Submission #923538

#	Time	Username	Problem	Language	Result	Execution time	Memory
923538		KiaRez	Kpart (eJOI21_kpart)	C++17	100 / 100	779 ms	3060 KiB

Main

/*
    IN THE NAME OF GOD
*/
#include <bits/stdc++.h>

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

using namespace std;

typedef long long ll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef long double ld;

#define F                                      first
#define S                                      second
#define Mp                                     make_pair
#define pb                                     push_back
#define pf                                     push_front
#define size(x)                                ((ll)x.size())
#define all(x)                                 (x).begin(),(x).end()
#define kill(x)		                           cout << x << '\n', exit(0);
#define fuck(x)                                cout << "(" << #x << " , " << x << ")" << endl
#define endl                                   '\n'

const int N = 3e5+23, lg = 18;
ll Mod = 1e9+7; //998244353;

inline ll MOD(ll a, ll mod=Mod) {a%=mod; (a<0)&&(a+=mod); return a;}
inline ll poww(ll a, ll b, ll mod=Mod) {
    ll ans = 1;
    a=MOD(a, mod);
    while (b) {
        if (b & 1) ans = MOD(ans*a, mod);
        b >>= 1;
        a = MOD(a*a, mod);
    }
    return ans;
}

int n, t, a[N], dp[N], cnt[N];
void solve() {
	vector<int> ans;
	int sum = 0;

	cin>>n;
	fill(dp, dp+N, 0);
	fill(cnt, cnt+N, 0);
	dp[0] = n+1;
	for(int i=1; i<=n; i++) {
		cin>>a[i];
		sum += a[i];
		for(int j=sum; j>=a[i]; j--) {
			dp[j] = max(dp[j], min(i, dp[j-a[i]]));
		}
		int sum2 = 0;
		for(int j=i; j>=1; j--) {
			sum2 += a[j];
			if(sum2%2==0 && dp[sum2/2] >= j) {
				cnt[i-j+1] ++;
			}
		}
	}
	for(int i=1; i<=n; i++) if(cnt[i] == n-i+1) ans.pb(i);
	cout<<size(ans)<<' ';
	for(auto it : ans) cout<<it<<' ';
	cout<<endl;
}
int main () {
	ios_base::sync_with_stdio(false), cin.tie(0);

	cin>>t;
	while(t--) {
		solve();
	}

	return 0;
}

• Subtask 1: 10 / 10
• Subtask 2: 20 / 20
• Subtask 3: 70 / 70