113 - Power of Cryptography

[40019JR]

[cpp]
This is my C++ answer for 113.
I got WA. However, I don't understand why. I thought it perfectly solves the problem. Can anyone help me?
#include <iostream>
#include <cmath>

using namespace std;

int main()
{
double n, k, p;

while (cin >> n >> p)
{

k = pow(p,(1/n));
cout << k << endl;
}
return 0;
} [/cpp]

[shamim]

Hi, I corrected the following line and got AC.

[cpp]cout << k << endl;[/cpp]

Most probably the default number digit after decimal is 6, so in UVA compiler your ans. for
2
16
would be
4.000000 and not 4.

Change it to the following line.
[cpp]printf("%.0lf\n",k); [/cpp]

[40019JR]

Thanks for your reply.
I corrected the line and sent it. However, I got following compiler errors.
Also, when I run it on VC, I get no error but the output is funny. I mean, this doesn't solve the problem.
So if you have any thought on that, I appreciate it.
02143378_24.c: In function `int main ()':
02143378_24.c:14: `printf' undeclared (first use this function)
02143378_24.c:14: (Each undeclared identifier is reported only once for
each function it appears in.)

[Almost Human]

the library iostream used by the judge's compiler doesn't support the printf function, so you have to include another library ( stdio.h )

I hope it helped you

[40019JR]

Thanks everyone. It worked.
But like how did you know that their compiler interpret it down to 6 decimal points and stuff like that?
Just out of curiosity.

[Almost Human]

I guess experience would be the greatest teacher in that case.

[Jer]

help me~~ plz~~

[cpp]
#include <iostream.h>
#include <math.h>
int main(void)
{
double p,n;
while(cin>>n>>p) {
cout<<pow(p,(double)1/n)<<endl;
}
return 0;
}[/cpp]

[pavelph]

1<=p<10^101

I think that double not support so big numbers.

[Krzysztof Duleba]

No, double is enough. I have a similar solution, but with two more lines at the begining of the main function:
[cpp] cout.setf(ios::fixed);
cout.precision(0); [/cpp]
I don't remember why I put them there, but there must have been a reason

[Jer]

No, double is enough. I have a similar solution, but with two more lines at the begining of the main function:
[cpp] cout.setf(ios::fixed);
cout.precision(0); [/cpp]
I don't remember why I put them there, but there must have been a reason

I fix the code
[cpp] cout<<(long)pow(p,(double)1/n)<<endl [/cpp]
and some answers will be decrease like 1234->1233
so i think the answers' form i calculate are wrong......

[Krzysztof Duleba]

So you finally got accepted?

[Jer]

Oh!!yes....~~thank you~~

[aditya]

i am trying to use the double data type but if i input a number having more than say 20 digits and then just print it ( no processing ) , i am not getting the same number that i had entered. after 20 digits the precision is lost.
could anyone pls suggest why it is so?

p.s. i have even tried long double

[Sayutee]

Your code is too much complex i think- the problem is not so much complex.

I got WA with double but AC with Long Double.

I used natural log instead of base 10.

k = exp(log(p)/n) is enough to solve it.

best of luck

[shamim]

I used double and got AC.

I think, the judge's data has no such case where precision will be lost.