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This is a very easy problem to solve. Just for practice!
Find the sum of the digits in the following string -
dfjksdj32423skfjsdfj78shdfjsh0hf3bsdjf3jfdsfj134343fbjdsfkjsdbfsjd017834sdfsdjfdsjfk972347jkhsfdjhsd34jhgha766afsdkfjdjfdfbsdj
Feel free to use any language...
Last edited by nsmgr8 (March 17, 2009 12:28 am)
<?php
$string = 'dfjksdj32423skfjsdfj78shdfjsh0hf3bsdjf3jfdsfj134343fbjdsfkjsdbfsjd017834sdfsdjfdsjfk972347jkhsfdjhsd34jhgha766afsd';
$size = str_split($string);
foreach($size as $key)
{
if(is_numeric($key))
$total = $total + $key;
}
echo $total;
?>
i think it's done 
Yes, that's well done. Now let's see how do you do it in Python.
s = 'dfjksdj32423skfjsdfj78shdfjsh0hf3bsdjf3jfdsfj134343fbjdsfkjsdbfsjd017834sdfsdjfdsjfk972347jkhsfdjhsd34jhgha766afsd'
print sum([int(i) for i in s if i in '1234567890'])Cool, no?
It's just plain English! If you read it loud, you'll read -
Print sum of int of i for i in s if i in one, two, three, four, five, six, seven, eight, nine, zero.
Last edited by nsmgr8 (March 18, 2009 9:23 pm)
wow 
, it's really great like English language. 
in c
#include<stdio.h>
#include<string.h>
int main()
{
char a[100];
int i,sum=0,L;
gets(a);
L=strlen(a);
for (i=0;i<L;i++)
{
if(a[i]>='0' && a[i]<='9')
sum=sum+a[i]-48;
}
printf("%d",sum);
return 0;
}
Once i wrote the code for detecting numbers between string and add....like..12sd1er7..sum should be 20. But, i think, i lost it. 
wrote that again
#include<stdio.h>
#include<string.h>
int main()
{
char a[100];
int i,sum=0,L,digit=0;
gets(a);
L=strlen(a);
for (i=0;i<L;i++)
{
if(a[i]>='0' && a[i]<='9')
{
digit=digit+a[i]-48;
if(a[i+1]>='0' && a[i+1]<='9')
digit=digit*10;
else
{
sum=sum+digit;
digit=0;
}
}
}
printf("%d",sum);
return 0;
}
it will find all "int" from a string and will add them.
is it ok sopnochari vaia?
mahdee, sum of integers is nice. Can you do it with float number now? I mean if you have "asd34.5hf23", then the result is 57.5.
Never use gets() for input. This is very dangerous. There are numerous thing can happen with it. And you cannot find the bug in your software if you use it.
i know the denger with that gets(). It holds the enter as an input and if i take input with scanf or somthing, then the first input i get, should be blank. I faced this problem and that took a lot time to find , what the bug was.
I ll try that float problem tonight
Here is a simpler solution to find the sum of integers in C:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main() {
int l, loc = 0, sum = 0;
char s[] = "dfjksdj32423skfjsdfj78shdfjsh0hf3bsdjf3jfdsfj134343fbjdsfkjsdbfsjd017834sdfsdjfdsjfk972347jkhsfdjhsd34jhgha766afsdkfjdjfdfbsdj";
char d[] = "1234567890";
l = strlen(s);
while(loc < l) {
loc += strcspn(s+loc, d);
sum += atoi(s+loc);
loc += strspn(s+loc, d);
}
printf("%d\n", sum);
return 0;
}
#include<stdio.h>
int main()
{
char a[100];
int i,L,count=1,j;
float sum=0,digit=0,point=0,sump=0;
for(i=0;i<100;i++)
{
scanf("%c",&a[i]);
if(a[i]=='\n')
break;
}
L=i; //strlen
for (i=0;i<L;i++)
{
if(a[i]>='0' && a[i]<='9')
{
digit=digit+a[i]-48;
if(a[i+1]>='0' && a[i+1]<='9')
digit=digit*10;
else
{
sum=sum+digit;
digit=0;
}}
if(a[i]=='.')
{
for(i=i+1;a[i]>='0' && a[i]<='9';i++)
{
point=point+a[i]-48;
if(a[i+1]>='0' && a[i+1]<='9')
{
point=point*10;
count++;
}
else
{ for(j=0;j<count;j++)
point=point/10;
sump=sump+point;
point=0;
count=1;
}
}
}
}
printf("%f",sum+sump);
return 0;
}
finally done 
he he he, Nasim vai's solution is so small. I did not tried in C 
i haven't used "stdlib.h" till 
. i think, i should start using it . 
good works:)
Dont know How it will look like.
int main(){
char c;
int sum=0;
while(scanf("%c",&c), c!='\n') sum += (c-'0')*(c>'0' && c<='9');
printf("Sum: %d\n",sum);
}
Last edited by shiplu (March 19, 2009 9:33 am)
wow.
Do (c>'0' && c<='9') (or such other conditions ) always return 1 or 0?
Yes, boolean operations always return 0 or 1.
Your float sum is also working. Nice!
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