Cod sursa(job #2922877)

Utilizator EasyTnsEasyTns EasyTns Data 10 septembrie 2022 14:17:34
Problema Range minimum query Scor 100
Compilator cpp-64 Status done
Runda Arhiva educationala Marime 1.84 kb
#include<vector>
#include<fstream>
#include<iostream>
using namespace std;
#define BUF 16384

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

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

public:
    InParser(const char* nume) {
        fin = fopen(nume, "r");
        buff = new char[BUF]();
        sp = BUF-1;
    }

    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 a[18][100002],b[100002];
int n,m,i,j,p, x,y,e,lung;
int main()
{
    freopen("rmq.out","w", stdout);
    InParser fin("rmq.in");

    fin>>n>>m;
    for( i=1;i<=n;i++)
        fin>>a[0][i];

    for(  p=1;(1<<p)<=n;p++)
        for( i=1;i<=n;i++)
        {
            a[p][i]=a[p-1][i];
            j=i+(1<<(p-1));
            if(j<=n&&a[p][i]>a[p-1][j])
                a[p][i]=a[p-1][j];
        }
    b[1]=0;
    for( i=2;i<=n;i++)
        b[i]=1+b[i/2];

    for( i=1;i<=m;i++)
    {
        fin>>x>>y;
        e=b[y-x+1];
        lung=(1<<e);
        printf("%d",a[e][x]>a[e][y-lung+1]?a[e][y-lung+1]:a[e][x]); printf("\n");
    }

}