11332 - Summing Digits

[apurba]

is there any algorithm for the problem? very east to understand but its really troubling.

[rio]

Straight ford algorithm. Just calculate.

PS. Please change the title to "11332 - Summing Digits"
----
Rio

[sclo]

is there any algorithm for the problem? very east to understand but its really troubling.

It is the most trivial problem from that problem set.

[shakil]

It's really easy.You can take input in string.And make it digit sum in long.
And next again find digit sum until it one digit.

[apurba]

that is what i am thinking about.............using string.

[apurba]

it is not so simple for all.

[S.M.ferdous]

If a number is divisible by 9 the sum of its digits succesively reduced to 9
or the sum of digits will be reduced to num mod 9.

As for xample 198 = 1+9+8 = 18 = 1+8 = 9 (as 198 mod 9 == 0)
agan 203 = 2+0+3 = 5 (203 mod 9 =5)

so in this problem one doesn't need to calculate any summation but simply mod operation by 9.

S.M.Ferdous

[RC's]

You're right..
There is no need to calculate it using a string.
Besides, your algorithm is very fast.
Thanks..

[turcse143]

using recursion algorithm it may be solved.
n=123456
f(n)=1+2+3+4+5+6=21
f1(n)=2+1=3
if(n<9)
return;

[Obaida]

My bad luck why every time it's me getting WA...
Can someone check this.

Code: Select all

Got Accepted

[helloneo]

My bad luck why every time it's me getting WA...
Can someone check this.

Try this input

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9
18
27
0

My output is

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9
9
9

[Obaida]

Thank you helloneo that was a misunderstanding. I got Accepted.

[lnr]

Code: Select all

This is a very is ploblem in this volume.

[mrmbdctg]

I am getting wrong answer with this code. Please help me

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Accepted

[Bluefin]

my algorithm is

Code: Select all

1. int sum = add every digit of n, since n is at most 2,000,000,000, sum has two digits at most.

2. int temp = sum/10 + sum %10

3. if temp < 10,  print temp
    else  print temp / 10 + temp % 10
 

this algorithm isn't very fast, but it works