C Basics

Master For Loops in C: Pattern Programs and Practical Logic Problems

In the previous section we discussed the basics of C for loops. Now let us take a look at some real-life application Problems.

1. Write a C Program to print the following pattern

*****

****

***

**

*

				
					#include <stdio.h>

int main() {
    int n = 5;

    for(int i = n; i >= 1; i--) {
        for(int j = 1; j <= i; j++) {
            printf("*");
        }
        printf("\n");
    }
    return 0;
}

// *****
// ****
// ***
// **
// *

Row Iteration (i)	Condition (i >= 1)	Inner Loop (j)	Inner Condition (j <= i)	Row Output
i = 5	True	1, 2, 3, 4, 5	1 <= 5 .........5 <= 5	*****
i = 4	True	1, 2, 3, 4	1 <= 4..........4 <= 4	****
i = 3	True	1, 2, 3	1 <= 3..........3 <= 3	***
i = 2	True	1, 2	1 <= 2..........2 <= 2	**
i = 1	True	1	1 <= 1	*
i = 0	False	-	-	Terminate

2. Write a C Program to print the Floyd’s Triangle

2 3

4 5 6

7 8 9 10

				
					#include <stdio.h>

int main() {
    int n = 4, num = 1;

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

// 1 
// 2 3 
// 4 5 6 
// 7 8 9 10

Row (i)	Inner (j)	Condition (j <= i)	num++	Row Output
1	1	1 <= 1	Print 1 then num becomes 2	1
2	1, 2	1 <= 2, 2<= 2	Print 2, 3 then num becomes 4	2, 3
3	1, 2, 3	1 <= 3,......3 <= 3	Print 4, 5, 6 then num becomes 7	4, 5, 6
4	1, 2, 3, 4	1 <= 4,......4 <= 4	Print 7, 8, 9, 10 then num becomes 11	7, 8, 9, 10

3. Write a C Program to print the pattern below

A B

A B C

A B C D

				
					#include <stdio.h>

int main() {
    int n = 4;

    for(int i = 1; i <= n; i++) {
        char ch = 'A';
        for(int j = 1; j <= i; j++) {
            printf("%c ", ch++);
        }
        printf("\n");
    }
    return 0;
}

// A 
// A B 
// A B C 
// A B C D

Row (i)	Inner (j)	ch Value	Action	Row Output
1	1	'A'	Print 'A', ch becomes 'B'	A
	2	'B'	2 <= 1 is False -> Reset ch
2	1	'A'	Print 'A', ch becomes 'B'	A B
	2	'B'	Print 'B', ch becomes 'C'
3	1	'A'	Print 'A', ch becomes 'B'	A B C
	2	'B'	Print 'B', ch becomes 'C'
	3	'C'	Print 'C', ch becomes 'D'
4	1	'A'	Print 'A', ch becomes 'B'	A B C D
	2	'B'	Print 'B', ch becomes 'C'
	3	'C'	Print 'C', ch becomes 'D'
	4	'D'	Print 'D', ch becomes 'E'

4. Write a C Program to print the below pattern

***

****

*****

				
					#include <stdio.h>

int main() {
    int n = 5;

    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= i; j++) {
            printf("*");
        }
        printf("\n");
    }
    return 0;
}

/ *
// **
// ***
// ****
// *****

Row (i)	Condition (i <= 5)	Inner (j)	Condition (j <= i)	Row Output
1	True	1	1 < = 1	*
2	True	1, 2	1 <= 2, 2 <= 2	**
3	True	1, 2, 3	1 <= 3, ......3 <= 3	***
4	True	1, 2, 3, 4	1 <= 4, ...... 4 <= 4	****
5	True	1, 2, 3, 4, 5	1 <= 5, .......5 <= 5	*****
6	False	-	-

5. Read n integers and count how many are positive, negative and zero.

				
					#include <stdio.h>

int main() {
    int n, num;
    int pos = 0, neg = 0, zero = 0;

    printf("Enter number of elements: ");
    scanf("%d", &n);

    printf("Enter several numbers including positive, negative and zero :");
    for(int i = 1; i <= n; i++) {
        scanf("%d", &num);

        if(num > 0)
            pos++;
        else if(num < 0)
            neg++;
        else
            zero++;
    }

    printf("Positive: %d\nNegative: %d\nZero: %d\n", pos, neg, zero);
    return 0;
}

// Enter number of elements: 5
// Enter several numbers including positive, negative and zero :5 -8 6 0 -7
//  Positive: 2
// Negative: 2
// Zero: 1

6. Write a C Program to print all prime numbers in a given range

				
					#include <stdio.h>
int main() {
    int a, b, flag;

    printf("Enter range: ");
    scanf("%d %d", &a, &b);

    for(int num = a; num <= b; num++) {
        if(num < 2) continue;

        flag = 0;
        for(int i = 2; i <= num / 2; i++) {
            if(num % i == 0) {
                flag = 1;
                break;
            }
        }

        if(flag == 0)
            printf("%d ", num);
    }
    return 0;
}

// Enter range: 5    80
// 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79

num	num / 2	Inner Loop (i)	Divisibility (num % i)	flag	Result
5	2	2	5 mod 2 = 1	0	Prime (Print 5)
6	3	2	6 mod 2 = 0	1	Not Prime (Break)
7	3	2, 3	7 mod 2 = 1, 7 7 mod 3 = 1	0	Prime (Print 7)
8	4	2	8 mod 2 = 0	1	Not Prime (Break)
9	4	2, 3	9 mod 2 = 1 9 mod 3 = 0	1	Not Prime (Break)
10	5	2	10 mod 2 = 0	1	Not Prime (Break)
11	5	2, 3, 4, 5	No divisors equal 0	0	Prime (Print 11)
12	6	2	12 mod 2 = 0	1	Not Prime (Break)
13	6	2, 3, 4, 5, 6	No divisors equal 0	0	Prime (Print 13)
14	7	2	14 mod 2 = 0	1	Not Prime (Break)

7. Write a C Program to print the multiplication table of the entered number.

				
					#include <stdio.h>
int main() {
    int num;

    printf("Enter the number: ");
    scanf("%d", &num);

    printf("\nMultiplication Table of %d\n", num);
    for(int j = 1; j <= 10; j++) {
        printf("%d x %d = %d\n", num, j, num * j);
    }
    
    return 0;
}

// Enter the number: 5

// Multiplication Table of 5
// 5 x 1 = 5
// 5 x 2 = 10
// 5 x 3 = 15
// 5 x 4 = 20
// 5 x 5 = 25
// 5 x 6 = 30
// 5 x 7 = 35
// 5 x 8 = 40
// 5 x 9 = 45
// 5 x 10 = 50