Submission #767043

#	Time	Username	Problem	Language	Result	Execution time	Memory
767043		YassineBenYounes	Secret (JOI14_secret)	C++17	0 / 100	400 ms	8316 KiB

secret

#include<bits/stdc++.h>
#include "secret.h"
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef double db;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define pbds tree<pair<int, int>, null_type, less<pair<int, int>>,rb_tree_tag, tree_order_statistics_node_update>
using namespace __gnu_pbds;
ll gcd(ll a , ll b) {return b ? gcd(b , a % b) : a ;} // greatest common divisor (PGCD)
ll lcm(ll a , ll b) {return (a * b) / gcd(a , b);} // least common multiple (PPCM)
int dx[8] = {1, 0, 0, -1, 1, 1, -1, -1};
int dy[8] = {0, 1, -1, 0, 1, -1, -1, 1};
#define endl "\n"
#define ss second
#define ff first
#define all(x) (x).begin() , (x).end()
#define pb push_back
#define vi vector<int>
#define vii vector<pair<int,int>>
#define vl vector<ll>
#define vll vector<pair<ll,ll>>
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pdd  pair<double,double>
#define vdd  vector<pdd>
#define speed ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
////////////////////Only Clear Code//////////////////////////

const int mx = 1005;
const int LOG = 25;
const int inf = 1e9+5;
const ll mod = 1e9+7;

int ans[mx][mx];

void build(int l,int r){
    if(r-l <= 1){
        return;
    }
    int md = (l+r)/2;
    for(int i = md-1; i >= l;i--){
        if(ans[i][md] == -1){
            ans[i][md] = Secret(ans[i][i+1], ans[i+1][md]);
            //cout << i << " " << md << endl;
        }
    }
    for(int i = md+1; i <= r;i++){
        if(ans[md][i] == -1){
            ans[md][i] = Secret(ans[i-1][i], ans[md][i-1]);
        }
    }
    build(l, md);
    build(md, r);
}

void Init(int N, int A[]){
    for(int i = 0; i <= N;i++){
        for(int j = 0; j <= N;j++){
            ans[i][j] = -1;
        }
    }
    for(int i = 0; i < N;i++){
        ans[i][i+1] = A[i];
    }
    build(0, N);
}

int Query(int L, int R){
    for(int i = L+1;i <= R;i++){
        if(ans[L][i] != -1 && ans[i][R+1] != -1){
            return Secret(ans[L][i], ans[i][R+1]);
        }
    }
    return ans[L][R+1];
}

/*
    NEVER GIVE UP!
    DOING SMTHNG IS BETTER THAN DOING NTHNG!!!
    Your Guide when stuck:
    - Continue keyword only after reading the whole input
    - Don't use memset with testcases
    - Check for corner cases(n=1, n=0)
    - Check where you declare n(Be careful of declaring it globally and in main)
*/

• Subtask 1: 0 / 100