113 - Power of Cryptography

[@mjad]

please help me

Code: Select all

#include<math.h>
#include<iostream>
using namespace std;
int main()
{
   long double f,s,t;
   //freopen("in.txt","r",stdin);
   while(scanf("%lf%lf",&f,&s)==2)
   {
      t=exp(log(s)/f);
      printf("%.0lf\n",t);
   }
   return 0;
}

[Hobby]

please :
printf("%ld",...)

[@mjad]

thanks hobby
but i did not get AC
it always WA
please help me, Specific way?

[shaon_cse_cu08]

There is no problem in outputting with "%.0f" .... Just take a look at the problem carefully...

"For all such pairs 1<=n<=200 and the other condition's ...."

Try this input....

-4
16
0
56
0
0

Ur program alsways generates... 1 or 0 for this cases.... there will be no output for this

To test ur input and output try "http://uvatoolkit.com/"...

Hope That works... And don't open new Thread....if there is already one.....PLZZZ

[zobayer]

You got WA because your solution is not correct, (I guess precision error)
For double data types, try to minimize the level of computation as more calculation will give you more errors.

[shaon_cse_cu08]

I Have already given its ans in ur previous tread and again u r opening a new tread... See ur previous post.... !!!

[Trarex]

Can someone say me about mistakes in my code? There should not be any problems with vars range (I've used float and double). So where can the problem hide?

Code: Select all

#include <iostream>
#include <math.h>
using namespace std;
int main()
{
    float n;
    double p;
    while (cin >> n >> p) cout << exp(log(p) / n) << endl;
    return 0;
}

Lately I modified the code in the next way:

Code: Select all

#include <iostream>
#include <math.h>
using namespace std;
int main()
{
    float n;
    double p;
    while (cin >> n >> p) cout << long(exp(log(p)/n)) << endl;
    return 0;
}

What is wrong?

[sir. manuel]

try with pow(p,1/n)

[Trarex]

WA ;(

Code: Select all

#include <iostream>
#include <math.h>
using namespace std;
int main()
{
    float n;
    double p;
    cout.precision(0);
    //while (cin >> n >> p) cout << long(exp(log(p)/n)) << endl;
    while (cin >> n >> p) cout << long(pow(p,1/n)) << endl;
    return 0;
}

[sir. manuel]

Code: Select all

#include<stdio.h>
#include<math.h>
int main()
{
double a,b;
while(scanf("%lf %lf",&a,&b)==2){
                                printf("%.0lf\n",pow(b,1/a));
                                } 
return 0;
}

Always use double!!!

[Trarex]

and your solution has been accepted?

I've tried this one, but still WA

Code: Select all

#include <iostream>
#include <math.h>
using namespace std;
int main()
{
    //float n;
    double n,p;
    cout.precision(0);
    //while (cin >> n >> p) cout << long(exp(log(p)/n)) << endl;
    while (cin >> n >> p) cout << long(pow(p,1/n)) << endl;
    return 0;
}

[foo]

Hi,
I just did problem '113 - Power of Cryptography', using the simple pow(p, 1/n) technique.
Though I don't quite see how this can be correct, as <= p <= 10^101 .
Isn't this too big for a double? For ex in java:
double d = Double.parseDouble("4357186184021382204544");
System.out.println(d); // outputs 4.357186184021382E21, major precision loss?

So shouldn't taking the square root of this number then give an incorrect answer?

thanks

[brianfry713]

double has enough precision for the I/O in the judge for this problem.

You are only taking the square root if n is 2.

[BlâckFoX]

Hii can anyone help me?

I submitted this ,but the judge kept saying i have wrong answer =( please help
#include <iostream>
#include <math.h>
using namespace std;

int main(){
double n, p;
double ans;
while(cin >> n >> p){
ans= pow(p,1/n);
cout << round(ans) << endl;
}
return 0;
}

[brianfry713]

Try printing only the integer portion instead of using round().