Submission #317750

#	Time	Username	Problem	Language	Result	Execution time	Memory
317750		soroush	Bali Sculptures (APIO15_sculpture)	C++14	100 / 100	244 ms	6272 KiB

sculpture

/*
#pragma GCC optimize("O2")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx,avx2,sse,sse2,fma,tune=native")
//*/
#include <bits/stdc++.h>

using namespace std;

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

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const ll maxn  = 2100;
const ll mod =1e9+7;
const ld PI = acos((ld)-1);

#define int ll
#define pb push_back
#define endl '\n'
#define dokme(x) cout << x , exit(0)
#define migmig ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define ms(x , y) memset(x , y , sizeof x)
ll pw(ll a, ll b, ll md = mod){ll res = 1;while(b){if(b&1){res=(a*res)%md;}a=(a*a)%md;b>>=1;}return(res);}

int n;
int a[maxn];
int l , r;

void sub1(){
    ll ans = 1e18;
    for(int i = 0 ; i < (1 << (n - 1)) ; i ++)
        if(__builtin_popcount(i)+1 >= l and  __builtin_popcount(i)+1 <= r){
        vector < ll > vec;
        vec.clear();
        vec.pb(0);
        for(int j = 0 ; j < n ; j ++){
            vec.back()+=a[j];
            if(i & (1 << j))
                vec.pb(0);
        }
        ll orr = 0;
        for(auto x : vec)
            orr|=x;
        ans = min(ans , orr);
    }
    dokme(ans);
}

int mark[200][200][600];

bool m[200][200];

int ans;

void dfs(int i , int k , int Or){
    if(i == n and k >= l and k <= r)
        ans = min(ans , Or);
    mark[i][k][Or] = 1;
    int sum = 0;
    for(int j = i+1 ; j <= n ; j ++){
        sum += a[j - 1];
        if(!mark[j][k+1][Or|sum])
            dfs(j, k + 1 , Or | sum);
    }
}


void sub2(){
    ans = 1e9;
    dfs(0 , 0 , 0);
    dokme(ans);
}

ll dp[maxn];

ll solve(){
	ans = (1LL << 60LL)-1;
	for(int i = 59 ; i >= 0 ; i --){
		ms(dp , 63), dp[0] = 0;
		ans ^= (1LL << i);
		for(int j = 0 ; j < n ; j ++){
			ll sum = 0;
			for(int k = j ; k < n ; k ++)
				sum += a[k], dp[k+1] = min(dp[k+1] ,(ans - (sum|ans))?dp[k+1] : dp[j]+1);
		}
		if(dp[n] > r)	
			ans += (1LL << i);
	}
	return(ans);
}

ll sub4(){
	ans = (1LL << 60LL)-1;
	for(int i = 59 ; i >= 0 ; i --){
		ans ^= (1LL << i);
		ms(m , 0);
		m[0][0] = 1;
		for(int i = 0 ; i < n ; i ++)
			for(int j = 0 ; j < r ; j ++ ){
					ll sum = 0;
					for(int k = i ; k < n ; k ++)
						sum += a[k] , 
						m[k+1][j+1] |= (m[i][j] && !((ans - (sum|ans))));
					
			}
		bool ok = 0;
		for(int i = l ; i <= r ; i ++) ok|=m[n][i];
		if(!ok)ans+=(1LL << i);
	}
	return(ans);
}

int32_t main(){
    migmig;
    cin >> n >> l >> r;
    for(int i = 0 ; i < n ; i ++)
        cin >> a[i];
    if(n <= 20){
        sub1();
    }
    if(n <= 50 and r <= 20 and *max_element(a , a + n) <= 10){
        sub2();
    }
    if(l == 1){
		dokme(solve());
	}
	//sub4
	cout << sub4();
    return(0);
}

• Subtask 1: 9 / 9
• Subtask 2: 16 / 16
• Subtask 3: 21 / 21
• Subtask 4: 25 / 25
• Subtask 5: 29 / 29