OJ Board

Posted: **Sat Oct 21, 2006 11:03 pm**

Try this out.. it worked with me:

cout<<setiosflags(ios::fixed)<<setprecision(0)<<pow(p, 1.0/n)<<endl;

Also this way works too:

Code: Select all

k = pow(p, 1.0/n); printf("%.0lf\n",k);

With both ways n, p, k are defined as double. Let know if it worked with you.

Edit: Edited the codes.. forgot something to mention.

Posted: **Wed Jan 03, 2007 2:51 am**

I did send the exercise 113 and i dont know why.
Here is the code...

Code: Select all

#include <iostream>
#include <cmath>

using namespace std;

int main(){
    double k,p,temp;
    while(cin>>k>>p){
        temp=(1.0)/k;
        cout << floor(pow(p,temp))<<endl;
    }
    return 0;
}

And i too submit is code, but i had been WA.

Code: Select all

#include <iostream>
#include <cmath>

using namespace std;

int main(){
    double k,p,temp;
    while(cin>>k>>p){
        temp=(1.0)/k;
        cout << pow(p,temp)<<endl;
    }
    return 0;
}

Thanks.

Posted: **Wed Jan 03, 2007 6:09 am**

Your program cannot solve the last case of the sample input. Consider using log.

Posted: **Wed Jan 03, 2007 6:22 am**

Try again using

Code: Select all

	cout.setf(ios::fixed); 
	cout.precision(0);

at very beginning of main function

Posted: **Sun Jan 07, 2007 4:28 pm**

Dear group,
My code is wrong, I know that very well. The code just simply hangs.. Can any one help.. Thnaks..

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#include<stdio.h>
#include<math.h>
#include<stdlib.h>

int main(void)
{
	 long long k,                
			 n,          
			   result,   
			   p;      
					   
    unsigned int       i;      
    					
    						
      i = pow(10,101);          			
      
      /*Give p and n we need to find k, which is the n pow k */
    						        
	     scanf("%lld %lld",&n,&p);
	     for(k =1 ; k < pow(10,9); k ++)
	     {
		     result = pow(n,k);
		     if(result == p)
		     {
			     printf("%lld\n",k);
			     	     		  
     		 }
     	
     		 if(p > i)
	     		    exit(0);	 
 	     }
 	     
		
	return 0;
}

Posted: **Sun Jan 14, 2007 6:52 am**

Since the nos are very large here, it is sufficient to take the root and store it in a long double. Just do the following:

1. You need to keep taking input as long as the judge keeps giving them. For that, use

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  while(scanf("%lf %lf",&n,&p)==2)
  {

   // the processing of each pair of input
  }

2. Inside the while loop, simply calculate p to the power 1.0/n, floor the result and print with "%0.lf\n"

Hope this helps

Posted: **Wed Jan 17, 2007 5:13 pm**

I thing double is enough to solve this problem

Posted: **Thu Jul 05, 2007 6:07 am**

Code: Select all


Thanks stubbscroll
GOT AC.

Posted: **Thu Jul 05, 2007 3:44 pm**

Change long double to double and don't forget to remove your code if it works.

Posted: **Mon Jul 09, 2007 1:54 am**

god! this is a frightening bad problem! i remember that i was very frustrated with this one!

Posted: **Thu Jan 31, 2008 1:13 am**

Good. Thanks. I got AC

Posted: **Sun Oct 25, 2009 7:54 am**

I have got acc before. But now I get T/E. Why

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Removed After A/C

Posted: **Thu Oct 29, 2009 1:27 am**

@ noor vai

have a look at the way you are taking the input

it should be while(scanf("%lf %lf",&b,&a)==2)

hope this kids advice helps LOL

Posted: **Wed Nov 04, 2009 9:39 am**

Thanks

Posted: **Thu Jul 22, 2010 1:09 pm**

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;
}

OJ Board

113 - Power of Cryptography

113 Somebody can Help me? WA

113, WA (beginner)

113:

Re: 113- power of cryptography T\E

Re: 113- power of cryptography - What is the algorithm?

Re: 113- power of cryptography - What is the algorithm?

113 whats my problem? please help me