multiply(r1, r2);
}


bts divide(vi pol, int d) {
	bts res;
	int ln = min(sz(pol), MX);
	for(int i = 0; i < ln; i++) {
		if(pol[i] > 0) {
			pol[i + d] -= pol[i];
			pol[i] = 0;
			res[i] = 1;
		}
	}

	return res;
}


void solve() {
	int n;
	cin >> n;
	vi arr(n);
	forn(i, n)
	cin >> arr[i];
	sort(all(arr));

	forn(i, n) {
		pols[i].resize(arr[i] + 1);
		pols[i][0] = pols[i][arr[i]] = 1;
	}

	bts s;
	s[0] = 1;
	int total = 0;
	bool odd = true, even = true;
	for(int a: arr) {
		s |= s << a;
		total += a;
		if(a % 2)
			even = false;
		else
			odd = false;
	}

	if(total % 2 || s[total / 2] == 0 || (!even && !odd))
	{
		cout << 0;
		return;
	}

	vi pol = mult(0, n - 1);

	bts ans;
	ans.flip();
	forn(i, n) {
		bts c = divide(pol, arr[i]);

		int d = (total - arr[i] - (odd ? 1 : 0)) / 2;

		ans &= c >> d;
	}

	vi vans;
	forbn(i, 1, MX) {
		if(ans[i]) {
			vans.pb(2 * i - (odd ? 1 : 0));
		}
	}

	cout << sz(vans) << '\n';
	for(int el: vans)
		cout << el << ' ';
}


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

//	freopen("triangles.in", "r", stdin);
//	freopen("triangles.out", "w", stdout);

	int t = 1;
//	cin >> t;
	while(t--) {
		solve();
	}
}

• Subtask 1: 6 / 6
• Subtask 2: 7 / 7
• Subtask 3: 0 / 15
• Subtask 4: 0 / 16
• Subtask 5: 0 / 21
• Subtask 6: 0 / 35