给定一个非常大的数字，任务是编写一个程序来计算其平方。
例子：
Input: 9999
Output: 99980001
9999*9999 = 99980001
Input: 45454545
Output: 2066115661157025
45454545*45454545 = 2066115661157025
幼稚的方法：幼稚的方法是通过将数字乘以自身来计算平方。但是在C++中，如果输入的数字很大，则结果平方将溢出。
高效的方法：一种有效的方法是将数字存储为字符串，然后对两个大数字进行乘法运算。
下面是上述方法的实现：
C++
// C++ program to multiply two numbers
// represented as strings.
#include
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 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
// sanjeev2552Python3
# 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 chitranayalC#
// 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输出：
206611570247933883884297520661157025想要从精选的最佳视频中学习并解决问题，请查看有关从基础到高级C++的C++基础课程以及有关语言和STL的C++ STL课程。要完成从学习语言到DS Algo等的更多准备工作，请参阅“完整面试准备课程” 。