Program design of extra Large number addition C language

#include  <stdio.h> #include  <string.h>int main (Int argc, const char  * argv[])  {        char one[100],two[100],sum[100 ];        int temp=0,lenth,lenthtwo,i,lenthofsum;   &NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;SCANF ("%s", one);     getchar ();//read carriage return character    &NBSP;&NBSP;SCANF ("%s", and;        lenthtwo =  (int) strlen ( Both);        if  (strlen () >strlen (one)//always let one store longer strings     {        char stringtemp[100];         strcpy (Stringtemp, two);         strcpy (Two, one);         strcpy (one, stringtemp) ;     }        lenth =  (int) (strlen (one)-strlen (one));         for  (i= (int) strlen (both) -1; i>=0 ; i--)      {        sum[i+lenth] = two[i] + one[i+lenth ]-' 0 ' +temp;                if   (sum[i+lenth]>57)  {            sum[i +lenth] = sum[i+lenth] - 10;             temp = 1;        }         else        {             temp = 0;        }     }    for  (i=lenth-1; i>=0; i--)  {         sum[i] = one[i]+temp;        if  (sum[i]> )         {             sum[i] = sum[i] -10;             temp = 1;        }         else        {             temp = 0;        }     }    lenthofsum = strlen (sum);     if   (temp==1)     {//the highest bit may be near-bit, resulting in a number longer than the length of the String one, so you need to move all the numbers back one, before adding a          for  (int i=0; i<lenthofsum; i++)          {            sum[lenthofsum-i] =  sum[lenthOfsum-i-1];        }         sum[0] =  ' 1 ';     }    printf ("%s\n", sum);     return 0;} Results Demo//3032832398719038781904783904783294873248783493194394132041834//                 323819473948324891247129084723412348713403284//---  ----------------------------------------------------------//3032832398719039105724257853108186120377868216606742845445118

This article is from the Web Learning Summary blog, so be sure to keep this source http://8947509.blog.51cto.com/8937509/1572062

Program design of extra Large number addition C language