soalGOJEK.rtf

functon f (a){ var i = 0; var s = 0; while(i
var  = * var c = 0 while ( , i-1) if([] <=   /([] <= a)) { c = c + 1 } =-1 } reurn c } 1%0'310 4ncaugh 56na7rror3 9llegal reurn sae:en functon foo(#a##i#*){ var  = * var c = 0 while ( , i-1) { if([] <=   /([] <= a)) { c = c + 1 } =-1 } reurn c } une!ne foo([11#10#10#"#10#1"#'0#10#$#11]#%#1%#&#) '

var  = * var c = 0 while ( , i-1) if([] <=   /([] <= a)) { c = c + 1 } =-1 } reurn c } 1%0'310 4ncaugh 56na7rror3 9llegal reurn sae:en functon foo(#a##i#*){ var  = * var c = 0 while ( , i-1) { if([] <=   /([] <= a)) { c = c + 1 } =-1 } reurn c } une!ne foo([11#10#10#"#10#1"#'0#10#$#11]#%#1%#&#) '

foo([11#10#10#"#10#1"#'0#10#$#11]#10#'0#0#2)  foo([11#10#10#"#10#1"#'0#10#$#11]#%#1%##&) 0 foo([11#10#10#"#10#1"#'0#10#$#11]#'0#10#0#2) 0 foo([11#10#10#"#10#1"#'0#10#$#11]##$#%#%) 1 '001 '001 functon g(sring sr){ var i = 0 sring newsr = >> while (i < sr.lengh - 1) { newsr = newsr + sr[i+1] i=i+1 } reurn newsr } '0131 4ncaugh 56na7rror3 4ne8ece ient!er functon g(sr){ var i = 0 sring newsr = >> while (i < sr.lengh - 1) { newsr = newsr + sr[i+1]

i=i+1 } reurn newsr } '01$3& 4ncaugh 56na7rror3 4ne8ece ient!er functon g (sr){ var i = 0 sring newsr = >> while (i < sr.lengh - 1) { newsr = newsr + sr[i+1] i=i+1 } reurn newsr } '01%3& 4ncaugh 56na7rror3 4ne8ece ient!er functon g (sr){ var i = 0 var newsr = >> while (i < sr.lengh - 1) { newsr = newsr + sr[i+1] i=i+1 } reurn newsr } une!ne

functon f(sr){ if(sr.lengh == 0){ reurn >> } else if (sr.lengh == 1) { reurn sr } else { reurn f(g(sr)) + sr[0] } } une!ne functon h(n# sr){ while (n /= 1){ if (n ? ' == 0) { n = n@' } else { n = &An + 1 } sr = f(sr) } reurn sr } une!ne functon 8ow(#6){ if (6 == 0){ reurn 1

} else { reurn  A 8ow (#6-1) } } une!ne h(1#>fruis>) >fruis> h('#>fruis>) >sturf> h("#>fruis>) >sturf> h(8ow('#1000000000000000#>fruis>) '&%31 4ncaugh 56na7rror3 :issing ) aBer argu:en lis h(8ow('#1000000000000000#>fruis>)) ''131 4ncaugh Cange7rror3 ai:u: call sac siDe eceee a 8ow (
8ow E ''131 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13"

8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13"







h(8ow('#2%&10"000"00000$#>fruis>)) ''131 4ncaugh Cange7rror3 ai:u: call sac siDe eceee a 8ow (







8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" 8ow E ''13" h(8ow('#1000000000000000)#>fruis>)) '31 4ncaugh 56na7rror3 4ne8ece oen ) h(8ow('#1000000000000000)#>fruis>) ''131 4ncaugh Cange7rror3 ai:u: call sac siDe eceee a 8ow (
a 8ow (







8ow E ''13" h(8ow('#100)#>fruis>) >fruis> h(8ow('#')#>fruis>) >fruis> h(8ow(1#')#>fruis>) >fruis> h(8ow(1#"000)#>fruis>) >fruis> h(8ow(1#1)#>fruis>) >fruis> h(8ow(1#$)#>fruis>) >fruis> h(8ow(1#0)#>fruis>) >fruis> h(8ow('#0)#>fruis>) >fruis>

Some people have been studying the following problem: given two numbers, A and B, how many numbers from A to B, inclusive, are divisible by another number K. For example, if A is , B is !, and K is ", then there are " numbers that satisfy this: ", #, and $. %elp them by programming a solution to this problem& 'nput: (he first line is the number of test cases (. )ach test case has three numbers A, B, K, each on their own line given in that order. *utput: For each test case, output one line of the form +Case

C: X

-without the uotes/, where 0 is the case

number -starting from /, and 1 is the number of integers between A and B, inclusive, that are divisible by K. 0onstraints:  2 ( 2 !!  2 A 2 B 3 !!!!  2 K 3 !!!! Sample 'nput: 2 1 10 3 8 20

soalGOJEK.rtf

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