113 - Power of Cryptography

[rickyok]

Hi, I'm a newby here
Please help me on this case
I can easly make a function for search k for input :
2
16
3
27

But it goes to error when i got input like
7
4357186184021382204544

Any ideas how i can handle very2 large integer
(Without using float/ double) coz i've tried double
and have no used against very large int.

Thank you
PLease reply

[ithamar]

Implement your own routines to work with BigNumbers. Its not that hard . Strings is the answer ...

[skylander]

you do not need to use string arithmetic, just some formula and a double will do the job...

[rickyok]

Well i've tried the string arithmathic but i failed
Becaused is not so simple to put a number into string and calculate a divide operation.

To Skylander:
Can you explain to me a little bit about that "Some formula and a double"

Thank you

[C8H10N4O2]

This entire problem is just one math equation. If I tell you the equation, then there is NO problem:p But here's a hint, think about logarithms...

[norton]

Hi.
I have a problem with program no. 113. I posted it for many times but I always get compiler eror as the answer. I don't see what's wrong with the source code cause it works perfectly on my computer.
Here is the source code and the compiler's messages i received from the judge:

"@begin_of_source_code"

/* @JUDGE_ID: 10427NY 113 C */

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

long koren(char stev1[], char stev2[]);

int main(void)
{

char stev1[3];
char niz[102];
char stev2[102];
int i = 0;

while (scanf("%s", niz) == 1) {
i++;
if(i % 2 == 1) strcpy(stev1, niz);
else {
strcpy(stev2, niz);
printf("%ld\n", koren(stev1, stev2));
}
}
return 0;
}

long koren(char stev1[], char stev2[]) {

int j, mesta, eksp = 0;
long vsota = 0;
double osnova = 0;

for(j = 0; j < strlen(stev1) - 1; j++) {
eksp += (stev1[j] - '0') * pow(10, strlen(stev1) - 1 - j);
}
eksp += (stev1[j] - '0');

mesta = strlen(stev2) - 1;

if(mesta < {
for(j = 0; j < mesta; j++) {
osnova += (stev2[j] - '0') * pow(10, mesta - j);
}
osnova += (stev2[j] - '0');
mesta = 0;
}

else {
for(j = 0; j < 8; j++) {
vsota += (stev2[j] - '0') * pow(10, 8 - j);
}
vsota += (stev2[j] - '0');
osnova = (double)vsota / pow(10, ;
}
return (long)(pow(10, (log10(osnova) + mesta) / eksp) + 0.5);
}
"@end_of_source_code"



Here are the compiler error messages:

/usr/lib/crt1.o: In function `_start':
/usr/lib/crt1.o(.text+0x18): undefined reference to `main'
collect2: ld returned 1 exit status

--
Can anyone help please?
Thanks.

[Picard]

strange. the judge compiled it for me (but wrong answer). maybe your mailer screw it up? btw this is a linker error after compiling, for some reason the main() function was not found.

[taj79]

I want some hint...
i m unable to read the sample data(which one is really big....even with double )
Plz tell me what should I do?

[visser]

Do you really need that much precision to get the correct answer?

[taj79]

can u give me some tips how to attack the problem?if i can't read the number by scanf using a double variable how will i start?

[Picard]

just read the older topics about P113
(this is what you should do first before creating a new topic...)

[NONAME_SUST]

HELLO,
The problem has a very simple solution. After viewing the problem you may say How easy!!! You know that c++ has a library named <<math.h>> . We will use this library to solve the problem. Here it is:

long double s = pow (p,1/n);
printf("%lf",s);

And thats it.........

[taj79]

But my problem is reading p when it is big.

Code: Select all

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

main()
{int d;
 long double p;
 scanf("%f",p);
 printf("%0.0f",p);
}

For the sample input I m failing to read the input.....[/code]

[Ivor]

When you read double you must use %lf instead of %f. The latter will only read a float value. For long double I suppose you should use %Lf.
Not sure myself.

Ivor

[taj79]

The problem is still there whether i use %lf or %Lf

I m unable to read the long sample input.