Hackerrank - Drawing Book - Pavol Pidanič

Hackerrank – Popis problému

Celý popis zadania sa nacháza – Hackerrank.

Riešenie

pre zistenie počtu otočení z predu –

$(p / 2)$

počet otočení zo zadu – najprv si zistím, či je

$n$

párne, ak je potrebujem si určiť stranu na pravej strane knihy a tá je vždy nepárna – musím si urobiť

$n' = n + 1$

; počet otočení

$(n' - p) / 2$

Vytvoril som riešenie v týchto programovacích jazykoch:

Scala
Java
Javascript

Všetky riešenia sú dostupné aj na mojom GitHub profile.

Java

import java.util.*;

public class Solution {

    public static void main(String[] args) {
        Scanner stdin = new Scanner(System.in);
        int n = stdin.nextInt();
        int p = stdin.nextInt();

        int fromFront = p;
        int fromBack = n;
        if(n % 2 == 0) {
            fromBack = n + 1;
        }
        int result = Math.min(fromFront / 2, (fromBack - p) / 2);
        System.out.println(result);

        stdin.close();
    }
}

import java.util.*;

public class Solution {

public static void main(String[] args) {

Scanner stdin = new Scanner(System.in);

int n = stdin.nextInt();

int p = stdin.nextInt();

int fromFront = p;

int fromBack = n;

if(n % 2 == 0) {

fromBack = n + 1;

}

int result = Math.min(fromFront / 2, (fromBack - p) / 2);

System.out.println(result);

stdin.close();

}

Java

import java.util.*;

public class Solution {

    public static void main(String[] args) {
        Scanner stdin = new Scanner(System.in);
        int n = stdin.nextInt();
        int p = stdin.nextInt();

        int fromFront = p;
        int fromBack = n;
        if(n % 2 == 0) {
            fromBack = n + 1;
        }
        int result = Math.min(fromFront / 2, (fromBack - p) / 2);
        System.out.println(result);

        stdin.close();
    }
}

import java.util.*;

public class Solution {

public static void main(String[] args) {

Scanner stdin = new Scanner(System.in);

int n = stdin.nextInt();

int p = stdin.nextInt();

int fromFront = p;

int fromBack = n;

if(n % 2 == 0) {

fromBack = n + 1;

}

int result = Math.min(fromFront / 2, (fromBack - p) / 2);

System.out.println(result);

stdin.close();

}

Javascript

'use strict';

const processData = input => {
    const [n, p] = input.split('\n').map(i => parseInt(i));

    const fromFront = p;
    let fromBack = n;
    if(n % 2 == 0) {
        fromBack = n + 1;
    }
    const result = parseInt(Math.min(fromFront / 2, (fromBack - p) / 2));
    console.log(result);
};

process.stdin.resume();
process.stdin.setEncoding("ascii");

let _input = "";
process.stdin.on("data", input => _input += input);
process.stdin.on("end", () => processData(_input));

'use strict';

const processData = input => {

const [n, p] = input.split('\n').map(i => parseInt(i));

const fromFront = p;

let fromBack = n;

if(n % 2 == 0) {

fromBack = n + 1;

}

const result = parseInt(Math.min(fromFront / 2, (fromBack - p) / 2));

console.log(result);

};

process.stdin.resume();

process.stdin.setEncoding("ascii");

let _input = "";

process.stdin.on("data", input => _input += input);

process.stdin.on("end", () => processData(_input));

Pavol Pidanič

Dokážem na 10 prstoch napočítať do 1023

Hackerrank – Drawing Book

Hackerrank – Popis problému

Riešenie

Java

Java

Javascript

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