OJ Board

AC.... Finally Thx to kp,mamun,helloneo...

kp wrote:Let C = A^B,

lg(C) = B*lg(A), lg - decimal logarithm lg(x) = ln(X)/ln(10).
First three digits of C are first three digits of mantissa of lg(C).

That's how I got AC (not sure about mathematical correctness).

kp: I really don't understand,why first three digits of C are first three digits of mantissa of lg(C)?? For example,
lg(123456)= 5.091512... where is 123 ???
or can someone else explain it to me ??

It's actually antilog of first 3 digits of mantissa of log(c).

kp wrote:Let C = A^B,

lg(C) = B*lg(A), lg - decimal logarithm lg(x) = ln(X)/ln(10).
First three digits of C are first three digits of mantissa of lg(C).

That's how I got AC (not sure about mathematical correctness).

how do u calulate mantissa in c++?

It's not a question of C++. It's simple maths.

Of a number, say 12.589, .589 is the mantissa. So you can calculate mantissa of a vlaue, x

mantissa(x) = x - floor(x)

You can use frexp() function in C/C++.

I also calculated the first three digits using <math.h> and got accepted, but I still have some doubts about precision.
How can we be sure that a 100-million digit number that starts off "123999999999999999999999999..." is not printed as "124..." without doing bigint calculus?

ohhhh. Sorry for the silly question

one last question
How to calculate this antilog?

If you do log calculation in base 10 ie. log10() then use pow(10,x) for antilog and if base is e ie. log(), then use exp().

thanks

I need some help for the leading part..
what is the problem with the following code snippet :

double dd=(double)k*log(mm);
char buf1[30];
sprintf(buf1,"%.Lf",exp(dd));

I have got WA

I did the first three with fast exponentiation, I just kept moving the decimal point so it wouldn't go into infinity. Never occured to me that I could've used logs

Please help me,I have read all posts here, but have no idea..
My code is:

little joey wrote:I also calculated the first three digits using <math.h> and got accepted, but I still have some doubts about precision.
How can we be sure that a 100-million digit number that starts off "123999999999999999999999999..." is not printed as "124..." without doing bigint calculus?

I think the 4th to 15th digit can be considered as some sort of pseudo-random number, then the probability that all these digits being 9 is exponentially small.

Darko wrote:I did the first three with fast exponentiation, I just kept moving the decimal point so it wouldn't go into infinity.

For the trailing part I used the Square-and-multiply method .
How to use the same for the leading .. I really can't understand