programming c

[vquad]

salam otai2 sume..saya nak mitak pencerahan soalan nie.. xnampk langsung cmne nak buat...


Our coins come in denominations of 1 cent, 5 cents, 10 cents, 20 cents, 50 cents and 1 dollar. Assuming that there are unlimited number of coins, devise a coin-change algorithm to compute the minimum number of coins required to make up a particular amount? For example, for 46 cents you need 4 coins: two 20-cent coins, one 5-cent coin, and one 1-cent coin.


Present your algorithm using pseudocode and flowchart.

 

[christmas]

em.. utk minimize kan syiling utk pulangkan baki duit tu..

bahagi dengan amount tu dengan 1dollar, kemudian , 50sen, pastu 20sen... 10sen , 5sen and seterusnye 1sen

guna mathematic je utk kira mcmane nk bg system ni keluarkan brape byk 1dollar.. brape byk 50sen tu.. n so on...

 

[erro_sennin]

cm ne kot..

def count( n, m ):
if n == 0:
return 1
if n < 0:
return 0
if m < 0 and n >= 1:
return 0

return count( n, m - 1 ) + count( n - S[m], m )



http://www.algorithmist.com/index.php/Coin_Change

 

[Ahmad07]

ni pakai bahagi, amik nombor bulat kat depan, pastu pakai function modulus utk tengok brape balance... bermula dengan bahagi coin terbesar, check sekiranya modulus > 0 then terus dengan coin yg lebih kecik...

 

[Ahmad07]

cm ne kot..

def count( n, m ):
if n == 0:
return 1
if n < 0:
return 0
if m < 0 and n >= 1:
return 0

return count( n, m - 1 ) + count( n - S[m], m )



http://www.algorithmist.com/index.php/Coin_Change

kene define dulu variable utk setiap jenis duit tu...

 

[firahz]

Senang je ni..
ni teori yg plg mudah:

1.00
0.50
0.10
0.05
0.01

yg ni sebagai base ko ikut susunan nilai yg besar kepada yg terkecil.

Katakan dia enter 87cent.

1st: 87cent ditolak dgn base terbesar 1.00. Bila ditolak dpt -ve kan? so skip kpd base kedua terbesar 0.50..ngam bila ditolak masih ada balance 37cent.Bila ada balance dia loop blk utk tolak dgn base terbesar semula.

2nd: katakan 37cent. tolak dgn 1.00 ada balance -ve. tolak dgn 0.50 pon ada -ve.Bila tolak pulak dgn base 0.10 dpt balance +ve. Selepas tu loop semula utk tolak dr base besar blk.Sampailah nilai enter 87cent tu jd 0cent.

3rd: setiap yg ditolak ada balance tu perlu direkodkan jumlah kuantiti - set pada variable je semasa looping...
Contoh 0.50cent (kuantiti 1), 0.10 (kuantiti 3), 0.05cent (kuantiti 1), 0.01 (kuantiti 2).

Good luck!

 

[vquad]

thanks kepada yang menolong...

 

[yunuskk]

sebab saya free petang ini ini codenya dalam c

#include <stdio.h>
void makechange(int, int*);

int main(void)
{
    int i = 0;
    int money = 0;
    int change[6] = {0};
    int noofcoin = 0;
    char *moneytype[6] = {"Satu Ringgit","50 sen", "20 sen", "10 sen", "5 sen", "1 sen"};
  
    printf("Please enter the amount of money (in cents): ");  
    scanf("%d", &money);
     
    
    makechange(money, change);
    for (i=0; i < 6; i++) 
    {
        if (change[i] > 0)
        {
            noofcoin = noofcoin + change[i];
            printf("%d %s\n", change[i], moneytype[i]);
        }
    }
    printf("No of coins %d\n", noofcoin);
    return 0;
}

void makechange(int moneytotal, int* result)
{
    int i = 0;
    int moneyvalues[6] = {100, 50, 20, 10, 5, 1};
 
    for (i = 0; i < 6; i++) 
    {
        result[i] = moneytotal / moneyvalues[i];
        moneytotal = moneytotal % moneyvalues[i];
    }
}

 

[christmas]

sebab saya free petang ini ini codenya dalam c

#include <stdio.h>
void makechange(int, int*);

int main(void)
{
    int i = 0;
    int money = 0;
    int change[6] = {0};
    int noofcoin = 0;
    char *moneytype[6] = {"Satu Ringgit","50 sen", "20 sen", "10 sen", "5 sen", "1 sen"};
  
    printf("Please enter the amount of money (in cents): ");  
    scanf("%d", &money);
     
    
    makechange(money, change);
    for (i=0; i < 6; i++) 
    {
        if (change[i] > 0)
        {
            noofcoin = noofcoin + change[i];
            printf("%d %s\n", change[i], moneytype[i]);
        }
    }
    printf("No of coins %d\n", noofcoin);
    return 0;
}

void makechange(int moneytotal, int* result)
{
    int i = 0;
    int moneyvalues[6] = {100, 50, 20, 10, 5, 1};
 
    for (i = 0; i < 6; i++) 
    {
        result[i] = moneytotal / moneyvalues[i];
        moneytotal = moneytotal % moneyvalues[i];
    }
}

ini yg paling terbaek ~