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AUTOMATA FIX FINAL PUSH

Q1.

The function searchBST accepts the two arguments -root ,node representing the root node of the tree and key an integer represents value to be searched respectively.It supposed to return 'T'if key is found in the binary search Tree otherwise it returns '0'

int searchBST(node *root, int key)

{

if(root==NULL)

return 0;

if(root value == key)

return 1;

if(key>=root value)

return searchBST(root right,key);

else

return searchBST(root left,key);

}


{

if(root==NULL)

return 0;


return 1;

if(key<=root value)


else


}


{

if(root==NULL)

return 0;


return 1;

if(key==root value)


else


}

Q2.

The Function searchkeyindex accepts two arguments -list_head a node representing the head the node of the linked list and key an integer representing the key value that must be searched in the linked list it is suppose to return list of integers representing the index values of key in the linked list if the key value is present in the linked list otherwise it returns list of length of 1 with the content '-1'

boundedarray* searchkeyindex(node **list_head, int key)

{

boundedarray* res = (boundedarray*)malloc(sizeof(boundedarray))

Int flag=0,index=-1,count_index=0;

node* current = *list_head;

int *digit = (int *) malloc (100*sizeof(int));

while(current next!= NULL)

{

index++;

if(currentdata == key)

{

digit[count_index]=index;

count_index++;

flag=1;

}

else

current = currentnext

}

if(flag==1)

{

resarr = (int *) malloc (sizeof(int)*count_index);

for(i=0;i

res arr[i] = digit[i];

ressize++; }

return res; }

else {

resarr = (int *) malloc (1*sizeof(int));

resarr[0]=-1;

ressize++;

return res; }


{






{

index++;


{


count_index;

flag=1;

}

else


}

if(flag==1)

{


for(i=0;i


ressize++; }

return res; }

else {


resarr[0]=-1;

ressize++;

return res; }


{






{

index++;


{

digit[count_index++]=index;

count_index; // not given

flag=1;

}

else


}

if(flag==1)

{


for(i=0;i


ressize++; }

return res; }

else {


resarr[0]=-1;

ressize++;

return res; }

Q3.

The function method oddsum accept three arguements -Start,End and list .An integer representing the starting and Ending index range list of values respectively .The function oddsum is supposed to return the sum of odd numbers in given range of values.

Sample InputSample Output

1

6

1

2

3

4

5

6

9

Time Limit: - ms Memory Limit: - kb Code Size: - kb

Q4.

The function CalculateGeneralHCF accepts the two arguments len,arr.an integer represents the length of the array and list of integers respectively.It suppose to calculate the and return the maximum element of the input array.Another function calculateHCF(int a,int b)return the HCF of two input numbers a and b.


10

1

2

3

4

5

6

7

8

9

10

9


Q5.

The function findindex() accepts the three arguements -size,list and key,an integer represents the size of the list ,list of integers ,an integer represents the value to search from the given list respectively.The function/metod findindex is supposed to return the index of the key value if the key is found in the list else return '-1'


5

10

20

30

40

50

30

2


Q6.

The function Evenmultiplication accept the two arguments size and list ,an integer represents the size of the list and a list of integers .The function/method Evenmultiplication is supposed to return an integer representing multiplicative value of all even integers present.


4

10

11

12

13

14

120


Q7.

The function CalculateGeneralMax Accepts the 2 arguments len and arr.Integer represents the length of array and list of integers respectively.Suppose to calculate and return the maximum element in the input array.The another function CalculateMax(int a,int b) returns the maximum element in the input array.


5

1

2

3

4

5

5


Q8.

Debug the code such that the result is the product of maximum 2 of 3 inputs:

int multiplyNumber(int a,int b,int c)

{

int result,min,max,mid;

max=((a>b)&&(a>c))&&((b>c)&&(b>a));

min=((a

mid=((a+b+c)-(min+max));

printf("%d %d",max,min);

result=(max*mid);

return result;

}

max=a>b&&a>c&&b>c&&b>a;

min=a

mid=a+b+c-min+max;

max=((a>b)&&(a>c))||((b>c)&&(b>a));

min=((a


max=((a>b)||(a>c))&&((b>c)||(b>a));

min=((a


max=((a>=b)&&(a>=c))?a:((b>=c)&&(b>=a))?b:c;

min=((a<=b)&&(a<=c))?a:((b<=c)&&(b<=a))?b:c;


Q9.

Print the following pattern of n lines

where n (26>=n>=0)


3

a

ab

abc


Q10.

The function Countdigits(int num) is supposed to return the remainder when the input argument num is divided by the number of digits in num.

int Countdigits(int num)

{

int count=0;

while(num!=0)

{

num=num/10;

count=count+1;

}

return (num%count);

}

Input Format

Single line (Integer value)

Output Format

Single line (Integer)


45

1


54

0


Q11.

Debug the code such that the code prints the count of repeated element in the array

Input Format

First line is a integer - number of elements in the array - N

Next N lines is the elements of the array

Output Format

Single line of Integer (count)


5

1

2

2

3

5

1


Q12.

Analyze the following code and debug the syntax errors.

Find the greatest of three numbers.

#include

int main()

{

int num1, num2, num3;

scanf(“%d %d %d”, &num1,&num2,&num3);

if (num1 > num2) && (num1 > num3)

{

printf(“%d”, num1);

}

elseif(num2>num3)

{

printf(“%d”, num2)

}

else

{

printf(“%d”, num3);

}

return 0;

}

Input Format

3 integers

Output Format

Greatest integer


30

20

10

30


5

6

8

8


Q13.

The function getarraysum(int *arr,int len) is supported to calculation and return the sum of elements of the input array arr of length len(len>0).

The function compiles successfully but fails to return the desired result due to logical errors

int getarraysum(int *arr,int len)

{

int sum=0'

for(i=0;i

{

sum=arr[i];

}

return sum;

Your task is to fix the logical error and run the code

Input Format

The first line of input contains no.of elements in the array.

Second line contains the elements of the array.

Output Format

Sum of the elements.


6

10 20 30 40 50 60

210


3

10 20 30

60


Q14.

Given a program to print the binary equivalent of the given number. The code has logical error in the line below.Debug the program and execute it.

return (num * 2) + 10 / binary_conversion(num / 2);

Input Format

Positive integer

Output Format

Corresponding binary number


4

100


9

1001


Q15.

Analyze the code and find the logical error in the below code.

#include

int main () {

int i, j, n ;

scanf("%d", &n);

for(i=1; i

{

for(j=i;j

{

printf("%d", i);

}

printf("\n");

}

}

Input Format

Positive integer.

Output Format

Corresponding pattern


5

1111

222

33

4


7

111111

22222

3333

444

55

6


Q16.

Given a program to print a pyramid pattern.

But the code has logical errors in it. Debug the code and execute the program.

for (c = 1; c <= row - 1; c++)

printf("*");

Input Format

Positive integer

Output Format



4

*

***

*****

*******


6

*

***

*****

*******

*********

***********


Q17.

Given a C program to find the sum of the elements in an array.

However, the given code has syntax/logical errors. Check for the errors and execute the program such that it works for all input values

#include

#include

int main()

{

int *a,i,n,sum=0;

scanf("%d",&n);

a=(int*)malloc(n*sizeof(int));

for(i=0;i

{

scanf("%d",a+i);

}

for(i=0;i

{

sum = sum *(*(a-i));

}

printf("The sum of the elements in the array is %d",sum);

return 0;

}

Input Format

Input consists of n+1 integers. The first integer corresponds to ‘n’ , the size of the array. The next ‘n’ integers correspond to the elements in the array. Assume that the maximum value of n is 15.

Output Format

Refer sample output for details.


5

3

2

2

1

6

The sum of the elements in the array is 14


3

1

2

3



Q18.

Given a program that asks a user for their birth year encoded as two digits (like "62") and for the current year, also encoded as two digits (like "99"). The program is to correctly write out the users age in years.

The code is given below, but it has logical errors in it. Debug the code so as to pass all the test cases.

#include

int main()

{

int curr_year,b_year,age;

scanf("%d",&b_year);

scanf("%d",&curr_year);

if(curr_year>b_year)

{

age=(curr_year) +(b_year);

printf("Your age is %d",age);

}

else if(curr_year

{

age=(curr_year+100)+b_year;


}

return 0;

}

Input Format

Input consists of 2 integers. The first integer corresponds to the last 2 digits of the birth year. The second integer corresponds to the last 2 digits of the current year.

Output Format

Refer sample input and output for further formatting specifications.


27

07

Your age is 80


99

21

Your age is 22


Q19.

The function printTable(int num)is supposed to print the first ten multiples of the multiplication table of the input number num.

The function compiles fine but fails to return the desired result for some test cases.

Your task is to fix the program so that it passes all the test cases.

void printTable(int num)

{

int value = 0;

for(int i=1;i<10;i++)

{

value = num * i ;

printf(“%d”, value);

}

}

Input Format

Positive Integer.

Output Format

Corresponding multiples of the integer.


2

2

4

6

8

10

12

14

16

18

20


10

10

20

30

40

50

60

70

80

90

100


Q20.

The function calculateMatrixSum(int **matrix, int m, int n) accepts a two-dimensional array matrix of dimensions m, n as an input and returns the sum of odd elements whose ith and jth index are the same.(m=n=3)

The function compiles fine but fails to return the desired result for some test cases.

Your task is to debug the program so that it passes all the test cases.

int calculateMatrixSum(int **matrix, int m, int n)

{

int i, j, sum=0,row = m, column = n;

if((row > 0) && (column > 0)){

for(i=0;i

sum=0;

for(j=0;j

if(i==j){

if(matrix[i][j]/2 != 0)

sum += matrix[(i)(j);;

}

}

} return sum;

}

else

return sum;

}

Input Format

3 X 3 matrix.

Output Format

Print the corresponding output.


1 1 1 1 1 1 1 1 1

3


1 2 3 4 5 6 7 8 9

15


Q21.

Given a program to check if a string is palindrome or not. The code has logical errors in it. Debug the code and execute the program.

while (text[length] = '\0')

length++;

end = length / 1;

middle = length*2;

Input Format

String

Output Format

Corresponding output.


mom

Palindrome.


string

Not a palindrome.


Q22.

Analyze the following and debug the errors.

#include

int main()

{

int number;

//printf("Enter a number \n");

scanf("%d", &number);

if (number >= 0);

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

else

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

}

Input Format

Integer

Output Format



9

9 is a positive number


-3

-3 is a negative number


Q23.

Analyze the code below and debug the program .

#include

int main()

{

int number;

//printf("Enter an integer: ");


// True if the number is perfectly divisible by 2

if(number % 2 == 0);

printf("%d is even.", number);

else

printf("%d is odd.", number);

return 0;

}

Input Format

Positive integer.

Output Format



15

15 is odd.


18

18 is even.


Q24.

Analyze the following code and complete the missing code.

#include

int add(int,int);

int main()

{

int a,b,c;

scanf("%d %d",&a,&b);

c=add(a,b);

printf("%d",c);

return 0;

}

int add(int a,int b)

{

// missing code

}

Input Format

Enter two numbers

Output Format

Print the sum of two numbers


10

2

12


5

6

11


Q25.

Given a program to find if a number is positive or negative. The code has syntax errors in it.Analyze the following and debug the errors.

scanf('%d', number);

Input Format

Integer

Output Format



9



-3



Q26.

Analyze the code and fill the missing lines inorder to pass all the test cases.

#include

#include

int computeSum(int n)

{

// missing code

}

int main()

{

int n;

scanf("%d",&n);

computeSum(n);

printf("The sum of digits in %d is %d",n,computeSum(n));

return 0;

}

Input Format

Input is a non-negative integer.

Output Format

Refer sample input and output for formatting specifications.


123

The sum of digits in 123 is 6


1240



Q27.

Analyze and debug the code to execute it.

#include

int main()

{

int n, reversedInteger = 0, remainder, originalInteger;

printf("Enter an integer: ");

scanf("%d", &n);

originalInteger = n;

// reversed integer is stored in variable

while( n!=0 )

{

remainder = n*10;

reversedInteger = reversedInteger*10 + remainder;

n /= 10;

}

// palindrome if orignalInteger and reversedInteger are equal

if (originalInteger != reversedInteger)

printf("%d is a palindrome.", originalInteger);

else

printf("%d is not a palindrome.", originalInteger);

return 0;

}

Input Format

Positive integer

Output Format

Corresponding output


505

505 is a palindrome.


103

103 is not a palindrome.


Q28.

Given a program to rearrange positive and negative numbers in an array. Analyze the code and complete the missing code to execute the program.

void rearrange_alternate_positions(int arr[], int n)

{

// missing code

}

Input Format

First line contains the size of array, followed by the elements of the array.

Output Format

Original array and the rearranged array.


5

1

-2

-3

4

-5

Original array : 1 -2 -3 4 -5

Rearranged array : -2 -3 -5 1 4


4

-1

-8

9

6

Original array : -1 -8 9 6

Rearranged array : -1 -8 9 6


Q29.

Analyze the code and complete the missing code.

#include

void swap(int *a, int *b);

int main()

{

int m,n;

scanf("%d %d",&m,&n);

swap(&m, &n);

}

void swap(int *a, int *b)

{

// missing code

}

Input Format

2 integers.

Output Format

Swapped integers.


10

20

20

10


30

40

40

30


Q30.

Given a program to find the maximum element in an array using recursion. Fill the missing lines of code so that the program passes through all the test cases.

#include

#include

int max=0;

int findMaxElement(int *arr,int size)

{

// missing code

}

int main()

{

int *arr,size,i;

scanf("%d",&size);

arr=(int *)malloc(size*sizeof(int));

for(i=0;i

{

scanf("%d",(arr+i));

}

findMaxElement(arr,size);

printf("Maximum element in the array is %d",findMaxElement(arr,size));

return 0;

}

Input Format

First line of input contains size of the array.

The next lines contain the elements of the array.

Output Format



4

2

3

5

6

Maximum element in the array is 6


5

62

35

21

45

100



Q31.

Given a program to find the average of two numbers using pointers.

The code has logical/syntax errors in the lines below. Debug the program and execute it.

sum=sum + (p+i);


2

5

6

5.50


2

25

25

25.00


Q32.

Given a program to print the following pattern.

1

1 1

1 1 2

The code has syntax/logical errors in the lines below. Debug the program and execute it.

printf('%d\t',c);

Input Format

Positive integer.

Output Format

Corresponding pattern.


3

1

11

112


5

1

11

112

1123

11235


Q33.

Given a program to check if a number is perfect number or not. The code has logical error in it. Debug the program and execute it.

rem = number/ i;

if (rem != 0)

{

sum = sum++;

}

Input Format

Positive integer

Output Format



25

Entered Number is not a perfect number


6

Entered Number is perfect number


Q34.

Given a program that determines the price of a movie ticket. The program asks for the customer's age and for the time on a 24-hour clock (where noon is 12.00 and 4:30 PM is 16.30). The show timings are 10.15, 13.30, 18.00 and 22.00. The normal adult ticket price is 8.00,howevertheadultmatineepriceis8.00,howevertheadultmatineepriceis5.00. Adults are those over 13 years. The normal children's ticket price is 4.00,howeverthechildren′smatineepriceis4.00,howeverthechildren′smatineepriceis2.00.

The code below has logical errors. Debug the code so as to pass through all the test cases.

#include

int main()

{

int age;

float time,time1;

scanf("%d",&age);

scanf("%f",&time);

time1=13.30;

if(age>13)

{

if(time!=time1)

{

printf("The ticket price is $5.00");

}

else

{


}

}

else if(age<=13)

{

if(time!=time1)

{


}

else

{


}

}

return 0;

}

Input Format

The first input is an integer which corresponds to the age. The second integer is a float which corresponds to the show timing.

Output Format

Refer sample input and output for further formatting specifications.


10

10.15

The ticket price is $4.00


Q35.

A train of compartments are reserved only with the seat numbers like digits 4 and 7 such as 4, 7, 44, 47, 74,...... 44744,.. etc. Given a ticket K, we need to find K-th seat number in the train. The code is given below, fill the missing code such that it passes through all the test cases.

#include

int main()

{

int i,j,k,t,n,a[10000];

scanf("%d",&t);

for(j=0;j

{ scanf("%d",&n);

a[1]=4;

a[2]=7;

for (int i=3; i<=n; i++)

{

// missing code

}

printf("%d\n",a[n]);

}

}

Input Format

The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case contains an integer k.

Output Format

Print in a new line the Kth number in the series.


3

1

2

3

4

7

44


5

1

3

5

7

9

4

44

74

444

474


Q36.

The function checkGrade(int score) is supposed to return a students grade when the students test score is passed to it as an argument(o<=sore<=100)

Given a particular score a grade is calculated as per the following table

Score Grade

1.score>=90 A

2 .76<=score<=90 B

3 .61<=score<=75 C

4 score<=60 D

The function compiles successfully but fails to return the desired result due to logical errors.

Your task is to debug the program to pass all the test cases.

char checkGrade(int score)

{

if(score<=60)

return ‘D’;

else if((61<=score)||(score<=75))

return ‘C’;

else if((76<=score)||(score<=90))

return ‘B’;

else

return ‘A’;

}

Input Format

Positive integer

Output Format



95

A


90

B


Q37.

Given a program to find the numbers divisible by 5 in the given range. Also find the sum of the numbers. But the code has syntax errors in it.Debug the code and execute the program.

if (i % 5 == 0)

{

printf('%3d ', &i);

count++;

sum = sum + i;

}

Input Format

Range to find out the numbers divisible by 5

Output Format

First line contains the numbers divisible by 5 in the given range.

Second line contains the sum of numbers divisible by 5.


1

20

5 10 15

30


21

40

25 30 35

90


Q38.

The seats are arranged in the lab in a rectangular fashion with m rows and n columns. The students are seated in the lab according to their roll numbers in row wise order.

Given the roll number of a student, the code is given to find whether the student is seated in the first row or in the 1st or last column. But the code has syntax errors. Debug the code without changing the logic.

#include

int main()

{

int rows,columns,rnum,d,e;

scanf("%d",&rows);

scanf("%d",&columns);

scanf("%d",&rnum);

d=rows*columns;

e=d-rows;

if((rnum%rows==1)||(rnum<=rows)||(rnum>e)

printf("yes");

}

else

{

printf("no");

}

return 0;

}}

Input Format

Refer sample input and output.

Output Format

The output consists of a string --- “yes” or “no”.


5

10

41

yes


5

10

12

no


Q39.

Given a program to convert a number from binary to octal.The code has logical errors in it.Debug the code and execute it.

remainder = binarynum *10;

octalnum = octalnum + remainder / j;

j = j * 2;

binarynum = binarynum - 10;

Input Format

Binary number

Output Format

Octal number


1101

Equivalent octal value: 15


1001

Equivalent octal value: 11


Q40.

Given a program to put the even and odd elements of array in 2 separate arrays. The code has logical errors in the lines below. Debug the code and execute the program.

if (ARR[i] /2 == 0)

{

EAR[j] = ARR[i];

j--;

}

else

{

OAR[k] = ARR[i];

k++;

}

Input Format


Next lines contain the elements of array.


4

11

12

13

14

The odd elements are

11

13

The even elements are

12

14


5

2

4

3

1

6

The odd elements are

3

1

The even elements are

2

4

6


Q41.

Roy bought a new bike last Sunday. He needs to decide the number plate number now. Though he likes 0’s and 1’s most out of number system, he wants his number plate to start with 1. Lets help him in finding out how many unique permutations start with 1 when he is given with N 0’s and M 1’s.

The code is given below. Complete the missing lines of code to execute the program.Try printing the answer modulo (109+7).

#include

#define MOD 1000000007

long long fact(int n,int m)

{

// missing code

}

int main(void) {

// your code goes here

int t,N,M;

scanf("%d",&t);

while(t--)

{

scanf("%d %d",&N,&M);

long long ans=fact(N,M-1);

printf("%lld\n",ans);

}

return 0;

}

Input Format

First line contains T, the number of test cases.

Each test case consists of N and M separated by a space.

Output Format

For each test case, print the answer modulo (109+7).


2

1 2

2 1

2

1


2

1 1

2 3

1

6


Q42.

Me and my friend always play a game. For every integer X I give, my friend has to find the smallest positive integer Y such that X * Y ( X multiplied by Y) contains only 4s and 0s and starts with one or more 4s followed by zero or more 0s. For example, 404 is an invalid number but 4400, 440, and 444 are valid numbers.

If a is the number of 4s and b is the number of 0s, can you print the value of (2 x a) + b ?

The code is given below, but it has logical errors in it.Debug and execute the code.

#include

#include

#include

#include

int main() {

int t,x,i,ans,out=0,multiple=0,j,k,power,n4,n0;

scanf("%d", &t);

for(;t>0;t-=1) {

scanf("%d",&x);

power = 1;

multiple=0;

out=0;

k=1;

n0=1;

while(1) {

for(i=power,j=1;j<=k;i=i/10,j+=1) {

multiple += (4*i);

if((multiple%x)!=0) {

out=1;

break;

}

}

if(out==1) {

break;

}

power/=10;

k+=1;

multiple = 0;

n0+=1;

}

ans = (2*j) + n0-j;

printf("%d\n",ans);

}

return 0;

}

Input Format

The first line of input contains a single integer T, the number of test cases.

T lines follow, each line containing the integer X as stated above.

Output Format

For every X, print the output (2 x a) + b in a newline as stated in the problem statement.


2

2

10

2

3


3

4

4

10

2

2

3


Q43.

Analyze the following code and fill the missing lines to find if an element is present in the array or not.

#include

#include

int *createArray(int);

void readElement(int *,int);

void Array(int *,int);

int main()

{

int *a,n;

scanf("%d",&n);

a=createArray(n);

readElement(a,n);

Array(a,n);

return 0;

}

int *createArray(int n)

{

int *a;

a=(int *)malloc(n*sizeof(int));

return a;

}

void readElement(int *a,int n)

{

int i;

for(i=0;i

{

scanf("%d",a+i);

}

}

void Array(int *a,int n)

{

// missing lines

}

Input Format

Input consists of n+2 integers. The first integer corresponds to ‘n’ , the size of the array. The next ‘n’ integers correspond to the elements in the array. The last integer corresponds to ‘a’, the element to be searched.

Output Format

Refer sample output for details.


5

12

23

34

45

56

12

12 is present in the array


6

1

2

3

4

5

6

7

7 is not present in the array


Q44.

Analyze the code and complete the missing code to print the zigzag pattern of array.

#include

void swap(int *x, int *y)

{

// missing code

}

int main()

{

int n;

//printf(“\nEnter the number of elements : ” );

scanf("%d", &n);

int arr[n], i;

//printf(“\nInput the array elements : “);

for(i=0; i

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

zig_zag_fashion(arr, n);

for (int i=0; i

printf("%d", arr[i]);

printf("\n");

return 0;

}

Input Format


Second line contains the elements of the array

Output Format

Corresponding pattern.


6

1

2

3

4

5

6

132546


5

5

4

3

2

1

45231


Q45.

Given a code to print the sum of 'n' natural numbers. The code has logical errors in the lines below. Debug the code and execute it.

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

{

sum -= i;

}

Input Format

Positive integer

Output Format



22

253


14

105


Q46.

Given a program to replace a substring in a string. Complete the missing code and execute the program.

void replaceSubstring(char [],char[],char[])

{

// missing code

}

Input Format

First line of input contains a string.

Second line contains the substring of the given string.

Third line contains a new string which should be replaced with the substring.

Output Format



schoolbus

bus

van

schoolvan


housetown

house

home

hometown


Q47.

Given a program to find the index of equilibrium element in the given array. In an array equilibrium element is the one where the sum of all the elements to the left side is equal to the sum of all the elements in the right side. Complete the missing lines of code and execute the program.

int findEquilibriumIndex(int a[], int n)

{

// missing code

}

Input Format

First line contains the size of the array.

Next lines contain the elements of array.

Output Format



6

1

2

3

4

3

3

4


3

2

1

2

1


Q48.

Given a program to find the factors of a number. The program has logical errors in the lines below. debug the program and execute it.

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

{

if (number*i = 0)

{

printf("%d ",i);

}

}

Input Format

Positive integer

Output Format

Factors of the number


12

Factors of 12 are: 1 2 3 4 6 12


5

Factors of 5 are: 1 5


Q49.

Given a program to find the second smallest element of the array. But the code has logical/syntax errors in it. Debug the program and execute the code.

if(arr[i] < second_min &arr[i] == min)

Input Format

First line contains the size of the array.

The next lines are followed by the elements of the array.

Output Format

Second smallest element


3

12

3

13

12


5

10

12

13

14

11

11


Q50.

Given a program to find the nth term in the Fibonacci series using recursion.

Fill the missing code so as to pass through all the test cases.

Note that the first 2 terms in the Fibonacci Series are 0 and 1.

#include

#include

int fib(int n)

{

// missing code

}

int main()

{

int n;

scanf("%d",&n);

fib(n);

printf("The term %d in the fibonocci series is %d",n,fib(n));

return 0;

}

Input Format

Input consists of an integer.

Output Format



5

The term 5 in the fibonocci series is 3


3

The term 3 in the fibonocci series is 1


Q51.

Given a program to find the product of 2 numbers using recursion. The code has logical errors in the line below. Debug the code an execute the program.

return (a * product(a, b +1));

Input Format

2 positive integers.

Output Format

Product of two numbers


4

5

20


10

30

300


Q52.

Given a program to find the frequency of each element of array . But the code below has syntax errors in it. Debug the program and execute it.

scanf(“%d”, size);

Input Format


Next lines contain the elements of the array.

Output Format



5

1

1

2

3

4

1 occurs 2 times

2 occurs 1 times

3 occurs 1 times

4 occurs 1 times


3

1

1

1

1 occurs 3 times


Q53.

Nikki has a row of N white tiles indexed from 1 to N. This time, she's painting them blue!

The code is given to find the number of ways Nikki can paint certain tiles in green so that the indices of the blue tiles form an Arithmetic Progression. As this value can be quite large, your answer must be modulo 109 + 7.

Note: Nikki must paint at least 1 tile.

The code below has syntax errors. Debug the program so as to pass through all the test cases.

#include

#include

#include

#include

int main() {

int t;

scanf("%d",&t);

for(int index = 0;index < t;index++){

int N;

scanf("%d",&N);

int count = N;

for(int i = 0;i < N;i++);{

for(int j = i+1;j < N;j++);{

int d = j - i;

int tot = (N - i - 1)/d;

count += tot;

}

}

printf("%d\n",count);

}

return 0;

}

Input Format

The first line contains a single integer, T , denoting the number of test cases.

Each test case consists of a single line containing an integer, N, denoting the length of row of tiles.

Output Format

On a new line for each test case, print the number of ways Nikki can paint her white tiles blue so that the indices of the blue tiles form an Arithmetic Progression. Because this number can be quite large, your answer must be modulo (109 + 7).


3

1

2

3

1

3

7


5

3

6

9

8

7

7

33

86

65

48


Q54.

The function patternPrint(int n) supposed to print n number of lines in the following pattern

For n=4 the pattern should be:

1

1 1

1 1 1

1 1 1 1

The function complies successfully but fails to return the desired results due to logical errors Your task is to debug the program to pass all the test cases

Input Format

Integer

Output Format



4

1

11

111

1111


5

1

11

111

1111

11111


Q55.

Analyze the code and find out the syntax error without changing the logic.

#include

int main()

{

int a,b;

scanf("%d",&a);

scanf("%d",&b);

if(a==b)

{

printf("%d is equal to %d",a,b)

}

else if(a>b)

{

printf("%d is greater than %d",a,b);

}

if(a

{

printf("%d is less than %d",a,b);

}

}

Input Format

Input consists of 2 integers.

Output Format


The first two lines corresponds to the input and the third line corresponds the output.


6

9

6 is less than 9


100

2

100 is greater than 2


Q56.

In these days kids are introduced to computers at a very early age and in some schools the dictation test is conducted using computers. The teachers found it a bit difficult to evaluate these tests and they requested the school management to lessen their burden by automating this task. The 12th class students are learning C programming and they took up the task of automating the dictation evaluation. Here is the code for evaluation, but it has logical/syntax errors. Check the errors and debug the code.

#include

#include

int main()

{

char a[50],b[50];

scanf("%s",a);

scanf("%s",b);

if(strcpy(a,b)!=0)

{

printf("It is correct");

}

else

{

printf("It is wrong");

}

return 0;

}

Input Format

Input consists of a two strings. Assume that the maximum length of the string is 50 and it contains only alphabets.

Output Format



hello

hello

It is correct


super

superb

It is wrong


Q57.

Given a program to find Second largest Number in an Array . The code has logical errors in the lines below. Debug the code and execute the program.

else if(arr[i] < second ||arr[i] > first)

Input Format

First line of input is the size of the array.

The next lines are followed by the elements of the array.

Output Format



5

1

2

3

4

5

4


4

6

7

3

9

7


Q58.

#include

int main()

{

int number;



if (number >= 0)


else


}

Input Format

Input consists of a single character.

Output Format

Output consists of a string --- “Vowel” / “Consonant” / “Not an alphabet”



a

Vowel


Z

Consonant


Q59.

The code below has syntax errors. Debug the program to print the given pattern.(Refer to sample output)

#include

int main()

{

int i,j,n;

scanf("%d",&n);

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

{

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

{

if(j%2==1)

printf("1 ")

else

printf("0 ");

}

printf("\n");

}

return 0;

}

Input Format

Input consists of a single integer.

Output Format

Refer sample output. There is a trailing space at the end of each line.


5

1

1 0

1 0 1

1 0 1 0

1 0 1 0 1


4

1

1 0

1 0 1

1 0 1 0


Q60.

Analyze the code below and debug the syntax errors.

Find the factorial of a given number.

#include

int main()

{

long int fact = 1, n, i;

scanf(“%d”, &n);

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

{

fact = fact * i;

}

printf(“%d”, fact);

return 0;

}

Input Format

Positive Integer

Output Format

Factorial


5

120


8

40320


Answer Key & Solution

Section 1 - Automata Fix

Q1


{

if(root==NULL)

return 0;


return 1;

if(key>=root value)


else


}

Solution

No Solution

Q2


{






{

index++;


{


count_index++;

flag=1;

}

else


}

if(flag==1)

{


for(i=0;i


ressize++; }

return res; }

else {


resarr[0]=-1;

ressize++;

return res; }

Solution

No Solution

Q3Test CaseInputOutput

6

10

6

7

8

9

10

16

Weightage - 100Sample InputSample Output

1

6

1

2

3

4

5

6

9

Solution

Header

#include

int oddsum(int start, int end, int *list)

{

//write your code here

int sum=0;

int i=start;

while(i

{

if(list[i]%2!=0)

{

sum+=list[i];

}

i++;

}

return sum;

Footer

}

int main(void) {

int list[100],start,end,i,a;

scanf("%d",&start);

scanf("%d",&end);

for(i=start;i<=end;i++)

{

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

}

printf("%d",oddsum(start,end,list));

return 0;

}


5

20

15

20

15

10

10


10

1

2

3

4

5

6

7

8

9

10

9

Solution

Header

#include

int calculateHCF(int a, int b)

{

if(a==0)

return b;

return calculateHCF(b%a,a); }

int calculateGeneralHCF(int len, int *arr)

{


int result;

for (int i=0;i

result = calculateHCF(arr[i],arr[i+=1]);

return result;

Footer

}

int main(void) {

int list[100],size,i,a;

scanf("%d",&size);

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

{


}

printf("%d",calculateGeneralHCF(size,list));

return 0;

}


3

4

6

8

5

-1


5

10

20

30

40

50

30

2

Solution

Header

#include

int findindex(int size, int* list, int key)

{


for(int i=0;i

{

if(list[i]==key)

{

return i;

}}

return -1;

Footer

}

int main(void) {

int list[100],size,key,i,a;

scanf("%d",&size);

for(i=0;i

{


}

scanf("%d",&key);

printf("%d",findindex(size,list,key));

return 0;

}


3

12

14

15

168


4

10

11

12

13

14

120

Solution

Header

#include

int evenmulti(int size, int *list)

{


int em=1;

for(int i=0;i

{

if(list[i]%2 ==0)

{

em*=list[i];

}

}

return em;

Footer

}

int main(void) {

int list[100],size,key,i,a;

scanf("%d",&size);

for(i=0;i

{


}

printf("%d",evenmulti(size,list));

return 0;

}


5

1

10

20

30

40

40


5

1

2

3

4

5

5

Solution

Header

#include

int calculatemax(int a, int b)

{

if(a>b)

return a;

else

return b;

}

int calculateGeneralmax(int len, int *arr)

{


int maximum = arr[0];

for (int c = 0; c <=len ; c++)

{

maximum = calculatemax(maximum,arr[c]);

}

return maximum;

Footer

}

int main(void) {

int list[100],size,i,a;

scanf("%d",&size);

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

{


}

printf("%d",calculateGeneralmax(size,list));

return 0;

}

Q8

max=((a>=b)&&(a>=c))?a:((b>=c)&&(b>=a))?b:c;

min=((a<=b)&&(a<=c))?a:((b<=c)&&(b<=a))?b:c;


Solution

#include

int multiplyNumber(int a,int b,int c);

int main()

{

int x,y,z,s;

scanf("%d%d%d",&x,&y,&z);

printf("%d",multiplyNumber(x,y,z));

}

int multiplyNumber(int a,int b,int c)

{

int result,min,max,mid;

max=((a>=b)&&(a>=c))?a:((b>=c)&&(b>=a))?b:c;

min=((a<=b)&&(a<=c))?a:((b<=c)&&(b<=a))?b:c;


//printf("%d %d",max,min);

result=(max*mid);

return result;

}


5

a

ab

abc

abcd

abcde

Weightage - 50InputOutput

2

a

ab


3

a

ab

abc

Solution

Header

#include

void printCharacterPattern(int n);

int main()

{

int z;

scanf("%d",&z);

printCharacterPattern(z);

}

void printCharacterPattern(int n)

{

int i,j,value=1;

char ch='a';

char print;

for(i=0;i

{

print=ch;

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

{

printf("%c",print++);

}

printf("\n");

}

Footer

}


456

0


46598

3


45

1


54

0

Solution

Header

#include

int Countdigits(int num);

int main() {

int n,c;

scanf("%d",&n);

c=Countdigits(n);

printf("%d",c);

return 0;

}

int Countdigits(int num)

{

int count=0,tem;

tem=num;

while(num!=0)

{

num=num/10;

count=count+1;

}

return tem%count;

Footer

}


2

1

1

0


5

4

4

4

4

4

4


4

1

2

2

3

1


5

1

2

2

3

5

1

Solution

Header

#include

int sameElementCount(int *arr,int len);

int main() {

int len,a[50],i,c=0;

scanf("%d",&len);

for(i=0;i

{

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

}

c=sameElementCount(a,len);

printf("%d",c);

return 0;

}

int sameElementCount(int *arr,int len)

{

int i,count=0;

for(i=0;i

{

if((arr[i]%2==0)&&(arr[i]==arr[i+1]))

count++;

}

return count;

Footer

}


100

200

300

300


25

75

50

75


40

80

60

80


5

15

10

15


1

2

3

3


30

20

10

30


5

6

8

8

Solution

#include

int main()

{

int num1, num2, num3;

scanf("%d %d %d", &num1,&num2,&num3);

if ((num1 > num2) && (num1 > num3))

{

printf("%d", num1);

}

else if(num2>num3)

{

printf("%d", num2) ;

}

else

{

printf("%d", num3);

}

return 0;

}


5

10 10 10 10 10

50


3

11 22 33

66


5

120 240 360 480 600

1800


6

25 50 75 100 125 150

525


6

1470 2580 3690 7410 8520 9630

33300


6

10 20 30 40 50 60

210


3

10 20 30

60

Solution

#include

int getarraysum(int *arr,int len)

{

int sum=0;

int i;

for(i=0;i

{

sum=sum+arr[i];

}

return sum;

}

main()

{

int arr[5];

int len;

scanf("%d",&len);

for(int i=0;i

{

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

}

printf("%d",getarraysum(arr,len));

}


100

1100100


50

110010


10

1010


25

11001


14

1110


4

100


9

1001

Solution

Header

#include

int binary_conversion(int);

int main()

{

int num, bin;

//printf("Enter a decimal number: ");

scanf("%d", &num);

bin = binary_conversion(num);

printf(" %d\n", bin);

}

int binary_conversion(int num)

{

if (num == 0)

{

return 0;

}

else

{

return (num % 2) + 10 * binary_conversion(num / 2);

Footer

}

}


25

111111111111111111111111

22222222222222222222222

3333333333333333333333

444444444444444444444

55555555555555555555

6666666666666666666

777777777777777777

88888888888888888

9999999999999999

101010101010101010101010101010

1111111111111111111111111111

12121212121212121212121212

131313131313131313131313

1414141414141414141414

15151515151515151515

161616161616161616

1717171717171717

18181818181818

191919191919

2020202020

21212121

222222

2323

24


15

11111111111111

2222222222222

333333333333

44444444444

5555555555

666666666

77777777

8888888

999999

1010101010

11111111

121212

1313

14


17

1111111111111111

222222222222222

33333333333333

4444444444444

555555555555

66666666666

7777777777

888888888

99999999

10101010101010

111111111111

1212121212

13131313

141414

1515

16


10

111111111

22222222

3333333

444444

55555

6666

777

88

9


3

11

2


5

1111

222

33

4


7

111111

22222

3333

444

55

6

Solution

#include

int main () {

int i, j, n ;

scanf("%d",&n);

for(i=1; i

{

for(j=i;j

{

printf("%d", i);

}

printf("\n");

}

}


10

*

***

*****

*******

*********

***********

*************

***************

*****************

*******************


8

*

***

*****

*******

*********

***********

*************

***************


7

*

***

*****

*******

*********

***********

*************


3

*

***

*****


5

*

***

*****

*******

*********


4

*

***

*****

*******


6

*

***

*****

*******

*********

***********

Solution

Header

#include

int main()

{

int row, c, n, s;

//printf("Enter the number of rows in pyramid of stars you wish to see\n");

scanf("%d", &n);

s = n;

for (row = 1; row <= n; row++) // Loop to print rows

{

for (c = 1; c < s; c++) // Loop to print spaces in a row

printf(" ");

s--;

for (c = 1; c <= 2*row - 1; c++) // Loop to print stars in a row

printf("*");

Footer

printf("\n");

}

return 0;

}


6

9

8

7

6

5

4



6

100

200

150

250

300

350



5

32

25

14

98

74



10

10

20

30

40

50

60

70

80

90

100



4

12

24

36

48



7

3

6

9

12

15

18

21



2

0

1



0

1

2



3

50

100

150



5

22

44

66

88

112



5

3

2

2

1

6



3

1

2

3


Solution

#include

#include

int main()

{

int *a,i,n,sum=0;

scanf("%d",&n);

a=(int*)malloc(n*sizeof(int));

for(i=0;i

{

scanf("%d",a+i);

}

for(i=0;i

{

sum = sum +(*(a+i));

}

printf("The sum of the elements in the array is %d",sum);

return 0;

}


68

58

Your age is 90


75

10

Your age is 35


65

10

Your age is 45


97

12

Your age is 15


88

15

Your age is 27


27

07

Your age is 80


99

21

Your age is 22

Solution

#include

int main()

{

int curr_year,b_year,age;

scanf("%d",&b_year);

scanf("%d",&curr_year);

if(curr_year>b_year)

{

age=(curr_year) - (b_year);


}

else if(curr_year

{

age=(curr_year+100)-b_year;


}

return 0;

}


0

0

0

0

0

0

0

0

0

0

0


5

5

10

15

20

25

30

35

40

45

50


20

20

40

60

80

100

120

140

160

180

200


30

30

60

90

120

150

180

210

240

270

300


55

55

110

165

220

275

330

385

440

495

550


2

2

4

6

8

10

12

14

16

18

20


10

10

20

30

40

50

60

70

80

90

100

Solution

#include "stdio.h"

void printTable(int n);

int main(){

int n;

scanf("%d",&n);

printTable(n);

}

void printTable(int num)

{

int i,value = 0;

for(i=1;i<=10;i++)

{

value = num * i ;

printf("%d\n", value);

}

}


1 2 3 1 2 3 1 2 3

4


3 3 3 3 3 3 3 3 3

9


2 5 8 3 6 9 1 4 7

7


1 2 3 4 5 6 7 8 9

15


3 2 1 6 5 4 9 8 7

15


1 1 1 1 1 1 1 1 1

3


1 2 3 4 5 6 7 8 9

15

Solution

#include

#include

int calculatematrixsum(int **matrix,int m,int n);

int main(){

int m=3,n=3;

int **matrix;

int i;

matrix=(int **)malloc(sizeof(int));

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

matrix[i]=(int *)malloc(sizeof(int));}

int j;

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

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

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

int s=calculatematrixsum(matrix,m,n);

printf("%d",s);

}

int calculatematrixsum(int **matrix,int m,int n)

{

int i,j,sum=0,row=m,column=n;

if((row>0)&&(column>0)){

int sum=0;

for(i=0;i

for(j=0;j

{

if(i==j)

{

if(matrix[i][j]%2!=0)

sum+=matrix[i][j];

//printf("%d",sum);

}

}

}

return sum;

}

else

return sum;

}


indiana

Not a palindrome.


mtr

Not a palindrome.


hard

Not a palindrome.


madam

Palindrome.


rar

Palindrome.


mom

Palindrome.


string

Not a palindrome.

Solution

Header

#include

int main()

{

char text[100];

int begin, middle, end, length = 0;

scanf("%s",&text);

while (text[length] != '\0')

length++;

end = length - 1;

middle = length/2;

Footer

for (begin = 0; begin < middle; begin++)

{

if (text[begin] != text[end])

{

printf("Not a palindrome.\n");

break;

}

end--;

}

if (begin == middle)

printf("Palindrome.\n");

return 0;

}


100



-30



-50



10



3



9



-3


Solution

#include

int main()

{

int number;



if (number >= 0)


else


}


11000

11000 is even.


582

582 is even.


369

369 is odd.


20

20 is even.


30

30 is even.


15

15 is odd.


18

18 is even.

Solution

#include

int main()

{

int number;



// True if the number is perfectly divisible by 2

if(number % 2 == 0)

printf("%d is even.", number);

else

printf("%d is odd.", number);

return 0;

}


1200

1156

2356


89

98

187


33

56

89


12

15

27


41

14

55


10

2

12


5

6

11

Solution

#include

int add(int,int);

int main()

{

int a,b,c;

scanf("%d %d",&a,&b);

c=add(a,b);

printf("%d",c);

return 0;

}

int add(int a,int b)

{

int sum;

sum=a+b;

return sum;

}


100



-30



-50



10



3



9



-3


Solution

Header

#include

int main()

{

int number;



Footer

if (number >= 0)


else


}


145698



741235



145



20



520



851



369



581



654



753



123



1240


Solution

#include

#include

int computeSum(int n)

{

int s;

if(n%10==n)

{

return n;

}

s=n%10;

return s+computeSum(n/10);

}

int main()

{

int n;

scanf("%d",&n);

computeSum(n);

printf("The sum of digits in %d is %d",n,computeSum(n));

return 0;

}


321123



2552



52225



20



102



505



103


Solution

#include

int main()

{

int n, reversedInteger = 0, remainder, originalInteger;


scanf("%d", &n);

originalInteger = n;

// reversed integer is stored in variable

while( n!=0 )

{

remainder = n%10;

reversedInteger = reversedInteger*10 + remainder;

n /= 10;

}

// palindrome if orignalInteger and reversedInteger are equal

if (originalInteger == reversedInteger)

printf("%d is a palindrome.", originalInteger);

else

printf("%d is not a palindrome.", originalInteger);

return 0;

}


20

1

-2

3

4

5

-6

7

-8

9

10

-11

Original array : 1 -2 3 4 5 -6 7 -8 9 10 -11 0 0 0 0 0 0 0 1180438888 32617

Rearranged array : -2 -6 -8 -11 0 0 0 0 0 0 0 1 3 4 5 7 9 10 1180438888 32617


15

1

2

-3

-4

5

6

-7

8

9

-10

11

12

13

-14

-15

Original array : 1 2 -3 -4 5 6 -7 8 9 -10 11 12 13 -14 -15

Rearranged array : -3 -4 -7 -10 -14 -15 1 2 5 6 8 9 11 12 13


12

12

11

10

-9

-8

-7

6

5

-4

3

2

1

Original array : 12 11 10 -9 -8 -7 6 5 -4 3 2 1

Rearranged array : -9 -8 -7 -4 12 11 10 6 5 3 2 1


5

1

2

-3

-4

5

Original array : 1 2 -3 -4 5

Rearranged array : -3 -4 1 2 5


3

1

-2

3

Original array : 1 -2 3

Rearranged array : -2 1 3


5

1

-2

-3

4

-5

Original array : 1 -2 -3 4 -5

Rearranged array : -2 -3 -5 1 4


4

-1

-8

9

6

Original array : -1 -8 9 6

Rearranged array : -1 -8 9 6

Solution

Header

#include

#include

void rearrange_alternate_positions(int arr[], int n)

{

int temp, j;

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

{

temp = arr[i];

if (temp > 0)

continue;

j = i - 1;

while ( arr[j] > 0 && j >= 0)

{

arr[j + 1] = arr[j];

j--;

}

arr[j + 1] = temp;

}

Footer

}

int main()

{

int n,i;

//printf("\nEnter the number of elements : ");

scanf("%d",&n);

int arr[n];

//printf("\nInput the array elements : ");

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

{

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

}

printf("\nOriginal array : ");

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

{

printf("%d ",arr[i]);

}

printf("\n");

rearrange_alternate_positions(arr, n);

printf("\nRearranged array : ");

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

{

printf("%d ",arr[i]);

}

printf("\n");

return 0;

}


100

200

200

100


25

85

85

25


40

60

60

40


3

6

6

3


4

1

1

4


10

20

20

10


30

40

40

30

Solution

#include

void swap(int *a, int *b);

int main()

{

int m,n;

scanf("%d %d",&m,&n);

swap(&m, &n);

}

void swap(int *a, int *b)

{

int tmp;

tmp = *a;

*a = *b;

*b = tmp;

printf("\n%d \n%d", *a, *b);

}


10

99

98

100

110

65

452

78

84

96

120



6

32

42

52

62

72

92



7

110

210

310

810

410

510

610



6

12

15

22

10

9

8



4

100

1000

10

100



4

2

3

5

6



5

62

35

21

45

100


Solution

#include

#include

int max=0;

int findMaxElement(int *arr,int size)

{

if(size==0)

{

return max;

}

if(*(arr+(size-1))>max)

{

max=(*(arr+(size-1)));

//printf("max %d\n",max);

}

else

{

max=max;

//printf("max %d\n",max);

}

return findMaxElement(arr,(size-1));

}

int main()

{

int *arr,size,i;

scanf("%d",&size);

arr=(int *)malloc(size*sizeof(int));

for(i=0;i

{

scanf("%d",(arr+i));

}

findMaxElement(arr,size);

printf("Maximum element in the array is %d",findMaxElement(arr,size));

return 0;

}


2

2000

2500

2250.00


2

120

160

140.00


2

369

254

311.50


2

10

15

12.50


2

7

9

8.00


2

5

6

5.50


2

25

25

25.00

Solution

Header

#include

#include

int main()

{

int n,*p,sum=0,i;

float avg;

scanf("%d",&n);

p=(int *) malloc(n*2);

if(p==NULL)

{

printf("\nMEMORY ALLOCATION UNSUCCCESSFUL");

}

for(i=0;i

{

scanf("%d",(p+i));

}

for(i=0;i

sum=sum+*(p+i);

Footer

avg=(float)sum/n;

printf("\n %0.2f",avg);

}


10

1

11

112

1123

11235

112358

11235813

1123581321

112358132134

11235813213455


12

1

11

112

1123

11235

112358

11235813

1123581321

112358132134

11235813213455

1123581321345589

1123581321345589144


15

1

11

112

1123

11235

112358

11235813

1123581321

112358132134

11235813213455

1123581321345589

1123581321345589144

1123581321345589144233

1123581321345589144233377

1123581321345589144233377610


6

1

11

112

1123

11235

112358


2

1

11


3

1

11

112


5

1

11

112

1123

11235

Solution

Header

#include

#include

int main(){

int a=0,b=1,i,c,n,j;

scanf("%d",&n);

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

{

a=0;

b=1;

printf("%d\t",b);

for(j=1;j

{

c=a+b;

printf("%d\t",c);

Footer

a=b;

b=c;

}

printf("\n");

}

return 0;

}


100



28



58



10



4



25



6


Solution

Header

#include

int main()

{

int number, rem, sum = 0, i;

//("Enter a Number\n");


for (i = 1; i <= (number - 1); i++)

{

rem = number % i;

if (rem == 0)

{

sum = sum + i;

}

Footer

}

if (sum == number)

printf("Entered Number is perfect number");

else

printf("Entered Number is not a perfect number");

return 0;

}


89

18.00



12

13.30



13

18.00



55

22.00



35

10.15



10

10.15


Solution

#include

int main()

{

int age;

float time,time1;

scanf("%d",&age);

scanf("%f",&time);

time1=13.30;

if(age>13)

{

if(time==time1)

{


}

else

{


}

}

else if(age<=13)

{

if(time==time1)

{


}

else

{


}

}

return 0;

}


12

1

2

3

4

5

6

7

8

9

10

11

4

7

44

47

74

77

444

447

474

477

744

744


9

3

6

9

2

5

8

1

4

7

44

77

474

7

74

447

4

47

444


7

10

14

15

16

17

18

19

477

777

4444

4447

4474

4477

4744


4

0

1

2

3

0

4

7

44


3

2

4

6

7

47

77


3

1

2

3

4

7

44


5

1

3

5

7

9

4

44

74

444

474

Solution

#include

int main()

{

int i,j,k,t,n,a[10000];

scanf("%d",&t);

for(j=0;j

{

scanf("%d

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