Submission #470738

#	Time	Username	Problem	Language	Result	Execution time	Memory
470738		Carmel_Ab1	Peru (RMI20_peru)	C++17	0 / 100	1 ms	340 KiB

peru

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

//using namespace __gnu_pbds;
using namespace std;

typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef vector<pl> vpl;
typedef vector<ld> vld;
typedef pair<ld, ld> pld;

//typedef tree<ll, null_type, less_equal<ll>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
template<typename T>
ostream &operator<<(ostream &os, vector<T> &a) {
    os << "[";
    for (int i = 0; i < ll(a.size()); i++) { os << a[i] << ((i != ll(a.size() - 1) ? " " : "")); }
    os << "]\n";
    return os;
}

#define all(x) x.begin(),x.end()
#define YES out("YES")
#define NO out("NO")
#define out(x){cout << x << "\n"; return;}
#define GLHF ios_base::sync_with_stdio(false); cin.tie(NULL)
#define print(x){for(auto ait:x) cout << ait << " "; cout << "\n";}
#define pb push_back
#define umap unordered_map

#include "peru.h"
//#include "grader.cpp"

ll mod=1e9+7;
struct seg{

    int sz=1;
    vi t;
    seg(int n){
        while(sz<=2*n)
            sz*=2;
        t.resize(sz);
    }
    void update(int i,int v){
        t[i+=sz/2]=v;
        while(i){
            t[i/2]=max(t[i],t[i^1]);
            i/=2;
        }
    }
    int query(int l,int r){
        int ans=0;
        for(l+=sz/2,r+=sz/2 +1; l<r; l/=2,r/=2){
            if(l&1)ans=max(ans,t[l++]);
            if(r&1)ans=max(ans,t[--r]);
        }
        return ans;
    }
};
int solve(int n, int k, int* a){

    int dp[n];
    for(int i=0; i<n; i++)
        dp[i]=1e9;

    int mx=0;
    for(int i=0; i<k; i++)
        mx=max(mx,a[i]),dp[i]=mx;



    seg s(n);
    for(int i=0; i<n; i++)
        s.update(i,a[i]);
    for(int i=k; i<n; i++){
        dp[i]=a[i]+dp[i-1];
        dp[i]=min(dp[i],s.query(i-k+1,i)+dp[i-k]);
    }
    ll ans=0,p=1;
    for(int i=n-1; 0<=i; i--){
        ans=(ans+p*dp[i])%mod;
        p=(p*23)%mod;
    }
    return ans;

}

• Subtask 1: 0 / 18
• Subtask 2: 0 / 31
• Subtask 3: 0 / 51