Dado un número muy grande, la tarea es escribir un programa para calcular su cuadrado.
Ejemplos:
Entrada: 9999
Salida: 99980001
9999*9999 = 99980001Entrada: 45454545
Salida: 2066115661157025
45454545*45454545 = 2066115661157025
Enfoque ingenuo : un enfoque ingenuo es calcular los cuadrados multiplicando el número por sí mismo. Pero en C++, si la entrada es un número grande, el cuadrado resultante se desbordará.
Enfoque eficiente : un enfoque eficiente es almacenar el número como strings y realizar la multiplicación de dos números grandes .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to multiply two numbers
// represented as strings.
#include <bits/stdc++.h>
using namespace std;
// Multiplies str1 and str2, and prints result.
string multiply(string num1, string num2)
{
int n1 = num1.size();
int n2 = num2.size();
if (n1 == 0 || n2 == 0)
return "0";
// will keep the result number in vector
// in reverse order
vector<int> result(n1 + n2, 0);
// Below two indexes are used to find positions
// in result.
int i_n1 = 0;
int i_n2 = 0;
// Go from right to left in num1
for (int i = n1 - 1; i >= 0; i--) {
int carry = 0;
int n1 = num1[i] - '0';
// To shift position to left after every
// multiplication of a digit in num2
i_n2 = 0;
// Go from right to left in num2
for (int j = n2 - 1; j >= 0; j--) {
// Take current digit of second number
int n2 = num2[j] - '0';
// Multiply with current digit of first number
// and add result to previously stored result
// at current position.
int sum = n1 * n2 + result[i_n1 + i_n2] + carry;
// Carry for next iteration
carry = sum / 10;
// Store result
result[i_n1 + i_n2] = sum % 10;
i_n2++;
}
// store carry in next cell
if (carry > 0)
result[i_n1 + i_n2] += carry;
// To shift position to left after every
// multiplication of a digit in num1.
i_n1++;
}
// ignore '0's from the right
int i = result.size() - 1;
while (i >= 0 && result[i] == 0)
i--;
// If all were '0's - means either both or
// one of num1 or num2 were '0'
if (i == -1)
return "0";
// generate the result string
string s = "";
while (i >= 0)
s += std::to_string(result[i--]);
return s;
}
// Driver code
int main()
{
string str1 = "454545454545454545";
cout << multiply(str1, str1);
return 0;
}
Java
// Java program to multiply two numbers
// represented as strings.
class GFG
{
// Multiplies str1 and str2, and prints result.
public static String multiply(String num1, String num2)
{
int n1 = num1.length();
int n2 = num2.length();
if (n1 == 0 || n2 == 0)
return "0";
// will keep the result number in vector
// in reverse order
int[] result = new int[n1 + n2];
// Below two indexes are used to find positions
// in result.
int i_n1 = 0;
int i_n2 = 0;
// Go from right to left in num1
for (int i = n1 - 1; i >= 0; i--)
{
int carry = 0;
int n_1 = num1.charAt(i) - '0';
// To shift position to left after every
// multiplication of a digit in num2
i_n2 = 0;
// Go from right to left in num2
for (int j = n2 - 1; j >= 0; j--)
{
// Take current digit of second number
int n_2 = num2.charAt(j) - '0';
// Multiply with current digit of first number
// and add result to previously stored result
// at current position.
int sum = n_1 * n_2 + result[i_n1 + i_n2] + carry;
// Carry for next iteration
carry = sum / 10;
// Store result
result[i_n1 + i_n2] = sum % 10;
i_n2++;
}
// store carry in next cell
if (carry > 0)
result[i_n1 + i_n2] += carry;
// To shift position to left after every
// multiplication of a digit in num1.
i_n1++;
}
// ignore '0's from the right
int i = result.length - 1;
while (i >= 0 && result[i] == 0)
i--;
// If all were '0's - means either both or
// one of num1 or num2 were '0'
if (i == -1)
return "0";
// generate the result string
String s = "";
while (i >= 0)
s += Integer.toString(result[i--]);
return s;
}
// Driver code
public static void main(String[] args)
{
String str1 = "454545454545454545";
System.out.println(multiply(str1, str1));
}
}
// This code is contributed by
// sanjeev2552
Python3
# Python3 program to multiply two numbers
# represented as strings.
# Multiplies str1 and str2, and prints result.
def multiply(num1, num2):
n1 = len(num1)
n2 = len(num2)
if (n1 == 0 or n2 == 0):
return "0"
# Will keep the result number in vector
# in reverse order
result = [0] * (n1 + n2)
# Below two indexes are used to
# find positions in result.
i_n1 = 0
i_n2 = 0
# Go from right to left in num1
for i in range(n1 - 1, -1, -1):
carry = 0
n_1 = ord(num1[i]) - ord('0')
# To shift position to left after every
# multiplication of a digit in num2
i_n2 = 0
# Go from right to left in num2
for j in range(n2 - 1, -1, -1):
# Take current digit of second number
n_2 = ord(num2[j]) - ord('0')
# Multiply with current digit of first number
# and add result to previously stored result
# at current position.
sum = n_1 * n_2 + result[i_n1 + i_n2] + carry
# Carry for next iteration
carry = sum // 10
# Store result
result[i_n1 + i_n2] = sum % 10
i_n2 += 1
# Store carry in next cell
if (carry > 0):
result[i_n1 + i_n2] += carry
# To shift position to left after every
# multiplication of a digit in num1.
i_n1 += 1
# Ignore '0's from the right
i = len(result) - 1
while (i >= 0 and result[i] == 0):
i -= 1
# If all were '0's - means either both or
# one of num1 or num2 were '0'
if (i == -1):
return "0"
# Generate the result string
s = ""
while (i >= 0):
s += str(result[i])
i -= 1
return s
# Driver code
if __name__ == "__main__":
str1 = "454545454545454545"
print(multiply(str1, str1))
# This code is contributed by chitranayal
C#
// C# program to multiply two numbers
// represented as strings.
using System;
using System.Collections.Generic;
class GFG
{
// Multiplies str1 and str2,
// and prints result.
public static String multiply(String num1,
String num2)
{
int n1 = num1.Length;
int n2 = num2.Length;
if (n1 == 0 || n2 == 0)
return "0";
// will keep the result number in vector
// in reverse order
int[] result = new int[n1 + n2];
// Below two indexes are used to
// find positions in result.
int i_n1 = 0;
int i_n2 = 0;
int i = 0;
// Go from right to left in num1
for (i = n1 - 1; i >= 0; i--)
{
int carry = 0;
int n_1 = num1[i] - '0';
// To shift position to left after every
// multiplication of a digit in num2
i_n2 = 0;
// Go from right to left in num2
for (int j = n2 - 1; j >= 0; j--)
{
// Take current digit of second number
int n_2 = num2[j] - '0';
// Multiply with current digit of first number
// and add result to previously stored result
// at current position.
int sum = n_1 * n_2 +
result[i_n1 + i_n2] + carry;
// Carry for next iteration
carry = sum / 10;
// Store result
result[i_n1 + i_n2] = sum % 10;
i_n2++;
}
// store carry in next cell
if (carry > 0)
result[i_n1 + i_n2] += carry;
// To shift position to left after every
// multiplication of a digit in num1.
i_n1++;
}
// ignore '0's from the right
i = result.Length - 1;
while (i >= 0 && result[i] == 0)
i--;
// If all were '0's - means either both or
// one of num1 or num2 were '0'
if (i == -1)
return "0";
// generate the result string
String s = "";
while (i >= 0)
s += (result[i--]).ToString();
return s;
}
// Driver code
public static void Main(String[] args)
{
String str1 = "454545454545454545";
Console.WriteLine(multiply(str1, str1));
}
}
// This code is contributed by Princi Singh
Javascript
<script>
// Javascript program to multiply two numbers
// represented as strings.
// Multiplies str1 and str2, and prints result.
function multiply(num1,num2)
{
let n1 = num1.length;
let n2 = num2.length;
if (n1 == 0 || n2 == 0)
return "0";
// will keep the result number in vector
// in reverse order
let result = new Array(n1 + n2);
for(let i=0;i<result.length;i++)
{
result[i]=0;
}
// Below two indexes are used to find positions
// in result.
let i_n1 = 0;
let i_n2 = 0;
// Go from right to left in num1
for (let i = n1 - 1; i >= 0; i--)
{
let carry = 0;
let n_1 = num1[i].charCodeAt(0) - '0'.charCodeAt(0);
// To shift position to left after every
// multiplication of a digit in num2
i_n2 = 0;
// Go from right to left in num2
for (let j = n2 - 1; j >= 0; j--)
{
// Take current digit of second number
let n_2 = num2[j].charCodeAt(0) - '0'.charCodeAt(0);
// Multiply with current digit of first number
// and add result to previously stored result
// at current position.
let sum = n_1 * n_2 + result[i_n1 + i_n2] + carry;
// Carry for next iteration
carry = Math.floor(sum / 10);
// Store result
result[i_n1 + i_n2] = sum % 10;
i_n2++;
}
// store carry in next cell
if (carry > 0)
result[i_n1 + i_n2] += carry;
// To shift position to left after every
// multiplication of a digit in num1.
i_n1++;
}
// ignore '0's from the right
let i = result.length - 1;
while (i >= 0 && result[i] == 0)
i--;
// If all were '0's - means either both or
// one of num1 or num2 were '0'
if (i == -1)
return "0";
// generate the result string
let s = "";
while (i >= 0)
s += (result[i--]).toString();
return s;
}
// Driver code
let str1 = "454545454545454545";
document.write(multiply(str1, str1));
// This code is contributed by avanitrachhadiya2155
</script>
Producción:
206611570247933883884297520661157025
Publicación traducida automáticamente
Artículo escrito por AdityaVerma1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA