제출 #1343568

#제출 시각아이디문제언어결과실행 시간메모리
1343568aritro_Triple Peaks (IOI25_triples)C++20
0.34 / 100
2095 ms1960 KiB
#include<bits/stdc++.h>
using namespace std;

#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;

template <typename T> using ordset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

/*
less<T> can be changed to:
1. less_equal<T>
2. greater<T>

ordered_set functions:
1. *(os.find_by_order(x)) -> return the value at position x (0 base)
2. os.order_of_key(x) -> return the number of elements less than x
*/

typedef long long ll;
#define endl '\n'
#define pb push_back
#define ff first
#define ss second
#define all(a) a.begin(),a.end()

const int MOD=1000000007;

#define gcd(a,b) __gcd(a,b)
#define lcm(a,b) (a*(b/gcd(a,b)))

#define fios(); ios_base::sync_with_stdio(0);cin.tie(0);
#define fraction(x); cout.unsetf(ios::floatfield); cout.precision(x); cout.setf(ios::fixed,ios::floatfield);
#define file(); freopen("input.txt","r",stdin);freopen("output.txt","w",stdout);

int count_triples(vector<int32_t> a){
    int n=a.size();
    int ans=0;
    for(int i=0;i<n;i++){
        for(int j=i+2;j<n;j++){
            int d=j-i;
            if(a[i]+a[j]==d){
                int k1=i+a[i];
                if(k1<j&&k1<n&&a[k1]==a[j]) ans++;
                int k2=i+a[j];
                if(k2<j&&k2<n&&a[k2]==a[i]) ans++;
            }
            if(a[i]==d){
                int k=i+a[j];
                if(k<j&&k<n&&a[k]==d-a[j]) ans++;
            }
            if(a[j]==d){
                int k=i+(a[i]);
                if(k<j&&k<n&&a[k]==d-a[i]) ans++;
            }
        }
    }
    return ans;
}

vector<int32_t> construct_range(int32_t M, int32_t K){
    vector<int> ans(M,1);
    if(M>500) return ans;
    ans[0]=1,ans[1]=2;
    for(int i=3;i<M;i++){
        int mx=0,add=1;
        for(int cur=1;cur<=10;cur++){
            vector<int> v;
            v.insert(v.end(),ans.begin(),ans.begin()+i+1);
            v.pb(cur);
            int tem=count_triples(v);
            if(tem>mx) add=cur;
            mx=max(tem,mx);
        }
        ans[i]=add;
    }
    return ans;
}
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