Cod sursa(job #1951717)

Utilizator DysKodeTurturica Razvan DysKode Data 3 aprilie 2017 19:21:38
Problema Secventa Scor 100
Compilator cpp Status done
Runda Arhiva de probleme Marime 2.07 kb
#include <bits/stdc++.h>

using namespace std;

#define f first
#define s second

ifstream fin( "secventa.in" );
ofstream fout( "secventa.out" );

deque< int > dq;
int v[500010],i,j,n,m,k,ans,a;

class InParser {
private:
    FILE *fin;
    char *buff;
    int sp;

    char read_ch() {
        ++sp;
        if (sp == 4096) {
            sp = 0;
            fread(buff, 1, 4096, fin);
        }
        return buff[sp];
    }

public:
    InParser(const char* nume) {
        fin = fopen(nume, "r");
        buff = new char[4096]();
        sp = 4095;
    }

    InParser& operator >> (int &n) {
        char c;
        while (!isdigit(c = read_ch()) && c != '-');
        int sgn = 1;
        if (c == '-') {
            n = 0;
            sgn = -1;
        } else {
            n = c - '0';
        }
        while (isdigit(c = read_ch())) {
            n = 10 * n + c - '0';
        }
        n *= sgn;
        return *this;
    }

    InParser& operator >> (long long &n) {
        char c;
        n = 0;
        while (!isdigit(c = read_ch()) && c != '-');
        long long sgn = 1;
        if (c == '-') {
            n = 0;
            sgn = -1;
        } else {
            n = c - '0';
        }
        while (isdigit(c = read_ch())) {
            n = 10 * n + c - '0';
        }
        n *= sgn;
        return *this;
    }
};

int main()
{
    InParser fin( "secventa.in" );

    ans = -1000000000;
    fin>>n>>k;
    for( i = 1 ; i <= n ; i++ )
        fin>>v[ i ];
    for( i = 1 ; i < k ; i++ )
    {
        while( dq.size() && v[ i ] <= v[ dq.front() ] )
            dq.pop_front();
        dq.push_front( i );
    }
    for( i = k ; i <= n ; i++ )
    {
        while( dq.size() && v[ i ] <= v[ dq.front() ] )
            dq.pop_front();
        dq.push_front( i );
        if( v[ dq.back() ] > ans )
        {
            ans = v[ dq.back() ];
            a = i;
        }
        if( dq.back() == i - k + 1 )
            dq.pop_back();
    }
    fout<<a-k+1<<' '<<a<<' '<<ans;

return 0;
}