Dados dos enteros grandes en forma de strings A y B y su producto también en forma de string C tal que un dígito del producto se reemplaza con X , la tarea es encontrar el dígito reemplazado en el producto C.
Ejemplos:
Entrada: A = 51840, B = 273581, C = 1418243×040
Salida: 9
Explicación:
El producto de los enteros A y B es 51840 * 273581 = 14182439040. Al compararlo con C, se puede concluir que el dígito reemplazado fue 9 Por lo tanto, imprima 9.Entrada: A = 123456789, B = 987654321, C = 12193263111 × 635269
Salida: 2
Enfoque ingenuo: el enfoque más simple para resolver el problema dado es encontrar el producto de dos números enteros grandes A y B usando el enfoque discutido en este artículo y luego comparándolo con C para encontrar el dígito X faltante resultante .
Complejidad de tiempo: O((log 10 A)*(log 10 B))
Espacio auxiliar: O(log 10 A + log 10 B)
Enfoque eficiente: el enfoque anterior también se puede optimizar mediante el uso de las siguientes observaciones:
Supongamos que N es un número entero tal que N = a m a m-1 a m-2 …….a 2 a 1 a 0 donde a x representa el dígito x . Ahora, N se puede representar como:
=> N = a m * 10 m + a m-1 * 10 m-1 + ………… + a 1 * 10 + a 0Realizando la operación de módulo con 11 en la ecuación anterior:
=> N (mod 11) = a m * 10 m + a m-1 * 10 m-1 + ………… + a 1 * 10 + a 0 (mod 11)
=> N (mod 11)= a m (-1) m + a m-1 (-1) m-1 + ……….. + a 1 (-1) + a 0 (mod 11) [Ya que 10 ≡ -1 (mod 11)]
=> N (mod 11) = T (mod 11) donde T = a 0 –a 1 + a 2 …… + (-1) m a m
Por lo tanto, de la ecuación anterior, A * B = C se puede convertir en (A % 11) * (B % 11) = (C % 11) donde el lado izquierdo de la ecuación es un valor constante y el lado derecho El lado de la mano será una ecuación en 1 variable X que se puede resolver para encontrar el valor de X. Puede existir la posibilidad de que X pueda tener un valor negativo después de realizar el módulo con 11 , para ese caso considere el valor positivo de X.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the replaced digit
// in the product of a*b
int findMissingDigit(string a, string b,
string c)
{
// Keeps track of the sign of the
// current digit
int w = 1;
// Stores the value of a % 11
int a_mod_11 = 0;
// Find the value of a mod 11 for
// large value of a as per the
// derived formula
for (int i = a.size() - 1; i >= 0; i--) {
a_mod_11 = (a_mod_11 + w * (a[i] - '0')) % 11;
w = w * -1;
}
// Stores the value of b % 11
int b_mod_11 = 0;
w = 1;
// Find the value of b mod 11 for
// large value of a as per the
// derived formula
for (int i = b.size() - 1;
i >= 0; i--) {
b_mod_11 = (b_mod_11
+ w * (b[i] - '0'))
% 11;
w = w * -1;
}
// Stores the value of c % 11
int c_mod_11 = 0;
// Keeps track of the sign of x
bool xSignIsPositive = true;
w = 1;
for (int i = c.size() - 1; i >= 0; i--) {
// If the current digit is the
// missing digit, then keep
// the track of its sign
if (c[i] == 'x') {
xSignIsPositive = (w == 1);
}
else {
c_mod_11 = (c_mod_11
+ w * (c[i] - '0'))
% 11;
}
w = w * -1;
}
// Find the value of x using
// the derived equation
int x = ((a_mod_11 * b_mod_11)
- c_mod_11)
% 11;
// Check if x has a negative sign
if (!xSignIsPositive) {
x = -x;
}
// Return positive equivaluent
// of x mod 11
return (x % 11 + 11) % 11;
}
// Driver Code
int main()
{
string A = "123456789";
string B = "987654321";
string C = "12193263111x635269";
cout << findMissingDigit(A, B, C);
return 0;
}
Java
// Java program for the above approach
class GFG {
// Function to find the replaced digit
// in the product of a*b
public static int findMissingDigit(String a, String b, String c)
{
// Keeps track of the sign of the
// current digit
int w = 1;
// Stores the value of a % 11
int a_mod_11 = 0;
// Find the value of a mod 11 for
// large value of a as per the
// derived formula
for (int i = a.length() - 1; i >= 0; i--) {
a_mod_11 = (a_mod_11 + w * (a.charAt(i) - '0')) % 11;
w = w * -1;
}
// Stores the value of b % 11
int b_mod_11 = 0;
w = 1;
// Find the value of b mod 11 for
// large value of a as per the
// derived formula
for (int i = b.length() - 1; i >= 0; i--) {
b_mod_11 = (b_mod_11 + w * (b.charAt(i) - '0')) % 11;
w = w * -1;
}
// Stores the value of c % 11
int c_mod_11 = 0;
// Keeps track of the sign of x
boolean xSignIsPositive = true;
w = 1;
for (int i = c.length() - 1; i >= 0; i--) {
// If the current digit is the
// missing digit, then keep
// the track of its sign
if (c.charAt(i) == 'x') {
xSignIsPositive = (w == 1);
} else {
c_mod_11 = (c_mod_11 + w * (c.charAt(i) - '0')) % 11;
}
w = w * -1;
}
// Find the value of x using
// the derived equation
int x = ((a_mod_11 * b_mod_11) - c_mod_11) % 11;
// Check if x has a negative sign
if (!xSignIsPositive) {
x = -x;
}
// Return positive equivaluent
// of x mod 11
return (x % 11 + 11) % 11;
}
// Driver Code
public static void main(String args[]) {
String A = "123456789";
String B = "987654321";
String C = "12193263111x635269";
System.out.println(findMissingDigit(A, B, C));
}
}
// This code is contributed by saurabh_jaiswal.
Python3
# Python3 Program to implement the above approach
# Function to find the replaced digit
# in the product of a*b
def findMissingDigit(a, b, c):
# Keeps track of the sign of the
# current digit
w = 1
# Stores the value of a % 11
a_mod_11 = 0
# Find the value of a mod 11 for
# large value of a as per the
# derived formula
for i in range(len(a) - 1, -1, -1):
a_mod_11 = (a_mod_11 + w * (ord(a[i]) - ord('0'))) % 11
w = w * -1
# Stores the value of b % 11
b_mod_11 = 0
w = 1
# Find the value of b mod 11 for
# large value of a as per the
# derived formula
for i in range(len(b) - 1, -1, -1):
b_mod_11 = (b_mod_11 + w * (ord(b[i]) - ord('0'))) % 11
w = w * -1
# Stores the value of c % 11
c_mod_11 = 0
# Keeps track of the sign of x
xSignIsPositive = True
w = 1
for i in range(len(c) - 1, -1, -1):
# If the current digit is the
# missing digit, then keep
# the track of its sign
if (c[i] == 'x'):
xSignIsPositive = (w == 1)
else:
c_mod_11 = (c_mod_11 + w * (ord(c[i]) - ord('0'))) % 11
w = w * -1
# Find the value of x using
# the derived equation
x = ((a_mod_11 * b_mod_11) - c_mod_11) % 11
# Check if x has a negative sign
if (not xSignIsPositive):
x = -x
# Return positive equivaluent
# of x mod 11
return (x % 11 + 11) % 11
A = "123456789"
B = "987654321"
C = "12193263111x635269"
print(findMissingDigit(A, B, C))
# This code is contributed by divyeshrabadiya07.
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the replaced digit
// in the product of a*b
static int findMissingDigit(string a, string b,
string c)
{
// Keeps track of the sign of the
// current digit
int w = 1;
// Stores the value of a % 11
int a_mod_11 = 0;
// Find the value of a mod 11 for
// large value of a as per the
// derived formula
for (int i = a.Length - 1; i >= 0; i--) {
a_mod_11 = (a_mod_11 + w * ((int)a[i] - 48)) % 11;
w = w * -1;
}
// Stores the value of b % 11
int b_mod_11 = 0;
w = 1;
// Find the value of b mod 11 for
// large value of a as per the
// derived formula
for (int i = b.Length - 1;
i >= 0; i--) {
b_mod_11 = (b_mod_11
+ w * ((int)b[i] - 48))
% 11;
w = w * -1;
}
// Stores the value of c % 11
int c_mod_11 = 0;
// Keeps track of the sign of x
bool xSignIsPositive = true;
w = 1;
for (int i = c.Length - 1; i >= 0; i--) {
// If the current digit is the
// missing digit, then keep
// the track of its sign
if (c[i] == 'x') {
xSignIsPositive = (w == 1);
}
else {
c_mod_11 = (c_mod_11
+ w * ((int)c[i] - '0'))
% 11;
}
w = w * -1;
}
// Find the value of x using
// the derived equation
int x = ((a_mod_11 * b_mod_11)
- c_mod_11)
% 11;
// Check if x has a negative sign
if (xSignIsPositive == false) {
x = -x;
}
// Return positive equivaluent
// of x mod 11
return (x % 11 + 11) % 11;
}
// Driver Code
public static void Main()
{
string A = "123456789";
string B = "987654321";
string C = "12193263111x635269";
Console.Write(findMissingDigit(A, B, C));
}
}
// This code is contributed by ipg2016107.
Javascript
<script>
// JavaScript Program to implement
// the above approach
// Function to find the replaced digit
// in the product of a*b
function findMissingDigit(a, b,
c)
{
// Keeps track of the sign of the
// current digit
let w = 1;
// Stores the value of a % 11
let a_mod_11 = 0;
// Find the value of a mod 11 for
// large value of a as per the
// derived formula
for (let i = a.length - 1; i >= 0; i--) {
a_mod_11 = (a_mod_11 + w * (a[i].charCodeAt(0) - '0'.charCodeAt(0))) % 11;
w = w * -1;
}
// Stores the value of b % 11
let b_mod_11 = 0;
w = 1;
// Find the value of b mod 11 for
// large value of a as per the
// derived formula
for (let i = b.length - 1;
i >= 0; i--) {
b_mod_11 = (b_mod_11
+ w * (b[i].charCodeAt(0) - '0'.charCodeAt(0)))
% 11;
w = w * -1;
}
// Stores the value of c % 11
let c_mod_11 = 0;
// Keeps track of the sign of x
let xSignIsPositive = true;
w = 1;
for (let i = c.length - 1; i >= 0; i--) {
// If the current digit is the
// missing digit, then keep
// the track of its sign
if (c[i] == 'x') {
xSignIsPositive = (w == 1);
}
else {
c_mod_11 = (c_mod_11
+ w * (c[i].charCodeAt(0) - '0'.charCodeAt(0)))
% 11;
}
w = w * -1;
}
// Find the value of x using
// the derived equation
let x = ((a_mod_11 * b_mod_11)
- c_mod_11)
% 11;
// Check if x has a negative sign
if (!xSignIsPositive) {
x = -x;
}
// Return positive equivaluent
// of x mod 11
return (x % 11 + 11) % 11;
}
// Driver Code
let A = "123456789";
let B = "987654321";
let C = "12193263111x635269";
document.write(findMissingDigit(A, B, C));
// This code is contributed by Potta Lokesh
</script>
2
Complejidad de Tiempo: O(log 10 A + log 10 B)
Espacio Auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por sourashis69 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA