Simple addition of large numbers

Ideas:

Use a string to represent a large number, and then use a column-like vertical method to sum two numeric strings by bit.

123 +

24

_____

147

Specific implementation:

Char * BigNumberAddition (const char * number1, const char * number2) {if (! Number1 |! Number2) {return NULL;} int len1 = strlen (number1); int len2 = strlen (number2 ); // the result of adding two numbers is the number of digits of a large number or the number of digits of a large number + 1 int result_len = max (len1, len2) + 1; char * result = (char *) malloc (result_len + 1); memset (result, 0, result_len + 1); int index = result_len; -- index; -- len1; -- len2; while (len1> = 0 & len2> = 0) {// add result by bit [index] + = (number1 [len1]-'0 ') + (number2 [len2]-'0'); if (result [index]> = 10) {// carry result [index-1] + = result [index]/10; result [index] = result [index] % 10;} -- index; -- len1; -- len2;} while (len1> = 0) {result [index] + = (number1 [len1]-'0 '); if (result [index]> = 10) {// carry result [index-1] + = result [index]/10; result [index] = result [index] % 10;} -- len1; -- index;} while (len2> = 0) {result [index] + = (number2 [len2]-'0'); if (result [index]> = 10) {// carry r Esult [index-1] + = result [index]/10; result [index] = result [index] % 10;} -- len2; -- index ;} // remove the first 0 int gap = result [0] = 0? 1: 0; for (int I = 0; I <result_len; I ++) {result [I] = result [I + gap] + '0 ';} result [result_len-gap] = '\ 0'; return result ;}

Note:

1. Carry should be taken into account when adding one by one

2. Generally, two numbers are added together. For example, if the number of digits in number 1 is a and the number of digits in number 2 is B, the number of digits in the addition result may be max (a, B) or max (a, B) + 1. therefore, note that the first part of the add result may be 0.

3. When a string is used to represent a number, it is used to represent the 0 --> '0' of the number. Therefore, the character must be converted to the corresponding number during addition calculation.

 

Int _ tmain (int argc, _ TCHAR * argv [])
{
Char X number1 = "9999999 ";
Char X number2 = "9999999 ";
Char * result = BigNumberAddition (number1, number2 );
Cout <result;
Free (result );

Return 0;

}

 

Running result:

 

Simple addition of large numbers