제출 #1215257

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1215257		M_SH_O	Bootfall (IZhO17_bootfall)	C++20	100 / 100	664 ms	4908 KiB

bootfall

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

#define ll long long
#define ll1 long long
#define ull unsigned long long
#define dou long double
#define str string
#define vll vector<ll>
#define vi vector<int>
#define pll pair<ll, ll>
#define vpll vector<pll>
#define vbool vector<bool>
#define vstr vector<str>
#define vvll vector<vll>
#define pb push_back
#define pf push_front
#define endl "\n"
#define fr first
#define se second
// #define sortcmp(a) sort(a.begin(), a.end(), cmp)
#define sort(a) sort(a.begin(), a.end())
#define all(a) a.begin(), a.end()
#define reverse(a) reverse(a.begin(), a.end())
#define speed ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define INF 1000000000000000000
#define ordered_set tree<pll, null_type, less<pll>, rb_tree_tag, tree_order_statistics_node_update>

using namespace std;
using namespace __gnu_pbds;

mt19937 rng(1488);
ll randll(ll l, ll r){
    return uniform_int_distribution<ll>(l, r)(rng);
}

/*vector<map<ll, vll>> s;
vll p;

ll find(ll v) {
    if (p[v] == v) return v;
    return p[v] = find(p[v]);
}

void unite(ll a, ll b) {
    a = find(a);
    b = find(b);

    if (a == b) return;
    if (s[a].size() < s[b].size()) {
        p[a] = b;
        for (auto i : s[a]) {
            for (int j : i.se) {
                s[b][i.fr].pb(j);
            }
        }
        s[a].clear();
    }
    else {
        p[b] = a;
        for (auto i : s[b]) {
            for (int j : i.se) {
                s[a][i.fr].pb(j);
            }
        }
        s[b].clear();
    }
}

vll tree1;

void bt(ll v, ll tl, ll tr, vll& a) {
    if (tl == tr) {
        tree1[v] = a[tl];
        return;
    }

    ll tm = (tl+tr)/2;
    bt(v*2, tl, tm, a);
    bt(v*2, tm+1, tr, a);

    tree1[v] = tree1[v*2]+tree1[v*2+1];
}

ll get(ll l, ll r, ll v, ll tl, ll tr) {
    if (l <= tl && tr <= r) return tree1[v];
    if (tl > r || tr < l) return 0;

    ll tm = (tl+tr)/2;
    return get(l, r, v*2, tl, tm)+get(l, r, v*2+1, tm+1, tr);
}

void update(ll idx, ll val, ll v, ll tl, ll tr) {
    if (tl == tr) {
        tree1[v] = val;
        return;
    }

    ll tm = (tl+tr)/2;

    if (idx <= tm) update(idx, val, v*2, tl, tm);
    else update(idx, val, v*2+1, tm+1, tr);

    tree1[v] = tree1[v*2]+tree1[v*2+1];
}*/

int main() {
    ll MOD = 1e18+7;
    ll n;
    cin >> n;
    vll a(n);
    ll s = 0;
    for (int i =0; i < n; i ++) {
        cin >> a[i];
        s += a[i];
    }
    sort(a);

    if (s % 2 == 1) {
        cout << 0 << endl;
        return 0;
    }

    ll mx = 250000;

    vll dp(mx+7, 0);
    dp[0] = 1;

    for (int i = 0; i < n; i ++) {
        for (int j = mx-a[i]; j >= 0; j --) {
            dp[j+a[i]] += dp[j];
            dp[j+a[i]] %= MOD;
        }
    }
    /*for (int j = 1; j <= mx; j ++) {
        cout << dp[j] << ' ';
    }
    cout << endl;*/
    if (dp[s/2] == 0) {
        cout << 0 << endl;
        return 0;
    }

    vbool sm(mx+7, 1);
    for (int i = 0; i < n; i ++) {
        vll dp1(mx+7, 0);
        dp1[0] = 1;
        vbool us1(mx+7, 0);
        for (int j = 1; j <= mx; j ++) {
            if (j < a[i]) {
                dp1[j] = dp[j];
                if (dp1[j] != 0) us1[abs(s-a[i]-2*j)] = 1;
                //cout << dp1[j] << endl;
                continue;
            }
            dp1[j] = (dp[j]-dp1[j-a[i]]+MOD)%MOD;
            //cout << dp1[j] << ' ' << dp1[j-a[i]] << ' ' << j-a[i] << ' ' << dp[j] << ' ' << dp[j]-dp1[j-a[i]] << endl;
            if (dp1[j] != 0) us1[abs(s-a[i]-2*j)] = 1;
        }
        //cout << endl;
        for (int j = 1; j <= mx; j ++) {
            if (!us1[j]) sm[j] = 0;
        }
    }

    vll res;
    for (int i = 1; i <= mx; i ++) {
        if (sm[i]) res.pb(i);
    }

    cout << res.size() << endl;
    for (int i : res) {
        cout << i << ' ';
    }
    cout << endl;
}

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