UVa OJ Board

we know that triangle-area is | (x1*y2) + (x2*y3) + (x3*y1) - (x2*y1) + (x3*y2) + (x1*y3) | / 2 simply, area of all of trianlges is ( A_ij ) * ( x_i ) * ( y_j ) / 2 if you know that A_ij for ( 1 <= i , j <= n ) then, you can calculate the all of triangles with complexity O(n^2) addtionally, the A_ij...

helloneo wrote:Sometimes, just a small difference make a rounding error..
For example, if you got the value 1.2345499999999999, 0.0000000000001 is very important..

thanks for your reply.

helloneo wrote:See sohel's post..

what's the difference between 3.1415926 and 3.141593 ??

it is related in the requirement of problem ( 4 precious ? )

anyway, thanks for your advice, helloneo. ^^

could you check the my code ??

would you teach me what is wrong ?

The online judge should send you an email explaining the reason for compile error. Please check the mails. If you don't receive any mail with each submission, you have the option disabled. So edit your accounts' profile so that you receive the mails. oh my god ! i forget them. i was soooooooooooooo...

when i use the VC 6 or gcc 3.4.4 (cygwin), i did'nt find the any erorrs

just, some warnings are found.

but, the result(submission status) in online-judge is compile error.


could you try to compile the following code ?

I got correct answers for most of input datas in this board. my the submissions' status is 2006-08-27 17:32:56 Wrong Answer 0.139 528 0.139 sec. please, give me some critical input datas or .... if you can test my code, i'll send it to you. thanks. addtional, following datas are founded in this boar...

D[ l, s ] = { l : Length of expression, s : sum of expression } D[ l, s ] = SUM{ D[ l-1 , m ] } for m=0 to N and l <= K and, output is Since Larry is only interested in the last few digits of the answer, for each pair of numbers N and K, print a single number mod 1,000,000 on a single line.

scanf( const char *format, ... );

must be pointer variables like int* , char *, double*, in the "..."

so,

scanf("%lf", &p);

( 2 * x ) or ( 3 * y ) or ( 4 * z ) <= 1500 prepare. x = 1, y = 1, z = 1 and the STACK[0] = 1 first. STACK[1] number is smallest number in ( 2*x ), ( 3*y), ( 4*z ) . you can found it ( 2 * x ) when x = 1 ; it is smallest number ; so, STACK[1] = 2 . and next number is must be larger then 2 so, set x=...

Please give me the test cases.. input 22 (5(4(11(7()())(2()()))()) (8(13()())(4()(1()())))) 20 (5(4(11(7()())(2()()))()) (8(13()())(4()(1()())))) 10 (3 (2 (4 () () ) (8 () () ) ) (1 (6 () () ) (4 () () ) ) ) 5 () 0 () -1 (-1()()) 77 (77(1()())()) -77 (-77()()) -7 (1 (-6 (2 (1 ()()) (-4 ()(1 (1 ()())...

Well the code you posted and ryan posted differs in output. My code produces the output exactly that ryan gave. Btw galious, how is input for 3 a length of 1 . For n=3 my output is 0 3. For n=1 my output is 0 1. Your output is 1 1. In problem it is stated that after the elimination round.For n=2 it...

if you not solved this problem, then how about visit to follow page url.

http://isd.korea.ac.kr/~ajava/10774.php?n=23403

it programming by php script languagae....

got AC when use the your power function.

time is 0:06.000(CPU).