Dado un número entero n , la tarea es imprimir el número máximo de alfiles que se pueden colocar en un tablero de ajedrez nxn para que no haya dos alfiles que se ataquen entre sí. Por ejemplo, se pueden colocar un máximo de 2 alfiles de forma segura en un tablero de ajedrez de 2 x 2.
Ejemplos:
Entrada: n = 2
Salida: 2
Podemos colocar dos alfiles en fila.Entrada: n = 5
Salida: 8
Aproximación: Un alfil puede viajar en cualquiera de las cuatro diagonales. Por lo tanto podemos colocar alfiles si no está en ninguna diagonal de otro alfil. El máximo de alfiles que se pueden colocar en un tablero de ajedrez de n*n será de 2*(n – 1) .
- Coloca n alfiles en primera fila
- Coloque n-2 alfiles en la última fila. Solo dejamos dos esquinas de la última fila.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <iostream>
using namespace std;
// Function to return the maximum number of bishops
// that can be placed on an n * n chessboard
int numberOfBishops(int n)
{
if (n < 1)
return 0;
else if (n == 1)
return 1;
else
return 2 * (n - 1);
}
// Driver code
int main()
{
int n = 5;
cout << numberOfBishops(n);
return 0;
}
Java
// Java implementation of the approach
class gfg
{
// Function to return the maximum
// number of bishops that can be
// placed on an n * n chessboard
static int numberOfBishops(int n)
{
if (n < 1)
return 0;
else if (n == 1)
return 1;
else
return 2 * (n - 1);
}
// Driver code
public static void main(String[] args)
{
int n = 5;
System.out.println(numberOfBishops(n));
}
}
// This code is contributed by Mukul Singh.
Python3
# Python3 implementation of the # approach import math as mt # Function to return the maximum number # of bishops that can be placed on an # n * n chessboard def numberOfBishops(n): if (n < 1): return 0 elif (n == 1): return 1 else: return 2 * (n - 1) # Driver code n = 5 print(numberOfBishops(n)) # This code is contributed by # Mohit kumar 29
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the maximum number
// of bishops that can be placed on an
// n * n chessboard
static int numberOfBishops(int n)
{
if (n < 1)
return 0;
else if (n == 1)
return 1;
else
return 2 * (n - 1);
}
// Driver code
public static void Main()
{
int n = 5;
Console.Write(numberOfBishops(n));
}
}
// This code is contributed
// by Akanksha Rai
PHP
<?php
// PHP implementation of the approach
// Function to return the maximum number
// of bishops that can be placed on an
// n * n chessboard
function numberOfBishops($n)
{
if ($n < 1)
return 0;
else if ($n == 1)
return 1;
else
return 2 * ($n - 1);
}
// Driver code
$n = 5;
echo numberOfBishops($n);
// This code is contributed by Ryuga
?>
Javascript
<script>
// Javascript implementation of the approach
// Function to return the maximum
// number of bishops that can be
// placed on an n * n chessboard
function numberOfBishops(n)
{
if (n < 1)
return 0;
else if (n == 1)
return 1;
else
return 2 * (n - 1);
}
// Driver code
let n = 5;
document.write(numberOfBishops(n));
// This code is contributed by patel2127
</script>
Producción:
8
A continuación se muestra la implementación para valores más grandes de n :
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the difference of
// two big numbers as string
string subtract(string str1, string str2)
{
string res = "";
int n1 = str1.length();
int n2 = str2.length();
// To make subtraction easy
reverse(str1.begin(), str1.end());
reverse(str2.begin(), str2.end());
int carry = 0;
for (int i = 0; i < n2; i++) {
// Subtract digit by bdigit
int subst = ((str1[i] - '0')
- (str2[i] - '0') - carry);
if (subst < 0) {
subst = subst + 10;
carry = 1;
}
else
carry = 0;
// Change subst as character and
// add it to result string
res.push_back(subst + '0');
}
for (int i = n2; i < n1; i++) {
int subst = ((str1[i] - '0') - carry);
if (subst < 0) {
subst = subst + 10;
carry = 1;
}
else
carry = 0;
res.push_back(subst + '0');
}
// Reverse result to make it actual number
reverse(res.begin(), res.end());
return res;
}
string NumberOfBishops(string a)
{
if (a == "1")
return a;
else {
// Subtract 1 from number
a = subtract(a, "1");
// Reverse the string to make calculations easier
reverse(a.begin(), a.end());
int carry = 0;
// Multiply by 2
for (int i = 0; i < a.size(); i++) {
int tmp = a[i] - '0';
tmp *= 2;
tmp += carry;
a[i] = '0' + (tmp % 10);
carry = tmp / 10;
}
if (carry > 0)
a += ('0' + carry);
// Reverse the string to get actual result
reverse(a.begin(), a.end());
// Return result
return a;
}
}
// Driver code
int main()
{
string a = "12345678901234567890";
cout << NumberOfBishops(a) << endl;
return 0;
}
Java
// Java implementation of the approach
import java.util.*;
class GFG{
public static char[] reverse(char []str)
{
char[] temp = new char[str.length];
// Fill character array backwards with
// characters of the string
for(int i = 0; i < str.length; i++)
temp[str.length - i - 1] = str[i];
// Convert character array to string
// and return it
return temp;
}
// Function to return the difference of
// two big numbers as String
static char[] subtract(char[] str1, char[] str2)
{
String res = "";
int n1 = str1.length;
int n2 = str2.length;
// To make subtraction easy
str1 = reverse(str1);
str2 = reverse(str2);
int carry = 0;
for(int i = 0; i < n2; i++)
{
// Subtract digit by bdigit
int subst = ((str1[i] - '0') -
(str2[i] - '0') - carry);
if (subst < 0)
{
subst = subst + 10;
carry = 1;
}
else
carry = 0;
// Change subst as character and
// add it to result String
res = res + (subst);
}
for(int i = n2; i < n1; i++)
{
int subst = ((str1[i] - '0') - carry);
if (subst < 0)
{
subst = subst + 10;
carry = 1;
}
else
carry = 0;
res += (subst);
}
// Reverse result to make it actual number
char[] Res = res.toCharArray();
Res = reverse(Res);
return Res;
}
static char[] NumberOfBishops(char[] a)
{
if (new String(a) == "1")
return a;
else
{
// Subtract 1 from number
a = subtract(a, "1".toCharArray());
//Console.WriteLine(new String(a));
// Reverse the String to make
// calculations easier
a = reverse(a);
int carry = 0;
// Multiply by 2
for(int i = 0; i < a.length; i++)
{
int tmp = a[i] - '0';
tmp *= 2;
tmp += carry;
a[i] = (char)('0' + (tmp % 10));
carry = tmp / 10;
}
String A = new String(a);
if (carry > 0)
A += ('0' + carry);
char[] a1 = A.toCharArray();
// Reverse the String to get
// actual result
a1 = reverse(a1);
// Return result
return a1;
}
}
// Driver code
public static void main(String []args)
{
char[] a = "12345678901234567890".toCharArray();
System.out.println(new String(NumberOfBishops(a)));
}
}
// This code is contributed by pratham76
Python3
# Python3 implementation of the approach # Function to return the difference # of two big numbers as string def subtract(str1, str2): res = "" n1 = len(str1) n2 = len(str2) # To make subtraction easy, # reverse the strings str1 = str1[::-1] str2 = str2[::-1] carry = 0 for i in range(0, n2): # Subtract digit by bdigit subst = int(str1[i]) - int(str2[i]) - carry if subst < 0: subst = subst + 10 carry = 1 else: carry = 0 # Change subst as character and # add it to result string res += str(subst) for i in range(n2, n1): subst = int(str1[i]) - carry if subst < 0: subst = subst + 10 carry = 1 else: carry = 0 res += str(subst) # Reverse result to make it # actual number return res[::-1] def NumberOfBishops(a): if a == "1": return a else: # Subtract 1 from number a = subtract(a, "1") carry = 0 # Reverse the string to make # calculations easier. Convert the # string to list to manipulate it # as strings are immutable in python a = list(a[::-1]) # Multiply by 2 for i in range(0, len(a)): tmp = (int(a[i]) * 2) + carry a[i] = str(tmp % 10) carry = tmp // 10 # Convert the list back to string a = ''.join(a) if carry > 0: a += str(carry) # Reverse the string to get # actual result return a[::-1] # Driver code if __name__ == "__main__": a = "12345678901234567890" print(NumberOfBishops(a)) # This code is contributed # by Rituraj Jain
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the difference of
// two big numbers as string
static char[] subtract(char[] str1, char[] str2)
{
string res = "";
int n1 = str1.Length;
int n2 = str2.Length;
// To make subtraction easy
Array.Reverse(str1);
Array.Reverse(str2);
int carry = 0;
for (int i = 0; i < n2; i++)
{
// Subtract digit by bdigit
int subst = ((str1[i] - '0') - (str2[i] - '0') - carry);
if (subst < 0)
{
subst = subst + 10;
carry = 1;
}
else
carry = 0;
// Change subst as character and
// add it to result string
res = res + (subst);
}
for (int i = n2; i < n1; i++)
{
int subst = ((str1[i] - '0') - carry);
if (subst < 0)
{
subst = subst + 10;
carry = 1;
}
else
carry = 0;
res += (subst);
}
// Reverse result to make it actual number
char[] Res = res.ToCharArray();
Array.Reverse(Res);
return Res;
}
static char[] NumberOfBishops(char[] a)
{
if (new string(a) == "1")
return a;
else
{
// Subtract 1 from number
a = subtract(a, "1".ToCharArray());
//Console.WriteLine(new string(a));
// Reverse the string to make calculations easier
Array.Reverse(a);
int carry = 0;
// Multiply by 2
for (int i = 0; i < a.Length; i++)
{
int tmp = a[i] - '0';
tmp *= 2;
tmp += carry;
a[i] = (char)('0' + (tmp % 10));
carry = tmp / 10;
}
string A = new string(a);
if (carry > 0)
A += ('0' + carry);
char[] a1 = A.ToCharArray();
// Reverse the string to get actual result
Array.Reverse(a1);
// Return result
return a1;
}
}
// Driver code
static void Main()
{
char[] a = "12345678901234567890".ToCharArray();
Console.WriteLine(new string(NumberOfBishops(a)));
}
}
// This code is contributed by divyeshrabadiy07
Javascript
<script>
// Javascript implementation of the approach
function reverse(str)
{
let temp = new Array(str.length);
// Fill character array backwards with
// characters of the string
for(let i = 0; i < str.length; i++)
temp[str.length - i - 1] = str[i];
// Convert character array to string
// and return it
return temp;
}
// Function to return the difference of
// two big numbers as String
function subtract(str1,str2)
{
let res = "";
let n1 = str1.length;
let n2 = str2.length;
// To make subtraction easy
str1 = reverse(str1);
str2 = reverse(str2);
let carry = 0;
for(let i = 0; i < n2; i++)
{
// Subtract digit by bdigit
let subst = (parseInt(str1[i]) -
parseInt(str2[i]) - carry);
if (subst < 0)
{
subst = subst + 10;
carry = 1;
}
else
carry = 0;
// Change subst as character and
// add it to result String
res = res + (subst).toString();
}
for(let i = n2; i < n1; i++)
{
let subst = (parseInt(str1[i]) - carry);
if (subst < 0)
{
subst = subst + 10;
carry = 1;
}
else
carry = 0;
res += (subst).toString();
}
// Reverse result to make it actual number
let Res = res.split("");
Res = reverse(Res);
return Res;
}
function NumberOfBishops(a)
{
if (a == "1")
return a;
else
{
// Subtract 1 from number
a = subtract(a, "1");
//Console.WriteLine(new String(a));
// Reverse the String to make
// calculations easier
a = reverse(a);
let carry = 0;
// Multiply by 2
for(let i = 0; i < a.length; i++)
{
let tmp = parseInt(a[i]);
tmp *= 2;
tmp += carry;
a[i] = (tmp % 10).toString();
carry = Math.floor(tmp / 10);
}
let A = a.join("");
if (carry > 0)
A += ( carry).toString();
let a1 = A.split("");
// Reverse the String to get
// actual result
a1 = reverse(a1);
// Return result
return a1;
}
}
// Driver code
let a = "12345678901234567890".split("");
document.write(NumberOfBishops(a).join(""));
// This code is contributed by unknown2108.
</script>
Producción:
24691357802469135778
Publicación traducida automáticamente
Artículo escrito por Vivek.Pandit y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA