Addition and multiplication of large numbers of large numbers

Learning this blog article: https://www.cnblogs.com/-Ackerman/

First we look at the addition of large numbers

 

 

 Can be understood, first added together, then the carry, of course, we are all here with a string to store, look at the code to understand the

#include <. bits / STDC H ++>
 the using  namespace STD; 

String BigAdd ( String S1, String S2) 
{ 
    String Anser;
     int LEN1 s1.length = (); // Get the character length 
    int LEN2 = s2.length ();
     int num1 [ 100 ] = { 0 }, num2 [ 100 ] = { 0 }; // this can be solved preamble 0 
    for ( int I = 0 ; I <LEN1; I ++ ) 
        num1 [LEN1 -I- . 1 ] S1 = [I] - ' 0 '; // reverse order to add num1, that num1 is in accordance with a ten hundred. . . 
    for ( int I = 0 ; I <LEN2; I ++ ) 
        num2 [LEN2 -I- . 1 ] = S2 [I] - ' 0 ' ;
     int lenth = LEN1> LEN2? LEN1: LEN2;
     for ( int I = 0 ; I <lenth; I ++ ) 
    { 
        [I] num1 + = [I] num2; // you adder 
        num1 [I + . 1 ] + = num1 [I] / 10 ; // carry, after small bits into a large position 
        num1 [ I]% = 10 ; // save the remainder after carried to the small bit
    }
     IF (! Num1 [lenth]) // go preamble 0
 
    lenth-- ;
     for ( int I = lenth; I> = 0 ; i-- ) 
    { 
        Anser + num1 = [I] + ' 0 ' ; 
     } 
     return Anser ; // the results returned 
} 
 
int main () 
{ 
    String A, B;
     the while (CIN >> A >> B) 
    { 
        COUT << BigAdd (A, B) << endl; 
    } 
    return  0 ; 
}

If you can understand large numbers of addition, multiplication of large numbers and so it is in fact carry on the same principle, but need to understand how to calculate multiplication

I 98 * 23, for example, answer = 3 * 8 * 9 * 10 + 3 + 2 + 2 * 10 * 8 * 9 * 10 * 10, you can verify yourself I write right

Law of multiplication: j-th bit number that is obtained by multiplying the number of the i-th bit and a number of other, must be accumulated to the result of the i + j bits. Here i, j is from right to left, starting from 0 count.
That is: ans [i + j] = a [i] * b [j];

#include <. bits / STDC H ++>
 the using  namespace STD;
 #define MAX 100    
 int X [MAX + 10 ], Y [MAX + 10 ], Z [MAX * 2 + 10 ], I, J;
 String BigMultiply ( String A, String B) 
{ 
    int LEN1 = a.length ();
     int LEN2 = to b.length ();
     string Anser;
      for (J = 0 , I = len1- . 1 ; I> = 0 ; i--) // string characters into digital, and store reverse 
        X [J ++] = A [I] - ' 0' ; // here added a more convenient J is equivalent to the addition of large numbers of the above [len-i-1], see you readily appreciated that 
    for (J = 0 , I = len2- . 1 ; I> = 0 ; I- - ) 
        Y [J ++] = B [I] - ' 0 ' ;
     for ( int I = 0 ; I <LEN1; I ++ ) 
    { 
        for ( int J = 0 ; J <LEN2; J ++ ) 
        { 
            Z [I + J] + = X [I] * Y [J]; //  
        } 
    } 
    for ( int I = 0 ; I < 2 * MAX; I ++) 
    { 
        IF (Z [I]> = 10 ) // binary 
        { 
        Z [I + . 1 ] + = (Z [I] / 10 ); 
        Z [I] = Z [I]% 10 ; 
        } 
    } 
    for (I * MAX = 2 ; I> 0 ; i--)   // remove prefix 0 
    {
         IF (Z [I] == 0 )
             Continue ;
         the else 
            BREAK ; 
    } 
    for (; I> = 0 ; i--)   // reverse output 
        anser + = z [i] +'0';
    return anser;
} 

int main()
{
    string a,b;
    while(cin>>a>>b)
    {
        cout<<BigMultiply(a,b)<<endl;
    }
    return 0;
}