Write a program which can do sum of large numbers having up to 1000 digits
Solution by Arjun Suresh
#include <stdio.h> #include <string.h> #include <ctype.h> void print(char *a, int l); int read(char *a); int sum(char *a, char* b, char * c, int al, int bl); int main(void) { char a[1000], b[1000], c[1000]; //a and b hold the input numbers and c hold the output number. Each array entry is a digit int al, bl; //for storing the number of digits of the two input numbers printf("Enter the first number "); al = read(a); printf("Enter the second number "); bl = read(b); int l = sum(a, b, c, al, bl); printf("sum = "); print(c, l); return 0; } int sum(char *a, char* b, char * c, int al, int bl) { int l = al > bl? al:bl;//l stores the no.of digit of output which may become l+1 due to carry from MSB int i = l; char carry = 0; while(al > 0 && bl > 0) { char val = a[--al] + b[--bl] + carry; c[i--] = val % 10; carry = val / 10; } while(al > 0)//If a has more digits than b { char val = a[--al] + carry; c[i--] = val % 10; carry = val / 10; } while(bl > 0)//If b has more digits than a { char val = b[--bl] + carry; c[i--] = val % 10; carry = val / 10; } c[0] = carry;//Assigning the final carry return l+1;//Return the no. of digits of output } int read(char *a) { char c; int i = 0; do { c = getchar(); } while(isspace(c)); //Reading and discarding any whitespace char typed while(isdigit(c)) { a[i++] = c - 48; //getchar returns the ASCII. So, for 1 it returns 49. Subtracting 48, we get the actual int value entered via keyboard c = getchar(); } return i;//Return the no. of digits read } void print(char *a, int l) { int i; if(a[0] != 0) printf("%d", a[0]);//If final carry is 0, ignore it for(i = 1; i < l; i++) { printf("%d", a[i]); } printf("\n"); }