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。
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例如：
输入： 9999 输出： 99980001 9999*9999 = 99980001
输入： 45454545 输出： 2066115661157025 45454545*45454545 = 2066115661157025
天真的方法 ：一种简单的方法是计算平方，然后将数字与自身相乘。但是在C++中，如果输入是大数，则生成的平方将溢出。
有效的方法 ：一种有效的方法是将数字存储为字符串，并执行 两个大数的乘法 .
以下是上述方法的实施情况：
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> |
输出：
206611570247933883884297520661157025
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