Mini Cash Counter Application program in C

Question

Develop a simple cash counter application for all the scenarios possible during the exchange.

Logic

• Get the cash, put that into the counter(2’s or 10’s)
• Calculate the return amount in descending order, from 10’s then 5’s and so on. Because if the return amount is 5, you must give one 5 rupee not five 1 rupee right!
• Separate the given change from the cash in counter.

Algorithm

1. Get the initial balance of the counter.
2. Get the cost and given cash and calculate change amount.
3. Get the count of each rupee given by the customer (Ex : 10’s -> 1, 5’s->1)
4. Calculate the change to be given in descending order (10,5,2,1)
5. Update the change amount at each step.
6. Update the count of each rupee in cash counter accordingly.

Program

#include<stdio.h>

int main()
{   /*cash -> cash given by customer
      change-> return amount to be given
      one,two,five,ten-> number of rupees ex: one =2 means two one rupee coin is in counter
      */
    int one=0, two=0, five=0, ten=0;
    int one1,two1,five1,ten1,temp, i;
    int cash, change, cost,num;
    printf("enter intial count of each rupee");
    scanf("%d%d%d%d",&one,&two,&five,&ten);             //intial counter balance

    do
    {
        printf("enter cost and given cash");
        scanf("%d%d",&cost,&cash);
        change = cash- cost;                            //calculate amount to return
        printf("enter the count of  each rupee given in increasing order");
        scanf("%d%d%d%d",&one1,&two1,&five1,&ten1);     //count of each rupee given by customer
        /*adding count of rupee to the cash
          already present in counter*/
        one+=one1;
        two+=two1;
        five+=five1;
        ten+=ten1;
        if(change==0)
            printf("Thank you for shopping...");
        if(change<0)
            printf("Sorry more cash needed");           //cost exceeds the given cash
        /* calculating return amount in descending order from 10 to 1
           Ex: generally if we need to give 3 rupee, first we give
           two(1) and one(1)*/
        if(change>0)
        {
            if(ten!=0)
            {   /*to find number of 10's we can give, divide by 10
                 ex: 25/10 = 2 => num
                 */
                num = change/10;
                temp = ten- num;                        //subtracting from number of 10's already present in counter
                if(temp>0)
                 {
                     change = change -(10*num);         //Subtracting change from 10's given ex: 25-(10*2)
                     ten = temp;
                     printf("\nten : %d ", num);
                 }
                 /*if number of ten's is lesser than number of ten's to be given
                   temp = 2-4
                   in such cases give all the ten's to the customer present in counter
                   */
                 else{
                    change =change -(10*ten);
                    printf("\nten : %d ", ten);
                    ten= 0;                             //number of ten in counter is empty
                 }

            }
            if(five!=0&&change)                         //change must be > 0
            {
                num= change/5;
                temp = five - num;
                if(temp>0)
                {
                    change = change-(5*num);
                    five =temp;
                    printf("five : %d ",num);

                }
                else
                {
                    change = change -(5*five);
                    printf("\nfive : %d ",five);
                    five=0;

                }

            }
            if(two!=0&&change)
            {

                num= change/2;
                temp = two - num;
                if(temp>0)
                {
                    change = change - (2*num);
                    two = temp;
                    printf("two : %d ",num);
                }
                else
                 {
                     change = change - (2*two);
                     printf("two : %d ",two);
                     two =0;
                 }

            }
            if(one!=0&&change)
            {
                num = change/1;
                temp =one - num;
                if(temp>0)
                {
                    change = change - num;
                    one = temp;
                    printf("one : %d",num);
                }
                else
                {
                    change = change -(one);
                    printf("one : %d",one);
                    one =0;
                }
            }
            if(change>0)                                    //cannot give change
                printf("\ninsuffient change");
            else
                printf("\nThanks for coming");              //change =0
        }
        printf("\nCustomers waiting in line...enter 1 if yes 0 if no: ");
        scanf("%d",&i);
    /* loop continue if more customers are waiting in line*/
    }while(i);
return 0;
}

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