10346 - Peter's Smokes

[athlon19831]

please help me

#include "iostream.h"
int main(int argc, char* argv[])
{
long n,k;
long sum;
while(cin>>n>>k)
{
sum=n;
if(k>1)
{
while(n>=k)
{
sum+=n/k;
n=n/k;
}
cout<<sum<<endl;
}
}
return 0;
}

[Wei-Ming Chen]

If the input is
31 2

The output might be 61
But your output says 57


Becare of the problem:
31/2=15...1
15/2=7....1
7/2=3....1
3/2=1....1
so your output is 31+15+7+1=57
But there are five 1 left!!

[ishtiaq ahmed]

Actully i dont know why i am facing wrong ans
plz help me.i am a beginner in programmig
my code is

Code: Select all

#include<stdio.h>
void main()
{
	long long a,b,sum;
	while(scanf("%lld %lld",&a,&b)!=EOF)
	{
		if(b>1)
		{
			sum=a;
			while(a!=0)
			{
				a=a/b;
				sum= sum+a;
			}
			printf("%lld\n",sum);
		}
	}
}

[DP]

You can check this I/O:

Code: Select all

3 2

Output:

Code: Select all

5

1 1 1 =3
---
1 =1
1
----- =1
Total =5

[newton]

delete your code and check the link. you will get more information about peter problem.



http://online-judge.uva.es/board/viewto ... ight=10346






newton... simply the best

[KaDeG]

I have made the 10346 problem but got Wa. I searched the forums and found the smart formula: sumciggaret=n+(n-1)/(k-1); I did this and got AC. The problem is with my code, the first idea i had i got WA, the problem testcases works fine and also some other testecase i created from my AC program are OK.
But i still get WA, any ideas why would be helpful...

Code: Select all

#include <stdio.h>

main()  {
        int n, k, sum, nk;
        while((scanf("%d %d", &n, &k))==2)      {
                sum=n;
                nk=sum/k;
                sum+=nk;
                while(nk>=k)    {
                        nk/=k;
                        sum+=nk;
                }
                printf("%d\n", sum);
        }
}

Thanks

[Jan]

Try the cases...

Input:

Code: Select all

8492 142
15881 300
10677 119

Output:

Code: Select all

8552
15934
10767

Hope these help.

[KaDeG]

yeah, fixed and got AC
thanks

[asif_khan_ak_07]

I wrote the following code and got WA.I tried some test cases and the answers were correct.pls help me to find the mistake

Code: Select all

#include<stdio.h>

int main()
{
long double x,n,ans,r;

while((scanf("%Lf %Lf",&x,&n)==2))
{

if(x<n)
printf("%.0Lf\n",x);

else
{
	
ans=x;

r=x/n;
do
{
ans=ans+r;
r=r/n;

}while(r>=1);

printf("%.0Lf\n",ans);

}

}
return 0;
}

[saiful_sust]

Hi asif_khan
Try this case:

INPUT:

Code: Select all

12 3
1200 5
1200 1200
3 2
25326 3652
456987 326596

OUTPUT:

Code: Select all

17
1499
1201
5
25332
456988

[*] IMPOSSIBLE MEANS I M POSSIBLE /[*]

[kaysar_buet08]

where is my error!!

#include<stdio.h>
int n,a,b;
int main()
{
while((scanf("%d%d",&a,&b))!=EOF)
{
n=a;
while(a>=b)
{
a=a/b;
n+=a;
}
printf("%d\n",n);
}
return 0;
}

[khairul]

hi kaysar u must keep rest by modules operator.
suppose,
5 3
output should 7 not 6.
because 5+1+1=5+1+((1+2))/3=7

and also use unsigned int/long.
i think it would work and let me know...
thanks,
khairul

[imran_12]

I cant understand why i am getting WA..
plz help

Code: Select all

#include<stdio.h>
int main()
{
    int n,k,m,p;
    while(scanf("%d %d",&n,&k)==2)
    {
        m=n;
        if(k>1)
        {
            p=n%k;
            n=n/k;
            p=(p+n)/k;
            m=m+n+p;
        }
        printf("%d\n",m);
    }
}

[lighted]

Use search and post in existing thread. http://acm.uva.es/board/search.php?keyw ... f03df06864
Input

Code: Select all

1000 5

Acc Output

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1249

Your Output

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1240

You are doing rolling process one time using if operator. I do it as many as it is possible using loop.

[tomb_raider]

Why Wrong ANSWER MANNNNN !?!?!?!?!?!?

Code: Select all

#include <stdio.h>
int main()
{
    long long int a,b,x,y,c,sum1,sum2;
    while(scanf("%lld %lld",&a,&b)==2)
    {
        sum1=0,sum2=0,c=0;
        while(a>0)
        {
            x=a%b;
            sum2=x+sum2;
            if(sum2/b>0)
            {
                c++;
                sum2=sum2/b;
            }
            sum1=sum1+a;
            a=a/b;
        }
        sum1=sum1+c;
        printf("%lld\n",sum1);
    }
    return 0;
}