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Number Series with a Twist 2

Consider the following series: 1, 1, 2, 3, 4, 9, 8, 27, 16, 81, 32, 243, 64, 729, 128, 2187 …

This series is a mixture of 2 series – all the odd terms in this series form a geometric series and all the even terms form yet another geometric series. Write a program to find the Nth term in the series.

The value N in a positive integer that should be read from STDIN. The Nth term that is calculated by the program should be written to STDOUT. Other than value of n th term,no other character / string or message should be written to STDOUT. For example , if N=16, the 16th term in the series is 2187, so only value 2187 should be printed to STDOUT.

You can assume that N will not exceed 30.

 

#include<stdio.h>

int main()

{

   int i, n, a=1, b=1;

   printf("enter number : ");

   scanf("%d",&n);

   for(i=1;i<=n;i++)

   {

       if(i%2!=0)

       {

           a = a * 2;

       }

       else

       {    

            b = b * 3;

       }

   }

    if(n%2!=0)

       {

           printf("\n%d term of series is %d\t",n,a/2);

       }

       else

       {    

           printf("\n%d term of series is %d\t",n,b/3);

       }

return 0;

}

Consider the below series :

0, 0, 2, 1, 4, 2, 6, 3, 8, 4, 10, 5, 12, 6, 14, 7, 16, 8

This series is a mixture of 2 series all the odd terms in this series form even numbers in ascending order and every even terms is derived from the previous  term using the formula (x/2)

Write a program to find the nth term in this series.

The value n in a positive integer that should be read from STDIN the nth term that is calculated by the program should be written to STDOUT. Other than the value of the nth term no other characters /strings or message should be written to STDOUT.

For example if n=10,the 10 th term in the series is to be derived from the 9th term in the series. The 9th term is 8 so the 10th term is (8/2)=4. Only the value 4 should be printed to STDOUT.

You can assume that the n will not exceed 20,000.

 



#include<stdio.h>


int main()

{

    int i, n, a=0, b=0;

    printf("enter number : ");

    scanf("%d",&n);

    

    

    for(i=1;i<=n;i++)

    {

        if(i%2!=0)

        {

            if(i>1)

                a = a + 2;

        }

        else

        {

            b = a/2;

        }

    }


    if(n%2!=0)

    {

        printf("%d",a);

    }

    else

    { 

        printf("%d",b);

    }

    

    return 0;

}

 

String with a Twist

1. The program will receive 3 English words inputs from STDIN

  1. These three words will be read one at a time, in three separate line
  2. The first word should be changed like all vowels should be replaced by %
  3. The second word should be changed like all consonants should be replaced by #
  4. The third word should be changed like all char should be converted to upper case
  5. Then concatenate the three words and print them

Other than these concatenated word, no other characters/string should or message should be written to STDOUT

For example if you print how are you then output should be h%wa#eYOU.

You can assume that input of each word will not exceed more than 5 chars

#include <stdio.h>

#include <string.h>

int main()

{

  char a[10], b[10], c[10];

  int i,j;

  int x, y, z;


  scanf("%s",a);

  scanf("%s",b);

  scanf("%s",c);


  x = strlen(a);

  y = strlen(b);

  for(i=0;i<x;i++)

  {

      if(a[i]=='a'||a[i]=='e'||a[i]=='i'||a[i]=='o'||a[i]=='u')

      {

          a[i] = '%';

      }

  }

  for(j=0;j<y;j++)

  {

      if(b[j]=='b'||b[j]=='c'||b[j]=='d'||b[j]=='f'||b[j]=='g'||b[j]=='h'||b[j]=='j'||b[j]=='k'||b[j]=='l'||

         b[j]=='m'||b[j]=='n'||b[j]=='p'||b[j]=='q'||b[j]=='r'||b[j]=='s'||b[j]=='t'||b[j]=='v'||b[j]=='w'||

         b[j]=='x'||b[j]=='y'||b[j]=='z')

      {

          b[j] = '#';

      }

      

      if(b[j]=='B'||b[j]=='C'||b[j]=='D'||b[j]=='F'||b[j]=='G'||b[j]=='H'||b[j]=='J'||b[j]=='K'||b[j]=='L'||

         b[j]=='M'||b[j]=='N'||b[j]=='P'||b[j]=='Q'||b[j]=='R'||b[j]=='S'||b[j]=='T'||b[j]=='V'||b[j]=='W'||

         b[j]=='X'||b[j]=='Y'||b[j]=='Z')

      {

          b[j] = '#';

      }

  }

  z=0;

   while (c[z] != '\0') {

      if (c[z] >= 'a' && c[z] <= 'z')

       {

         c[z] = c[z] - 32;

      }

      z++;

   }

   printf("%s%s%s",a,b,c);

}

Addition of two numbers a Twist

1. Using a method, pass two variables and find the sum of two numbers.

Test case:

Number 1 – 20

Number 2 – 20.38

Sum = 40.38

There were a total of 4 test cases. Once you compile 3 of them will be shown to you and 1 will be a hidden one. You have to display an error message if numbers are not numeric.

 

 

#include<stdio.h>

addition(int x, float y)

{

    float ans;

    ans = (float)x + y;

    printf("Answer : %.2f",ans);

}

int main()

{

   int a;

   float b;

   printf("enter first number : ");

   scanf("%d",&a);

   printf("enter second number : ");

   scanf("%f",&b);

   addition(a, b);

}

 

by Expert (38,790 points)
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Prime Numbers with a Twist

Ques. Write a code to check whether no is prime or not. Condition use function check() to find whether entered no is positive or negative ,if negative then enter the no, And if yes pas no as a parameter to prime() and check whether no is prime or not?

  • Whether the number is positive or not, if it is negative then print the message “please enter the positive number”
  • It is positive then calls the function prime and checks whether the positive number is prime or not.

#include<stdio.h>

void prime(int n)

{

int c=0;

    for(int i=2;i<n;i++)

    {

        if(n%i==0)

        c = c+1;

    }

    if(c>=1)

        printf("%d is not a prime number",n);

    else

        printf("%d is a prime number",n);

}

void main()

{

int n;

printf("Enter no : "); //enter the number

scanf("%d",&n);

if(n<0)

    {

    printf("Please enter a positive integer");

    }

else

    prime(n);

}

Number Series with a Twist – 1

Find the 15th term of the series?

0,0,7,6,14,12,21,18, 28

Explanation : In this series the odd term is increment of 7 {0, 7, 14, 21, 28, 35 – – – – – – }

                        And even term is a increment of 6 {0, 6, 12, 18, 24, 30 – – – – – – }

#include <stdio.h>

int main()

{

   int i, n, a=0, b=0;

   printf("enter number : ");

   scanf("%d",&n);

   for(i=1;i<=n;i++)

  {

      if(i%2!=0)

     {

           a = a + 7;

     }

       else

      {    

           b = b + 6;

     }

  }

     if(n%2!=0)

     {

           printf("%d term of series is %d\t",n,a-7);

     }

      else

     {    

          printf("%d term of series is %d\t",n,b-6);

     }

return 0;

}

by Expert (38,790 points)
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Question 9

Problem Statement

FULLY AUTOMATIC VENDING MACHINE – dispenses your cuppa on just press of button. A vending machine can serve range of products as follows:

Coffee

  1. Espresso Coffee
  2. Cappuccino Coffee
  3. Latte Coffee

Tea

  1. Plain Tea
  2. Assam Tea
  3. Ginger Tea
  4. Cardamom Tea
  5. Masala Tea
  6. Lemon Tea
  7. Green Tea
  8. Organic Darjeeling Tea

Soups 

  1. Hot and Sour Soup
  2. Veg Corn Soup
  3. Tomato Soup
  4. Spicy Tomato Soup

Beverages

  1. Hot Chocolate Drink
  2. Badam Drink
  3. Badam-Pista Drink

Write a program to take input for main menu & sub menu and display the name of sub menu selected in the following format (enter the first letter to select main menu):

Welcome to CCD 

Enjoy your

Example 1:

  • Input:
    • c
    • 1
  • Output
    • Welcome to CCD!
    • Enjoy your Espresso Coffee!

Example 2:

  • Input
    • t
    • 9
  • Output
    • INVALID OUTPUT!

 

#include <stdio.h>

int main()

{


    char c[3][20]={"Espresso Coffee","Cappuccino Coffee","Latte Coffee"};


    char t[8][30]={"Plain Tea","Assam Tea","Ginger Tea","Cardamom Tea","Masala Tea","Lemon Tea","Green Tea","Organic Darjeeling Tea"};


    char s[4][20]={"Hot and Sour Soup","Veg Corn Soup","Tomato Soup","Spicy Tomato Soup"};


    char b[3][20]={"Hot Chocolate Drink","Badam Drink","Badam-Pista Drink"};


    char str[]="Welcome to CCD!\nEnjoy your ";


    char ch;


    int  item, i;


    scanf("%c",&ch);


    scanf("%d",&item);


    if(ch=='c')


    {


        for(i=0; i<3; i++)


        {


            if(item==i+1)


            {


                printf("Welcome to CCD!\nEnjoy your %s!",c[i]);


                break;


            }


        }


        if(i==3)


        {


            printf("INVALID OPTION!");


        }


    }


    else if(ch=='t')


    {


        for(i=0; i<8; i++)


        {


            if(item==i+1)


            {


                printf("Welcome to CCD!\nEnjoy your %s!",t[i]);


                break;


            }


        }


        if(i==8)


        {


            printf("INVALID OPTION!");


        }


    }


    else if(ch=='s')


    {


        for(i=0; i<4; i++)


        {


            if(item==i+1)


            {


                printf("Welcome to CCD!\nEnjoy your %s!",s[i]);


                break;


            }


        }


        if(i==4)


        {


            printf("INVALID OPTION!");


        }


    }


    else if(ch=='b')


    {


        for(i=0; i<3; i++)


        {


            if(item==i+1)


            {


                printf("Welcome to CCD!\nEnjoy your %s!",b[i]);


                break;


            }


        }


        if(i==3)


        {


            printf("INVALID OPTION!");


        }


    }


    else


    {


        printf("INVALID INPUT!");


    }


    return 0;


}

Question 10

 

Problem Statement

A doctor has a clinic where he serves his patients. The doctor’s consultation fees are different for different groups of patients depending on their age. If the patient’s age is below 17, fees is 200 INR. If the patient’s age is between 17 and 40, fees is 400 INR. If patient’s age is above 40, fees is 300 INR. Write a code to calculate earnings in a day for which one array/List of values representing age of patients visited on that day is passed as input.

Note:

  • Age should not be zero or less than zero or above 120
  • Doctor consults a maximum of 20 patients a day
  • Enter age value (press Enter without a value to stop):

Example 1:

  • Input
    20
    30
    40
    50
    2
    3
    14
  • Output
    Total Income 2000 INR

Note: Input and Output Format should be same as given in the above example.

For any wrong input display INVALID INPUT

Output Format

  • Total Income 2100 INR

 

age = []


for i in range(20):


    m = input()


    if m == "":


        break


    elif int(m) in range(0,120):


        age.append(int(m))


    else:


        print("INVALID INPUT")


        exit()


fees = 0


for i in age:


    if i < 17:


        fees+=200


    elif i <40:


        fees+=400


    else:


        fees+=300


print("Total Income {} INR".format(fees))

 

---------------------

 

Checking if a given year is leap year or not

Explanation:

To check whether a year is leap or not

Step 1:

  • We first divide the year by 4.
  • If it is not divisible by 4 then it is not a leap year.
  • If it is divisible by 4 leaving remainder 0 

Step 2:

  • We divide the year by 100
  • If it is not divisible by 100 then it is a leap year.
  • If it is divisible by 100 leaving remainder 0

Step 3:

  • We divide the year by 400
  • If it is not divisible by 400 then it is a leap year.
  • If it is divisible by 400 leaving remainder 0 

Then it is a leap year

 

#include<stdio.h>

int leapprog(int year)

{

//checking divisibility by 4

    if(year%4 == 0)

    {

//checking divisibility by 100

        if( year%100 == 0)

        {

//checking divisibility by 400

            if ( year%400 == 0)

                printf("%d, the year entered happens to be a leap year", year);

            else

                printf("%d is surely not a leap year", year);

        }

        else

            printf("%d, the year entered happens to be a leap year", year );

    }

    else

        printf("%d is surely not a leap year", year);

       return 0;

}

int main()

{

    int input_year, val;

    printf("Enter the year that you want to check"); //enter the year to check

    scanf("%d",&input_year);

val = leapprog(input_year);    

return 0;

}

 

by Expert (38,790 points)
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Question 7

Problem Statement

There are a total n number of Monkeys sitting on the branches of a huge Tree. As travelers offer Bananas and Peanuts, the Monkeys jump down the Tree. If every Monkey can eat k Bananas and j Peanuts. If the total m number of Bananas and p number of Peanuts are offered by travelers, calculate how many Monkeys remain on the Tree after some of them jumped down to eat.

At a time one Monkey gets down and finishes eating and goes to the other side of the road. The Monkey who climbed down does not climb up again after eating until the other Monkeys finish eating.

Monkeys can either eat k Bananas or j Peanuts. If for the last Monkey there are less than k Bananas left on the ground or less than j Peanuts left on the ground, only that Monkey can eat Bananas(<k) along with the Peanuts(<j).

Write code to take inputs as n, m, p, k, j and return  the number of Monkeys left on the Tree.

Where, n= Total no of Monkeys

     k= Number of edible Bananas by Single Monkey (Monkey that jumped down last may get less than k Bananas)

     j = Number of eatable Peanuts by single Monkey(Monkey that jumped down last may get less than j Peanuts)

     m = Total number of Bananas

     p  = Total number of Peanuts

Remember that the Monkeys always eat Bananas and Peanuts, so there is no possibility of k and j having a value zero

Example 1:

Input Values    

20

2

3

12

12

Output Values

Number of  Monkeys left on the tree:10

 

Note: Kindly follow  the order of inputs as n,k,j,m,p as given in the above example. And output must include  the same format  as in above example(Number of Monkeys left on the Tree:)

For any wrong input display INVALID INPUT

 

#include<stdio.h>

int main ()

{

   int n, k, j, m, p;

   float atebanana = 0.0, atepeanut = 0.0;

   scanf ("%d %d %d %d %d", &n, &k, &j, &m, &p);

   if (n < 0 || k < 0 || j < 0 || m < 0 || p < 0)

   {

       printf ("INVALID INPUT");

   }

   else

   {

       if (k > 0)

       {

           atebanana = (float) (m / k);

        m = m % k;

}

       if (j > 0)

{

           atepeanut = (float) (p / j);

        p = p % j;

}

       n = n - atebanana - atepeanut;

       if ((m != 0) || (p != 0))

    n = n - 1;

       printf ("Number of Monkeys left on the Tree:%d", n);

   }

   return 0;

}

Question 8

 

Problem Statement

Chain Marketing Organization has a scheme for income generation, through which its members generate income for themselves. The scheme is such that suppose A joins the scheme and makes R and V to join this scheme  then A is a Parent Member of R and V who are child Members. When any member joins the scheme then the parent gets a total commission of 10% from each of its child members.

Child members receive commission of 5% respectively. If a Parent member does not have any member joined under him, then he gets commission of 5%.

Take the name of the members joining the scheme as input.

Display how many members joined the scheme including parent member.Calculate the Total commission gained by each member in the scheme. The fixed amount for joining the scheme is Rs.5000 on which commission will be generated

SchemeAmount = 5000

Example 1: When there are more than one child members 

Input : (Do not give input prompts.Accept values as follows. )

Amit         //Enter parent Member as this

Y               //Enter Y if  Parent member has child members otherwise enter N

Rajesh,Virat //Enter names of child members of Amit in comma separated

Output:(Final Output must be in format given below.)

TOTAL MEMBERS:3

COMMISSION DETAILS

Amit: 1000 INR

Rajesh :250 INR

Virat: 250 INR

Example 2: When there is only one child member in the hierarchy

Input :

Amit

Y

Rajesh

Output:

Total Members: 2 

Commission Details

Amit: 500 INR

Rajesh: 250 INR

 

using namespace std;

int main()

{

    string par;

    cin >> par;

   string x;

   cin >> x;

    if (x == "N") {

        cout << "TOTAL MEMBERS:1\n";

        cout << "COMISSION DETAILS\n";

        cout << par << ":250 INR\n";

    } else {

        string child;

        cin >> child;

        vector<string>v;

        string temp = "";

        for (int i = 0; i < child.length(); i++) {

            if (child[i] == ',') {

                v.push_back(temp);

                temp = "";

            }

            else  if (child[i] != ' ')

                temp += child[i];

        }

        v.push_back(temp);

        cout << "TOTAL MEMBERS:" << v.size() + 1 << "\n";

        cout << "COMISSION DETAILS\n";

        cout << par << ":" << v.size() * 500 << " INR\n";

        for (auto a : v) {

            cout << a << ":" << "250 INR\n";

     }

  }

}

 

by Expert (38,790 points)
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Question 5

Problem Statement

We want to estimate the cost of painting a property. Interior wall painting cost is Rs.18 per sq.ft. and exterior wall painting cost is Rs.12 per sq.ft.

Take input as

1. Number of Interior walls

2. Number of Exterior walls

3. Surface Area of each Interior 4. Wall in units of square feet

Surface Area of each Exterior Wall in units of square feet

If a user enters zero  as the number of walls then skip Surface area values as User may don’t  want to paint that wall.

Calculate and display the total cost of painting the property

Example 1:

6

3

12.3

15.2

12.3

15.2

12.3

15.2

10.10

10.10

10.00

Total estimated Cost : 1847.4 INR

Note: Follow in input and output format as given in above example

#include<stdio.h>

int main()

{

int ni,ne,i=0;

float int_p=18,ext_p=12,cost=0,temp;

scanf("%d %d",&ni,&ne);

if(ni<0 || ne<0 )

{

     printf("INVALID INPUT");

}

else if(ni==0 && ne==0)

{

     printf("Total estimated Cost : 0.0");

}

else

{

     for(i=0;i<ni;i++)

     {

         scanf("%f",&temp);

         cost+= int_p*temp;

     }

     for(i=0;i<ne;i++)

     {

         scanf("%f",&temp);

         cost+= ext_p*temp;

     }

     printf("Total estimated Cost : %.1f",cost);

}

return 0;

}

 

Question 6

 

Problem Statement

A City Bus is a Ring Route Bus which runs in circular fashion.That is, Bus once starts at the Source Bus Stop, halts at each Bus Stop in its Route and at the end it reaches the Source Bus Stop again.

If there are n  number of Stops and if the bus starts at Bus Stop 1, then after nth Bus Stop, the next stop in the Route will be Bus Stop number 1 always.

If there are n stops, there will be n paths.One path connects two stops. Distances (in meters) for all paths in Ring Route is given in array Path[] as given below:

Path = [800, 600, 750, 900, 1400, 1200, 1100, 1500]

Fare is determined based on the distance covered from source to destination stop as  Distance between Input Source and Destination Stops can be measured by looking at values in array Path[] and fare can be calculated as per following criteria:

  • If d =1000 metres, then fare=5 INR
  • (When calculating fare for others, the calculated fare containing any fraction value should be ceiled. For example, for distance 900n when fare initially calculated is 4.5 which must be ceiled to 5)

Path is circular in function. Value at each index indicates distance till current stop from the previous one. And each index position can be mapped with values at same index in BusStops [] array, which is a string array holding abbreviation of names for all stops as-

“THANERAILWAYSTN” = ”TH”, “GAONDEVI” = “GA”, “ICEFACTROY” = “IC”, “HARINIWASCIRCLE” = “HA”, “TEENHATHNAKA” = “TE”, “LUISWADI” = “LU”, “NITINCOMPANYJUNCTION” = “NI”, “CADBURRYJUNCTION” = “CA”

Given, n=8, where n is number of total BusStops.

BusStops = [ “TH”, ”GA”, ”IC”, ”HA”, ”TE”, ”LU”, ”NI”,”CA” ]

Write a code with function getFare(String Source, String Destination) which take Input as source and destination stops(in the format containing first two characters of the Name of the Bus Stop) and calculate and return travel fare.

Example 1:

Input Values

ca

Ca

Output Values

INVALID OUTPUT

Example 2:

Input Values

NI

HA

Output Values

23.0 INR

Note: Input and Output should be in the format given in example.

Input should not be case sensitive and output should be in the format   INR

 

#include <bits/stdc++.h>

using namespace std;

int main() {

string s , d;

 cin>>s>>d;

 transform(s.begin(),s.end() , s.begin(),::toupper);

 transform(d.begin(),d.end() , d.begin(),::toupper);

 string arrs[8] = {"TH" , "GA", "IC" , "HA" , "TE", "LU" ,"NI","CA"};

 float arr[8]={800,600,750,900,1400,1200,1100,1500};

 float res=0;

 int st ,ed;

 for(int i=0;i<8;i++)

 {

  if(s==arrs[i])

  st=i;


  if(d==arrs[i])

  ed=i;

  }

  if(st==ed)

  {

   cout<<" INVALID INPUT";

   return 0;

  }

  else

  {

   int i=st+1;

   cout<<i;

     while(i!=ed+1)

     {

         res+=(arr[i]);

         i=(i+1)%8;

     }

     cout<<(ceil)(res*0.005);

     return 0;

}

}



by Expert (38,790 points)
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Question 4

 

Problem Statement

The Caesar cipher is a type of substitution cipher in which each alphabet in the plaintext or messages is shifted by a number of places down the alphabet.

For example,with a shift of 1, P would be replaced by Q, Q would become R, and so on.

To pass an encrypted message from one person to another, it is first necessary that both parties have the ‘Key’ for the cipher, so that the sender may encrypt and the receiver may decrypt it.

Key is the number of OFFSET to shift the cipher alphabet. Key can have basic shifts from 1 to 25 positions as there are 26 total alphabets.

As we are designing custom Caesar Cipher, in addition to alphabets, we are considering numeric digits from 0 to 9. Digits can also be shifted by key places.

For Example, if a given plain text contains any digit with values 5 and keyy =2, then 5 will be replaced by 7, “-”(minus sign) will remain as it is. Key value less than 0 should result into “INVALID INPUT”

Example 1:

Enter your PlainText: All the best

Enter the Key: 1

The encrypted Text is: Bmm uif Cftu

Write a function CustomCaesarCipher(int key, String message) which will accept plaintext and key as input parameters and returns its cipher text as output.

                        #include<bits/stdc++.h>

int main()

{

   char str[100];

    int key, i=0, left;

    printf("Enter your plain text : ");

    scanf("%[^\n]s",str);

    printf("Enter the key : ");

    scanf("%d",&key);

    if(key==0)

    {

        printf("INVALID INPUT");

    }

    else

    {

        while(str[i]!='\0')

        {

            //printf("%d\n", str[i]);

            if(str[i]>=48 && str[i]<=57)

            {

                if(str[i]+key<=57)

                {

                    str[i] = str[i] + key;

                }

                else

                {

                    left = (str[i] + key) - 57;

                    str[i] = 47 + left;

                }   

            }

            else if(str[i]>=65 && str[i]<=90)

            {

                if(str[i]+key<=90)

                {

                    str[i] = str[i] + key;

                }

                else

                {

                    left = (str[i] + key) - 90;

                    str[i] = 64 + left;

                }  

            }

            else if(str[i]>=97 && str[i]<=122)

            {

                if(str[i]+key<=122)

                {

                    str[i] = str[i] + key;

                }

                else

                {

                    left = (str[i] + key) - 122;

                    str[i] = 96 + left;

                } 

            }

            i++;

        }

        printf("The encrypted text is : %s",str);

  }

  return 0;

}

 

by Expert (38,790 points)
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Question 3

Problem Statement

A washing machine works on the principle of the Fuzzy System, the weight of clothes put inside it for washing is uncertain But based on weight measured by sensors, it decides time and water level which can be changed by menus given on the machine control area.  

For low level water, the time estimate is 25 minutes, where approximately weight is between 2000 grams or any nonzero positive number below that.

For medium level water, the time estimate is 35 minutes, where approximately weight is between 2001 grams and 4000 grams.

For high level water, the time estimate is 45 minutes, where approximately weight is above 4000 grams.

Assume the capacity of machine is maximum 7000 grams

Where approximately weight is zero, time estimate is 0 minutes.

Write a function which takes a numeric weight in the range [0,7000] as input and produces estimated time as output is: “OVERLOADED”, and for all other inputs, the output statement is

“INVALID INPUT”.

Input should be in the form of integer value –

Output must have the following format –

Time Estimated: Minutes

Example:

  • Input value

2000

  • Output value

Time Estimated: 25 minutes

Solution

 

#include<stdio.h>

void calculateTime(int n)

{

    if(n==0)

        printf("Time Estimated : 0 Minutes");

    else if(n>0 && n<=2000) 

        printf("Time Estimated : 25 Minutes"); 

    else if(n>2000 && n<=4000) 

        printf("Time Estimated : 35 Minutes"); 

    else if(n>4000 && n<=7000)

        printf("Time Estimated : 45 Minutes");

    else

        printf("INVALID INPUT");

}

int main()

{

    int machineWeight;

    scanf("%d",&machineWeight);

    calculateTime(machineWeight);

    return 0;

}

 

by Expert (38,790 points)
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Question 2

Selection of MPCS exams include a fitness test which is conducted on ground. There will be a batch of 3 trainees, appearing for the running test in track for 3 rounds. You need to record their oxygen level after every round. After trainees are finished with all rounds, calculate for each trainee his average oxygen level over the 3 rounds and select one with the highest oxygen level as the most fit trainee. If more than one trainee attains the same highest average level, they all need to be selected.

Display the most fit trainee (or trainees) and the highest average oxygen level.

Note:

  • The oxygen value entered should not be accepted if it is not in the range between 1 and 100.
  • If the calculated maximum average oxygen value of trainees is below 70 then declare the trainees as unfit with a meaningful message as “All trainees are unfit.
  • Average Oxygen Values should be rounded.

Example 1:

  • INPUT VALUES

95

92

95

92

90

92

90

92

90

 

 

  • OUTPUT VALUES
    • Trainee Number : 1
    • Trainee Number : 3

Note:

Input should be 9 integer values representing oxygen levels entered in order as

Round 1

  • Oxygen value of trainee 1
  • Oxygen value of trainee 2
  • Oxygen value of trainee 3

Round 2

  • Oxygen value of trainee 1
  • Oxygen value of trainee 2
  • Oxygen value of trainee 3

Round 3

  • Oxygen value of trainee 1
  • Oxygen value of trainee 2
  • Oxygen value of trainee 3

Output must be in the given format as in the above example. For any wrong input final output should display “INVALID INPUT”

#include <stdio.h>   

int main()  

{

int trainee[3][3];

int average[3] = {0};

int i, j, max=0;

for(i=0; i<3; i++)

{

     for(j=0; j<3; j++)

     {

         scanf("%d",&trainee[i][j]);

         if(trainee[i][j]<1 || trainee[i][j]>100)

         {

             trainee[i][j] = 0;

         }

     }

}

for(i=0; i<3; i++)

{

     for(j=0; j<3; j++)

     {

         average[i] = average[i] + trainee[j][i];

     }

     average[i] = average[i] / 3;

}

for(i=0; i<3; i++) { if(average[i]>max)

     {

         max = average[i];

     }

}

for(i=0; i<3; i++)

{

     if(average[i]==max)

     {

         printf("Trainee Number : %d\n",i+1);

     }

     if(average[i]<70)

     {

         printf("Trainee is Unfit");

     }

}

return 0;

}  

 

by Expert (38,790 points)
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Question 1

There is a JAR full of candies for sale at a mall counter. JAR has the capacity N, that is JAR can contain maximum N candies when JAR is full. At any point of time. JAR can have M number of Candies where M<=N. Candies are served to the customers. JAR is never remain empty as when last k candies are left. JAR if refilled with new candies in such a way that JAR get full.

Write a code to implement above scenario. Display JAR at counter with available number of candies. Input should be the number of candies one customer can order at point of time. Update the JAR after each purchase and display JAR at Counter.

Output should give number of Candies sold and updated number of Candies in JAR.

If Input is more than candies in JAR, return: “INVALID INPUT”

Given, 

N=10, where N is NUMBER OF CANDIES AVAILABLE

K =< 5, where k is number of minimum candies that must be inside JAR ever.

Example 1:(N = 10, k =< 5)

  • Input Value
    • 3
  • Output Value
    • NUMBER OF CANDIES SOLD : 3
    • NUMBER OF CANDIES AVAILABLE : 7

Example : (N=10, k<=5)

  • Input Value
    • 0
  • Output Value
    • INVALID INPUT
    • NUMBER OF CANDIES LEFT : 10
#include<stdio.h>   

int main()  

{

int n=10, k=5;

int num;

scanf("%d",&num);

if(num>=1 && num<=5)

{

     printf("NUMBER OF CANDIES SOLD : %d\n",num);

     printf("NUMBER OF CANDIES LEFT : %d",n-num);

}

else

{

     printf("INVALID INPUT\n");

     printf("NUMBER OF CANDIES LEFT : %d",n);

}

return 0;

} 

 

by Expert (38,790 points)

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