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Write a program which can do sum of large numbers having up to 1000 digits
  
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==={{Template:Author|Arjun Suresh|{{arjunweb}} }}===
  
 
<syntaxhighlight lang="c" name="largesum">
 
<syntaxhighlight lang="c" name="largesum">
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int sum(char *a, char* b, char * c, int al, int bl)
 
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 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;
 
         int i = l;
 
         char carry = 0;
 
         char carry = 0;
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         c[0] = carry;//Assigning the final carry
 
         c[0] = carry;//Assigning the final carry
 
return l+1;//Return the no. of digits of output
 
return l+1;//Return the no. of digits of output
 +
}
  
}
 
 
int read(char *a)
 
int read(char *a)
 
{
 
{
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return i;//Return the no. of digits read
 
return i;//Return the no. of digits read
 
}
 
}
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void print(char *a, int l)
 
void print(char *a, int l)
 
{
 
{

Latest revision as of 15:54, 17 June 2014

Write a program which can do sum of large numbers having up to 1000 digits

Solution by Arjun Suresh

<syntaxhighlight lang="c" name="largesum">

  1. include <stdio.h>
  2. include <string.h>
  3. 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"); } </syntaxhighlight>





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<syntaxhighlight lang="c" name="largesum">

  1. include <stdio.h>
  2. include <string.h>
  3. 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"); } </syntaxhighlight>





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