📘 Revision — Sheet 1

A full revision sheet mixing everything you’ve learned so far: arrays, nested loops, functions, pointers-style thinking, and 2D matrices. Try every problem on your own first, then peek at the solution.

🎯 What You’ll Practice

Nested loops and pattern printing
Writing and calling your own functions
Digit-by-digit number manipulation (%10, /10)
2D matrices: scan, print, add, subtract, multiply, transpose
Classic problems: Fibonacci, GCD, Strong numbers, Pythagorean triples

🧪 Exercises

Exercise 1: Print Unique Numbers (No Duplicates)

Task: Use a single-subscripted array to solve the following problem. Read in 20 numbers, each of which is between 10 and 100 inclusive. As each number is read, print it only if it’s not a duplicate of a number already read. Provide for the “worst case” in which all 20 numbers are different. Use the smallest possible array to solve this problem.

💡 Hint

Keep an array to store numbers you’ve already seen, and a variable index that tracks how many unique numbers you’ve stored so far
For each new number, use a flag variable set to 0, then loop through the stored numbers and set flag = 1 if you find a duplicate
If flag is still 0 after the inner loop, the number is unique → print it and add it to the array
The “smallest possible array” is of size 20 (worst case: all numbers different)

🟢 Click to Show Solution

int main(){

    int arr[20]; //[1,2,3]
    int index=0;
    int num;
    int flag;

    for (int i = 0; i < 20; i++)
    {
        scanf("%d",&num);
        flag=0;
        for (int j = 0; j < index; j++)
        {
            if (num==arr[j])
            {
                flag=1;
            }  
        }
        if (flag==0)
        {
            printf("%d ",num);
            arr[index]=num;
            index++;
        }        
    }
    return 0;
}

Exercise 2: Sum of Numbers by Digit Sum Range

Task: Given three numbers N, A, B. Print the summation of the numbers between 1 and N whose sum of digits is between A and B inclusive.

💡 Hint

Write a helper function sumOfDigits(num) that repeatedly does num % 10 to grab the last digit, adds it to a running sum, then does num / 10 to drop that digit
Loop from 1 to N, call sumOfDigits(i) on each number, and if the result is between A and B inclusive, add i to the total
Be careful: you’re adding i to the total, not the sum of its digits

🟢 Click to Show Solution

int sumOfDigits(int num){ 
    int sum=0;
    while (num!=0)
    {
        sum=sum+num%10;
        num=num/10;
    }
    return sum;  
}

int main(){

    int N,A,B,sum=0,sOfDigits;
    scanf("%d",&N);
    scanf("%d",&A);
    scanf("%d",&B);
    for (int i = 1; i <= N; i++)
    {
        sOfDigits = sumOfDigits(i);
        if(sOfDigits>=A && sOfDigits<=B){
            sum+=i;
        }
    }
    printf("%d",sum);
    return 0;
}

Exercise 3: Repeat a Symbol

Task: Given 3 lines of input described as follows:

First line contains a symbol S (*, /, +, -).
Second line contains a number N.
Third line contains N numbers.

For each number Xi in the N numbers, print a new line that contains the symbol S repeated Xi times.

💡 Hint

Use scanf("%c", &c) to read the character
Use a nested loop: outer loop runs N times (once per number), inner loop runs x times and prints the character
Don’t forget \n after each inner loop finishes

🟢 Click to Show Solution

int main(){

    char c;
    int N,x;
    scanf("%c",&c);
    scanf("%d",&N);
    for (int i = 0; i < N; i++)
    {
        scanf("%d",&x);
        for (int j = 0; j < x; j++)
        {
            printf("%c",c);
        }
        printf("\n");  
    }
    
    return 0;
}

Exercise 4: Lucky Array

Task: Given a number N and an array A of N numbers. Determine if the array is lucky or not.

Note: The array is lucky if the frequency (number of occurrences) of the minimum element is odd.

Examples:

[2,2,1,1,1] --> LUCKY       (min=1, appears 3 times, odd)
[6,6,7,6]   --> LUCKY       (min=6, appears 3 times, odd)
[1,2,3,4,1] --> NOT LUCKY   (min=1, appears 2 times, even)
[2,2,2,4,5] --> LUCKY       (min=2, appears 3 times, odd)

💡 Hint

Break the problem into 3 steps:
1. Find the minimum in the array (loop and compare)
2. Count how many times the minimum appears (another loop)
3. Check if the count is odd (count % 2 != 0) → LUCKY, otherwise NOT LUCKY

🟢 Click to Show Solution

int main(){

    // input: N, A[N]
    // steps:
    // 1. search for minimmum number "done"
    // 2. find frequency of the minimum number  "done"
    // 3. frequency is odd (lucky) or even (not lucky)? 
    // output: lucky or not lucky
    // EXAMPLES:
    // [2,2,1,1,1] --> lucky
    // [6,6,7,6]   --> lucky
    // [1,2,3,4,1] --> not lucky
    // [2,2,2,4,5] --> lucky

    int N;
    scanf("%d",&N);
    int A[N];
    for (int i = 0; i < N; i++)
    {
        scanf("%d",&A[i]);
    }
    int min=A[0];
    for (int i = 0; i < N; i++)
    {
        if (min>A[i]){
            min=A[i];
        } 
    }
    int c=0;
    for (int i = 0; i < N; i++)
    {
        if (min==A[i]){
            c++;
        } 
    }
    if (c%2==0)
    {
        printf("NOT LUCKY");
    }
    else
    {
        printf("LUCKY");
    }
    
    return 0;
}

Exercise 5: Swap Min and Max in Array

Task: Given a number N and an array A of N numbers. Print the array after doing the following operations:

Find the minimum number in these numbers.
Find the maximum number in these numbers.
Swap the minimum number with the maximum number.

💡 Hint

Don’t just track the min/max values — also track their indices (minIndex, maxIndex), because you need the indices to swap
Classic swap with a temp variable: temp = A[minIndex]; A[minIndex] = A[maxIndex]; A[maxIndex] = temp;
Then just print the array after the swap

🟢 Click to Show Solution

int main(){

    int N;
    scanf("%d",&N);
    int A[N];
    for (int i = 0; i < N; i++)
    {
        scanf("%d",&A[i]);
    }
    int min=A[0], minIndex=0;
    for (int i = 0; i < N; i++)
    {
        if (min>A[i])
        {
            min=A[i];
            minIndex=i;
        } 
    }
    int max=A[0], maxIndex=0;
    for (int i = 0; i < N; i++)
    {
        if (max<A[i])
        {
            max=A[i];
            maxIndex=i;
        } 
    }

    // [1,2,3,5,4]
    //  0,1,2,3,4   <-- index
    int temp=A[minIndex]; // 1
    A[minIndex]  = A[maxIndex];
    A[maxIndex]  = temp;

    for (int i = 0; i < N; i++)
    {
        printf("%d ",A[i]);
    }
    return 0;
}

Exercise 6: First 10 Fibonacci Numbers

Task: Write a C program to find and print the first 10 Fibonacci numbers using functions.

💡 Hint

Fibonacci sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ... — each number is the sum of the two before it
Start with n1 = 0 and n2 = 1, print them first
In a loop, calculate n3 = n1 + n2, print n3, then shift: n1 = n2; n2 = n3;
Loop 8 more times to get 10 total numbers (since you already printed 2)

🟢 Click to Show Solution

int main(){

    // 0  1  1  2 3 5 8 13 21
    // n1 n2 n3
 
    int n1=0, n2=1,n3;
    printf("%d %d ",n1,n2);
    // n3=n1+n2; then print n3
    // n1=1 n2=1
    // n3=1+1=2  then print n3
    // n1=1 n2=2
    // n3=1+2=3  then print n3
    for (int i = 0; i < 8; i++)
    {
        n3=n1+n2;
        printf("%d ",n3);
        n1=n2;
        n2=n3;
    }
    return 0;
}

Exercise 7: Strong Number

Task: Write a program to check whether a given number is a strong number or not.

Example: 1! + 4! + 5! = 1 + 24 + 120 = 145 → 145 is a strong number.

💡 Hint

Write a fact(num) function that returns the factorial using a while loop
Keep the original number in a variable (e.g. num) and a copy (e.g. n) to modify
Extract each digit with n % 10, compute fact() of that digit, add to sum, then n = n / 10
At the end compare num (original) with sum — equal → STRONG, otherwise NOT STRONG

🟢 Click to Show Solution

int fact(int num){ 
    int f=1;
    while (num!=0)
    {
        f=f*num;
        num--;
    }
    return f;
}

int main(){

    // 6!=6*5*4*3*2*1
    // 145 , 1!=1, 4!=4*3*2*1=24 , 5!=5*4*3*2*1=120
    // 120+24+1= 145

    // input: num
    // steps:
    // 1. bring each digit
    // 2. bring the factorial of each digit from step 1
    // 3. calculate the sum of the factorial calculated in step 2
    // 4. if (num==sum) --> strong else --> not strong
    //output: strong or not strong

    int num, sum=0;
    scanf("%d",&num);
    int n= num;

    while (n!=0)
    {
        sum=sum+fact(n%10);
        n=n/10;
    }
    if (num==sum){
        printf("STRONG");
    }
    else{
        printf("NOT STRONG");
    }
    return 0;
}

Exercise 8: GCD (Greatest Common Divisor)

Task: Write a program to find the GCD (greatest common divisor) of two numbers.

💡 Hint

The GCD of a and b is the largest number that divides both of them with no remainder
Loop i from 2 up to the smaller of a and b
If both a % i == 0 and b % i == 0, update gcd = i
At the end, gcd holds the largest common divisor (initialize it to 1, since 1 always divides anything)

🟢 Click to Show Solution

int main(){
    //a=12 b=24
    int a,b,gcd=1;
    scanf("%d %d",&a,&b);

    for (int i = 2; i <=a && i<=b ; i++)
    {
        if (a%i==0 && b%i==0)
        {
            gcd=i;
        }
    }
    printf("%d",gcd);
    return 0;
}

Exercise 9: Volume of a Cylinder

Task: Write a program to calculate the volume of a cylinder using user input. Read from the user the radius and height and output the volume. Consider π = 3.14.

Formula: volume = π × radius² × height

💡 Hint

Use double (with %lf in scanf) for the radius and height so fractional values work
r*r is squaring — no need for a pow function
Use %f in printf for the output

🟢 Click to Show Solution

int main(){

    double r,h,v;
    scanf("%lf%lf",&r,&h);
    v=3.14*r*r*h;
    printf("v = %f",v);
    return 0;
}

Exercise 10: Print All Factors of a Number

Task: Write a C program to find all factors of a number.

💡 Hint

A factor of n is any number i such that n % i == 0
Loop i from 1 to n inclusive; every time n % i == 0, print i
Factors of 12: 1 2 3 4 6 12

🟢 Click to Show Solution

int main(){

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

    for ( int i = 1; i <= n; i++)
    {
        if (n%i==0)
        {
            printf("%d ",i);
        }   
    }
    return 0;
}

Exercise 11: Search Function

Task: Write a function that searches for a specific integer in an array and returns 1 if it is found and 0 if not.

💡 Hint

Function signature: int search(int arr[], int size, int n)
Loop through the array — the moment you find arr[i] == n, return 1 immediately (no need to keep looping)
If the loop finishes without finding it, return 0 after the loop
sizeof(arr)/sizeof(arr[0]) gives the number of elements of an array declared in the same scope

🟢 Click to Show Solution

int search(int arr[], int size, int n){
    for (int i = 0; i < size; i++)
    {
        if (arr[i]==n)
        {
            return 1;
        }
    }
    return 0;
}

int main(){

    int arr[]={1,2,3,4,5,6,1,20};
    int size=sizeof(arr)/sizeof(arr[0]);
    int n=4;

    if (search(arr,size,n)==1)
    {
        printf("FOUND");
    }
    else{
        printf("NOT FOUND");
    }
    return 0;
}

Exercise 12: Series 1 + 11 + 111 + 1111 + …

Task: Write a program in C to find the sum of the series 1 + 11 + 111 + 1111 + ... up to n terms.

Test Data:

Input the number of terms : 5

Expected Output :
1 + 11 + 111 + 1111 + 11111
The Sum is : 12345

💡 Hint

Start with p = 1 (the first term). To get the next term: p = p * 10 + 1 → 1 → 11 → 111 → 1111 ...
Keep a running sum += p inside the loop
For the pretty output, print p+ for every term except the last one; for the last term (i == n-1), print just p without the +

🟢 Click to Show Solution

int main(){

    int n,sum=0,p=1;
    scanf("%d",&n);
    // 123 % 10 = 3
    // 123 / 10 = 12  // integer division
    // 123 * 10 = 1230

    // 1 + 11 + 111 + 1111
    for (int i = 0; i < n; i++)
    {
        sum= sum+p;
        if (i==n-1)
        {
            printf("%d",p);
        }
        else{
            printf("%d+",p);
        }
        p=p*10 +1;
    }
    printf("\nSum = %d",sum);
    return 0;
}

Exercise 13: Number Triangle Pattern

Task: Write a C program to display a pattern like a right angle triangle with numbers.

The pattern:

💡 Hint

Classic nested loop — outer loop for rows, inner loop for columns
On row i (0-indexed), print numbers from 1 to i+1
In the inner loop print j+1 (since j starts at 0), and \n after the inner loop

🟢 Click to Show Solution

int main(){

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

    for (int i = 0; i < n; i++) // rows
    {
        for (int j = 0; j <= i; j++) // cols
        {
            printf("%d",j+1);
        }
        printf("\n");
    }
    return 0;
}

Exercise 14: Multiplication Table 6–10

Task: Write a C program that prints the following output:

         6  7  8  9  10
36 42 48 54  60
42 49 56 63  70
48 56 64 72  80
54 63 72 81  90
60 70 80 90 100

💡 Hint

Print the header row manually with \t (tab) before 6 7 8 9 10
Then loop i from 6 to 10, and for each row print i followed by i*6, i*7, i*8, i*9, i*10
Use \t to line up the columns

🟢 Click to Show Solution

int main(){

    printf("\t6  7  8  9  10\n");
    for(int i=6;i<=10;i++)
    {
        printf("%d\t%d %d %d %d %d\n",i,i*6,i*7,i*8,i*9,i*10);
    }
    return 0;
}

Exercise 15: Print Unique Elements in an Array

Task: Write a program in C to print all unique elements in an array.

A unique element is one that appears exactly once in the array. Example: in {1,2,3,4,5,4,3} the unique elements are 1 2 5.

💡 Hint

Use a nested loop: outer loop picks arr[i], inner loop counts how many times arr[i] appears in the whole array
If count == 1 at the end of the inner loop, print arr[i]
Note this is different from Exercise 1 — here we print only truly unique elements, not the first occurrence of each

🟢 Click to Show Solution

int main(){
    // {1,2,3,4,5,4,3};
    // 1 2 5
    int  n,count;
    scanf("%d",&n);
    int arr[n];
    for (int i = 0; i < n; i++)
    {
        scanf("%d",&arr[i]);
    }

    for (int i = 0; i < n; i++)
    {
        count=0;
        for (int j = 0; j < n; j++)
        {
            if(arr[i]==arr[j]){
                count++;
            }
        }
        if (count==1)
        {
            printf("%d ",arr[i]);
        }        
    }
    return 0;
}

Exercise 16: Matrix Calculator (3×3)

Task: Write a program that accepts two (3 × 3) matrices from the user and then asks the user to select one operation:

Multiplication
Addition
Subtraction
Transpose

The program should print the output as a (3 × 3) matrix.

💡 Hint

Break the program into helper functions: scanMatrix, printMatrix, addMatrix, subMatrix, multMatrix, transposeMatrix — much cleaner than one giant main
Add/Sub: r[i][j] = m1[i][j] ± m2[i][j]
Transpose: r[i][j] = m1[j][i] (swap row and column)
Multiplication uses three nested loops: r[i][j] = sum of m1[i][z] * m2[z][j] for z from 0 to 2 — don’t forget to reset r[i][j] = 0 before that inner sum
Use switch on the user’s choice to call the right function

🟢 Click to Show Solution

void scanMatrix(int arr[3][3]){
    for (int i = 0; i < 3; i++)
    {
        for (int j = 0; j < 3; j++)
        {
            scanf("%d",&arr[i][j]);
        }
    }
}
void printMatrix(int arr[3][3]){
    for (int i = 0; i < 3; i++)
    {
        for (int j = 0; j < 3; j++)
        {
            printf("%d ",arr[i][j]);
        }
        printf("\n");
    }
}
void addMatrix(int m1[3][3],int m2[3][3],int r[3][3]){
    for (int i = 0; i < 3; i++) // rows
    {
        for (int j = 0; j < 3; j++) //cols
        {
            r[i][j]=m1[i][j]+m2[i][j];
        }
    }
}
void subMatrix(int m1[3][3],int m2[3][3],int r[3][3]){
    for (int i = 0; i < 3; i++) // rows
    {
        for (int j = 0; j < 3; j++) //cols
        {
            r[i][j]=m1[i][j]-m2[i][j];
        }
    }
}
void transposeMatrix(int m1[3][3],int r[3][3]){
    for (int i = 0; i < 3; i++) // rows
    {
        for (int j = 0; j < 3; j++) //cols
        {
            r[i][j]=m1[j][i];
        }
    }
}
void multMatrix(int m1[3][3], int m2[3][3],int r[3][3]){
    for (int i = 0; i < 3; i++) // rows
    {
        for (int j = 0; j < 3; j++) //cols
        {
            r[i][j]=0;
            for (int z = 0; z < 3; z++)
            {
                r[i][j]=r[i][j]+m1[i][z] * m2[z][j];
            }
            
        }
    }
}

int main(){
    //   0 1 2
    // 0{1,2,3}  {1,1,1} = {2,3,4}   1*1+2*2+3*3=14                   1-->0,0   2-->0,1  3-->0,2     1-->0,0   2-->1,0  3-->2,0
    // 1{1,2,3}  {2,2,2} = {3,4,5}
    // 2{1,2,3}  {3,3,3} = {4,5,6}
    //    m1       m2   =    r


    int m1[3][3];
    int m2[3][3];
    int r[3][3];
    int n;
    printf("For add press 1\n");
    printf("For sub press 2\n");
    printf("For mul press 3\n");
    printf("For transpose press 4\n");
    scanf("%d",&n); //2
    switch (n)
    {
    case 1:
        scanMatrix(m1);
        scanMatrix(m2);
        addMatrix(m1,m2,r);
        printMatrix(r);
        break;
    case 2:
        scanMatrix(m1);
        scanMatrix(m2);
        subMatrix(m1,m2,r);
        printMatrix(r);
        break;
    case 3:
        scanMatrix(m1);
        scanMatrix(m2);
        multMatrix(m1,m2,r);
        printMatrix(r);
        break;
    case 4:
        scanMatrix(m1);
        transposeMatrix(m1,r);
        printMatrix(r);
        break;
    default:
        printf("ERROR CHOICE!");
        break;
    }
    return 0;
}

Exercise 17: Pythagorean Triples

Task: A right triangle can have sides that are all integers. The set of three integer values for the sides of a right triangle is called a Pythagorean triple. These three sides must satisfy the relationship that the sum of the squares of two of the sides is equal to the square of the hypotenuse. Write a program to find all Pythagorean triples for side1, side2, and the hypotenuse all no larger than 30. Use a triple-nested for loop that simply tries all possibilities.

💡 Hint

Relationship: hyp² = side1² + side2²
Three nested loops: i, j, k each going from 1 to 30
Inside the innermost loop check if (k*k == i*i + j*j) then print the triple
Total iterations: 30 × 30 × 30 = 27000 — brute force, but it works

🟢 Click to Show Solution

// 3 3 6
// h = root(x^2 + y^2)
// h^2=x^2+y^2
// 18 =3^2 +3^2;
// 1 30
// 1 1 1
// 1 1 2
// 1 1 3
// .
// .
// 1 1 30
// 1 2 1
// 1 2 2

#include <math.h>
int main(){
    for (int i = 1; i <= 30; i++) // loop of side1
    {
        for (int j = 1; j <= 30; j++) // loop of side2
        {
            for (int k = 1; k <= 30; k++) // loop of hypotnous
            {
                if (k*k==(i*i)+(j*j))
                {
                    printf("s1:%d s2:%d hyp:%d is pythegorean triples\n",i,j,k);
                }        
            } 
        } 
    }   
    // number of total loops = 30 * 30 * 30 = 27000
    return 0;
}

Exercise 18: Maximum Element of Each Row

Task: Find the maximum element of each row in a matrix.

💡 Hint

For each row, reset max to arr[i][0] at the start of that row — don’t reuse the old row’s max!
Then loop through the columns of that row comparing against max
Print the max after each row’s inner loop finishes (not after the whole matrix)

🟢 Click to Show Solution

int main(){

    int arr[3][5]=
    {
        {10,2,3,4,4},
        {1,2,1,3,3},
        {3,2,5,7,8}
    };

    for (int i = 0; i < 3; i++) // rows
    {
        int max=arr[i][0];
        for (int j = 0; j < 5; j++) // cols
        {
            if (arr[i][j]>max)
            {
                max=arr[i][j];
            } 
        }
        printf("Max element in row number %d equal %d\n",i+1,max);
    }
    return 0;
}

Exercise 19: Sum of All Elements in a 4×3 Matrix

Task: Write a program that initializes a 4×3 matrix of integers with predetermined values and then computes the sum of all elements in the matrix.

💡 Hint

Initialize the 4×3 matrix directly in the declaration using nested {} braces
Classic nested loop: outer loop over 4 rows, inner loop over 3 columns
Keep a single sum variable and add arr[i][j] each iteration

🟢 Click to Show Solution

int main(){
    
    int arr[4][3]=
    {
        {1,2,2},
        {1,2,1},
        {3,2,5},
        {3,2,5}
    };
    int sum=0;
    for (int i = 0; i < 4; i++)
    {
        for (int j = 0; j < 3; j++)
        {
            sum=sum+arr[i][j];
        } 
    }
    printf("sum= %d",sum);
    
    return 0;
}

Exercise 20: Check If Two Matrices Are Equal

Task: Write a C program that checks if 2 matrices are equal or not.

💡 Hint

Two matrices are equal only if every corresponding element is the same
Write an isEqualMatrix function — loop through both matrices, and the moment you find any mismatch, return 0 immediately (early exit)
If the whole loop finishes without a mismatch, return 1
In main, call the function and print the result based on the return value

🟢 Click to Show Solution

void scanMatrix(int m1[3][3]){
    for (int i = 0; i < 3; i++)
    {
        for (int j = 0; j < 3; j++)
        {
            scanf("%d",&m1[i][j]);
        }
    }
}

int isEqualMatrix(int m1[3][3], int m2[3][3]){
    for (int i = 0; i < 3; i++)
    {
        for (int j = 0; j < 3; j++)
        {
            if (m1[i][j]!=m2[i][j])
            {
                return 0;
            }  
        }
    }
    return 1;
}

int main(){

    int m1[3][3];
    int m2[3][3];
    scanMatrix(m1);
    scanMatrix(m2);

    if(isEqualMatrix(m1,m2)==1){
        printf("EQUAL MATRIX");
    }
    else
    {
        printf("NOT EQUAL MATRIX");
    }
    

    return 0;
}

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