For CS Freshmen, Sophomores and Juniors of PUP

Hi guys! Kung may mga projects or case studies kayo na mahirap, you can post it here as well. Maybe some members of the forum can help you. Tulong tulong tayo mga kapatid..hehe.

nakOW! tamang-tama., freshie ako taz mei case study pa kami., patulong ah., please.,

sige Kathy, post mo lang dito yung problem at baka matulungan ka namin..

I. NUMBERS
A. Conversions
a.decimal-to-binary
b.decimal-to-octal
c.decimal-to-hexadecimal

B. ABUNDANT, DEFICIENT or PERFECT

Input a no. Determine if the number is ABUNDANT, DEFICIENT or PERFECT. ABUNDANT is said to be trhe sum of all factors of the number, except the number itself, is greater than the number. DEFICIENT is said to be the sum of all factors of the number, except the number itself, is less than the number. and PERFECT number is said to be the sum of all factors of the number, except the number itself, is equal to the number.

INPUT: 6
OUTPUT: Perfect (Factors are 1,2,3 and sum is 6)
INPUT: 9
OUTPUT: Deficient (Factors are 1,3 and sum is 4)
INPUT: 12
OUTPUT: Abundant (Factors are 1,2,3,4,6 and sum is 15)

C. Armstrong
(input a positive no. Determine if the number is armstrong or not. Armstrong no. is the sum of the cube of each digit from the number.
INPUT: 153
OUTPUT: Armstrong (1^3 + 5^3 + 3^3 = 1+ 125 + 27 =153)
INPUT: 143
OUTPUT: Not Armstrong (1^3 + 4^3 + 3^3 = 1+ 64 + 27 =92)

II. STRING
A. Palindrome
B. Longest Word
INPUT:patulong naman please
OUTPUT:patulong
(kapag dalawa o higit pa ung magkaparehas ng haba ng characters.. lahat un lalabas .. )
C. Occurence
INPUT:patulong naman
OUTPUT:
a = 3
p = 1
t = 1
u = 1
l = 1
o = 1
n = 3
m = 1
space = 1




sa conversion mei nasimulan na..

ung palindrome tapos na.,

ako na bahala sa B. ABUNDANT, DEFICIENT or PERFECT at C. Armstrong

phelp naman sa iba., pls




PM nio n lng po ako sa YM

friendship_kathy_24



tnx ulit


promise., tinatry ko din gawin.,

bigyan nio na lang ako idea kung paano gagawin., ako gagawa.,

Nosebleed to Kathy a.. hehehe..

Susubukan ko gawin ngayon. Sensya na, sobrang busy talaga. Pero wala kong pasok ngayong monday!

callbookol wrote:

Nosebleed to Kathy a.. hehehe..

Susubukan ko gawin ngayon. Sensya na, sobrang busy talaga. Pero wala kong pasok ngayong monday!

mas nosebleed nga ako eh., T_T tulo pa sipon ko., ahihi., kahit nilalagnat+headache., dahil sa case study hnd ako makapagpahinga., haha!

sa conversion, tingin ko division ln un trick jan.

to binary
1. divide by 2
2. place remainder in a string(even if the remainder is 0).
3. repeat steps 1 and 2 but stop when the quotient is already 0.(in step 2, after placing the remainder in the string, move the index of the string to the next one.)
3.x. given is 6, string is binary and the starting index is 0. (step 1)divide 6 by 2 and you'll get the quotient 3, remainder 0. (step 2)the string would hold the remainder in the starting index([0]) so it would be "0"(binary[0] = '0'). increase the index to 1. (repeating 1 and 2)divide the answer(which is 3) to 2 again you'll get quotient 1 and 1 as the remainder. the string would now hold "01"(binary[1] = '1'). increase the index to 2. (repeating 1 and 2) and again divide the answer(1) by 2, you'll get 0 as the quotient and 1 as the remainder. the string would now hold "011"(binary[2] = '1'). increase the index to 3. now that tha quotient is 0, this is where you stop.
4. reverse the order of the string and you'll get the binary form of the decimal.
4.x. if you'll do the example of 3.x. the string binary would hold "011". reverse its order and you'll get "110" which is the binary of the given 6.

to octal and hexadecimal, they have the same steps just change the divisor, although hexadecimal has an added tweak where you need to change the remainder to A-F once they reach 10-15.

kuya seph,. salamat,

yiPEeE! natapos na sa wakas ang madugong pakikipaglaban sa programming., nxt sem ulit! hahaha! nyok., off-topic tong post ko? hehehe