我读过很多博客和代码示例，但我试图通过一种目前逻辑上不工作的方式来实现Karatsuba乘法。 只有个位数的数字乘法是有效的，但是任何长于1的数字都显示完全错误的答案。 它应该能够接受长数量的输入，但我不允许使用长int存储类型。 另外，它的意思是，能够乘以不同的基数（1-10），但我不确定如何做，所以一开始我只是尝试基数10。 这是我到目前为止的代码:

#include <iostream>
#include <string>
#include <algorithm>
#include <cmath>
#include <sstream> 
int compareSize(string integer1, string integer2)
{
    int length = 0;
    if (integer1.length() > integer2.length())
    {
        //allocating int 1's length as final length if bigger than int 2
        length = integer1.length();
    }
    else if (integer2.length() > integer1.length())
    {
        //allocating int 2's length as final length if bigger than int 1
        length = integer2.length();
    }
    else
    {
        length = integer1.length();
    }

    return length;
}

int multiplication( string I1, string I2, int B){
    
    int length = compareSize(I1, I2);

    //converting the strings into integers
    stringstream numberOne(I1);
    int digitOne = 0;
    numberOne >> digitOne;

    stringstream numberTwo(I2);
    int digitTwo = 0;
    numberTwo >> digitTwo;

    //checking if numbers are single digits
    if ( (10 > digitOne) || (10 > digitTwo) ) {
        //cout<<(digitOne * digitTwo)<<endl;
        return (digitOne * digitTwo);
    }

    int size = ( length % 2) + (length / 2);

    int power = pow(10, size);

    int a = digitOne / power;   
    int c = digitTwo / power;   
    int b = digitOne - (a * power);  
    int d = digitTwo - (c * power);  

    int p2 = b * d;   
    int p0 = a * c;   
    int p1 = (b + a) * (d + c);
    //int final = p1 - p2 - p0;
    int sum = ((a*d) + (b*c)) ;

    int p = (p0*(pow(10,length))) + (sum * (pow(10,length)) + p2);
    //int p = ( p2 * (pow(10,size*2)) + ( p1 - (p2+p0)) * pow(10,size) + p0 ) * 100;
    // int p = (p2 * (long long)(pow(10, 2 * size))) + p0 + ((p1 - p0 - p2) * power);
    return p; 
}

int main(){

    string int1, int2;
    int base;
    cout<<"Enter I1, I2 and B: ";
        cin>> int1 >> int2 >> base;
    
    cout<<" "<<multiplication(int1, int2, base)<<endl;

    return 0;
    
}