38613220 C Pattern Codes

8/3/2019 38613220 C Pattern Codes

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Number problems in C# Write C program to print the following pattern:

1 22 333 4444 55555

Program: view source print? 01 02 03 04 05 06 07 08 09 10 11 12 13 } for (j = 1;j <= i; j++) { /* %2d ensures that the number is printed in two spaces for alig

nment and the numbers are printed in the order. */ printf("%2d", i); #include<stdio.h> int main() { int i, j, k, c = 5; for (i = 1; i <= 5; i++) { /* k is takenfor spaces */ for (k = 1; k <= c; k++) { /* blank space */ printf(" ");

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14 15 16 17 18 19 20 } }

} printf("\n"); /*c is decremented by 1 */ c--;

return 0;

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Output:

1 22 333 4444 55555

Explanation: Here ‘i’ loop is used for printing the numbers in the respective rows and ‘k’ loop is used for providing spaces. ‘j’ loop prints the numbers. ‘c’ is decrementedr numbers to be displayed in alternate columns.

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# Write C program to print the following pattern:

1

8/3/2019 38613220 C Pattern Codes


121 12321 1234321 123454321

Program: view source print? 01 02 03 04 05 06 07 08 09 } printf("%2d", j); } #include<stdio.h> int main() { /* c taken for columns */ int i, j, c = 9, m, k; for(i = 1; i <= 5; i++) { /* k is used for spaces */ for (k = 1; k <= c; k++) { printf(" ");

10 for (j = 1; j <= i; j++) { for (m = j - 2; m > 0; m--) { 11 12 13 14 15 16 1718 } printf("\n"); /* c is decremented by 2 */ c = c - 2;

/* %2d ensures that the number * is printed in two spaces * for alignment */ printf("%2d", m);

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19 20 21 }

} return 0;

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Output:

1 121 12321 1234321 123454321

Explanation: Here ‘i’ loop is used for printing numbers in rows and ‘k’ loop is used forproviding spaces. ‘j’ loop is used for printing numbers in increasing order. ‘m’ loop is used for printing numbers in reverse order.

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# Write a C program to display the following format:

-----ab -----15 24

8/3/2019 38613220 C Pattern Codes


33 42 51 ------

Program: view source print? 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 } printf("----------"); #include<stdio.h> int main() { int i = 1, j = 5; printf("----------\n"); printf("a \t b \n"); printf("----------\n"); /* logic: whileloop repeats * 5 times i.e. until * the condition i<=5 fails */ while (i <= 5) {/* i and j value printed */ printf("%d \t %d\n", i, j); /* i and j value increm

ented by 1 for every iteration */ i++; j--;

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19 20 }

return 0;

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Output:

-----ab -----15 24 33 42 51 ------

Explanation: Here, ‘i’ is initialized to least value 1 and ‘j’ initialized to highest value 5. We keep incrementing the i’ value and decrementing the ‘j’ value until the condition fails. The value is displayed at each increment and at each decrement. Back to top

# Write a C program to display the following format:

-------no. sum --------

8/3/2019 38613220 C Pattern Codes


1 1 2 3 3 6 4 10 5 15 --------

Program: view source print? 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 #include<stdio.h> int main() { int num = 1, sum = 0; printf("-----------\n"); printf("num \t sum\n"); printf("-----------\n"); /* while loop repeats 5 times * i.e.until the condition * num <= 5 fails */ while (num <= 5) { sum = sum + num; printf("%d \t %d\n", num, sum); /* num incremented by 1 * for every time * the loop

is executed */ num++;

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17 18 19 20 }

} printf("-----------"); return 0;

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Output:

-------no. sum -------1 1 2 3 3 6 4 10 5 15 --------

Explanation: In the above program we have taken two variables ‘num’ and ‘sum’. ‘num’ is usto check the condition and to display the numbers up to 5. ‘sum’ is used to add thenumbers which are displayed using variable ‘num’. The ‘sum’ value is initialized to zero. sum is added to the numbers which are incremented by ‘i’ and displayed.

10 most challanging pattern problems in C1. Write a C program to print the following pattern:

8/3/2019 38613220 C Pattern Codes


* * * * * * * * * *

2. Write a C program to print the following pattern:

* ***

* ***

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


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

*

*

*

*

*


8/3/2019 38613220 C Pattern Codes


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


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


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


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

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


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

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***

***

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


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

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

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


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

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


* * * * * * * * * *

Program: view source print? 01/* This is a simple mirror-image of a right angletriangle */ 02 03#include <stdio.h> 04int main() { 05 char prnt =

*

;

8/3/2019 38613220 C Pattern Codes


06 int i, j, nos = 4, s; 07 for (i = 1; i <= 5; i++) { 08for (s = nos; s >= 1; s--) { // Spacing factor 09 printf(" ");

10 } 11 for (j = 1; j <= i; j++) { 12 printf("%2c", prnt);

13 } 14 printf("\n"); 15 --nos; // Controls the spacing factor 16 } 17 return 0;18}

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2. Write C program to print the following pattern:

* ***

* ***

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

Program:

8/3/2019 38613220 C Pattern Codes


view source print? 01#include<stdio.h> 02int main() { 03 char prnt =

*

; 04 inti, j, k, s, c = 1, nos = 9; 05 for (i = 1; c <= 4; i++) { 06 // As we want to print the columns in odd sequence viz. 1,3,5,.etc

07 if ((i % 2) != 0) { 08 for (j = 1; j <= i; j++) {

09 printf("%2c", prnt); 10} 11for (s = nos; s >= 1; s--) { //The spacing factor

12 13 14 15 16 17 18 19 20 21 22 23 24 if (c == 4 && s == 1) { break; } printf(""); } for (k = 1; k <= i; k++) { if (c == 4 && k == 5) { break; } printf("%2c",prnt); } printf("\n"); nos = nos - 4; // controls the spacing factor

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25

++c;

26 } 27 } 28 return 0; 29}

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3. Write C program to print the following pattern:

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

*

*

*

*

*

Program: view source print? 01#include<stdio.h>

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02int main() { 03 char prnt =

*

; 04 int i, j, k, s, p, r, nos = 7; 05 06 for (i = 1; i <= 5; i++) { 07 for (j = 1; j <= i; j++) { 08 if ((i % 2) != 0 && (j %2) != 0) { 09printf("%3c", prnt); 10} 11else if ((i % 2) == 0 && (j % 2) == 0) {12printf("%3c", prnt); 13} 14else { 15printf(" "); 16} 17} 18for (s = nos; s >=1; s--) { // for the spacing factor 19 printf(" ");

20 } 21 for (k = 1; k <= i; k++) { //Joining seperate figures 22if (i == 5 && k

== 1) { 23 continue; 24} 25if ((k % 2) != 0) { 26printf("%3c", prnt); 27}

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28else { 29printf(" "); 30} 31} 32printf("\n"); 33nos = nos - 2; // space control 34} nos = 1; // remaining half.. 35for (p = 4; p >= 1; p--) { 36 for (r = 1; r<= p; r++) { 37if ((p % 2) != 0 && (r % 2) != 0) { 38printf("%3c", prnt); 39} 40else if ((p % 2) == 0 && (r % 2) == 0) { 41printf("%3c", prnt); 42} 43else { 44printf(" "); 45} 46} 47for (s = nos; s >= 1; s--) { 48 printf(" ");

49 } 50 for (k = 1; k <= p; k++) { 51 52 53 if ((k % 2) != 0) { printf("%3c", pr

nt); }

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54else { 55 56 printf(" "); }

57 } 58 nos = nos + 2; // space control 59 printf("\n"); 60 } 61 return 0; 62}

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Explanation: This can be seen as an inverted diamond composed of stars. It can b

e noted that the composition of this figure follows sequential pattern of consecutive stars and spaces.

In case of odd row number, the odd column positions will be filled up with ‘*’, elsea space will be spaced and vice-versa in case of even numbered row.

In order to achieve this we will construct four different right angle triangles

aligned as per the requirement. Back to top


8/3/2019 38613220 C Pattern Codes


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

Program: view source print? 01#include<stdio.h> 02int main() { 03 char prnt =

*

; 04 int i, j, s, nos = 0; 05 for (i = 9; i >= 1; (i = i - 2)) { 06 for (s = nos; s >= 1; s--) { 07 printf(" ");

08 } 09 for (j = 1; j <= i; j++) { 10 11 12 13 if ((i % 2) != 0 && (j % 2) != 0)

{ printf("%2c", prnt); } else { printf(" ");

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14

}

15 } 16 printf("\n"); 17 nos++; 18 } 19 nos = 3; 20 for (i = 3; i <= 9; (i = i +2)) { 21for (s = nos; s >= 1; s--) { 22 printf(" ");

23 } 24 for (j = 1; j <= i; j++) { 25 26 27 28 29 30 if ((i % 2) != 0 && (j % 2)!= 0) { printf("%2c", prnt); } else { printf(" "); }

31 } 32 nos--; 33 printf("\n"); 34 } 35 return 0; 36}

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8/3/2019 38613220 C Pattern Codes



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

Program: view source print? 01#include<stdio.h> 02int main() { 03 char prnt =

*

; 04 int i, j, k, s, nos = 4; 05 for (i = 1; i <= 5; i++) { 06for (s = nos; s >= 1; s--) { 07 printf(" ");

08 }

8/3/2019 38613220 C Pattern Codes


09 for (j = 1; j <= i; j++) { 10 printf("%2c", prnt);

11 } 12 for (k = 1; k <= (i - 1); k++) { 13if (i == 1) { 14} 15printf("%2c", prnt); 16} 17 printf("\n"); nos--; 18} 19 nos = 1; 20for (i = 4; i >= 1; i--) { 21for (s = nos; s >= 1; s--) { 22 printf(" "); continue;


26 } 27 for (k = 1; k <= (i - 1); k++) { 28 printf("%2c", prnt);

29 } 30 nos++; 31 printf("\n"); 32 } 33 nos = 3; 34 for (i = 2; i <= 5; i++) {

8/3/2019 38613220 C Pattern Codes


35if ((i % 2) != 0) { 36for (s = nos; s >= 1; s--) { 37 38 39 40 41 printf(" ");} for (j = 1; j <= i; j++) { printf("%2c", prnt); }

42 } 43 if ((i % 2) != 0) { 44 45 printf("\n"); nos--;

46 } 47 } 48 return 0; 49}



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

8/3/2019 38613220 C Pattern Codes


** *

Program: view source print? 01/* 02 03 04*/ 05#include <stdio.h> 06// This function controls the inner loop and the spacing 07// factor guided by the outer loopindex and the spacing index. 08int triangle(int nos, int i) { 09 char prnt =

*

; 10 int s, j; 11 for (s = nos; s >= 1; s--) { 12 printf(" "); 13 } 14 for (j =1; j <= i; j++) { 15 printf("%2c", prnt); 16 } 17 return 0; 18} 19 20int main()

{ //The inner loop // Spacing factor This can be seen as two right angle triangles sharing the same base which is modified by adding few extra shifting spaces

8/3/2019 38613220 C Pattern Codes


21 int i, nos = 5; 22 //draws the upper triangle 23 for (i = 1; i <= 4; i++) { 24 triangle(nos, i); 25 nos++; 26 printf("\n"); } 27nos = 7; //Draws the lower triangle skipping its base. 28for (i = 3; i >= 1; i--) { 29 int j = 1; 30 triangle(nos, i); // Inner loop construction 31 nos = nos - j; 32 printf("\n"); 33 } 34return 0; 35} // Spacing factor //Inner loop construction // Increments the spacing factor



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

8/3/2019 38613220 C Pattern Codes


** ***

** ***

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

Program: view source print? 01#include <stdio.h> 02 03int main() { 04 char prnt

=

*

; 05 int i, j, k, s, nos = -1; 06 for (i = 5; i >= 1; i--) { 07 for (j = 1;j <= i; j++) { 08 printf("%2c", prnt); 09} 10for (s = nos; s >= 1; s--) { 11 printf(" ");

12 } 13 for (k = 1; k <= i; k++) { 14 15 16 17 if (i == 5 && k == 5) { continue;} printf("%2c", prnt);

18 }

8/3/2019 38613220 C Pattern Codes


19 nos = nos + 2; 20 printf("\n"); 21 } 22 nos = 5; 23 for (i = 2; i <= 5; i++){ 24 for (j = 1; j <= i; j++) { 25 printf("%2c", prnt); 26} 27for (s = nos; s >=1; s--) { 28 printf(" ");

29 } 30 for (k = 1; k <= i; k++) { 31 32 33 34 if (i == 5 && k == 5) { break; }printf("%2c", prnt);

35 } 36 nos = nos - 2; 37 printf("\n"); 38 } 39 return 0; 40}


8/3/2019 38613220 C Pattern Codes



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

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

Program: view source print? 01#include <stdio.h> 02int main() { 03 char prnt =

*

; 04 int i, j, k, s, sp, nos = 0, nosp = -1; 05 for (i = 9; i >= 3; (i = i - 2)) { 06 for (s = nos; s >= 1; s--) { 07 printf(" ");

08 } 09 for (j = 1; j <= i; j++) { 10printf("%2c", prnt); 11}

8/3/2019 38613220 C Pattern Codes


12for (sp = nosp; sp >= 1; sp--) { 13 printf(" ");

14 } 15 for (k = 1; k <= i; k++) { 16 if (i == 9 && k == 1) { 17continue; 18} 19printf("%2c", prnt); 20} 21nos++; 22nosp = nosp + 2; 23printf("\n"); 24} 25nos =4; 26for (i = 9; i >= 1; (i = i - 2)) { 27 for (s = nos; s >= 1; s--) { 28 printf(" ");


32 } 33 nos++; 34 printf("\n"); 35 } 36 37 return 0;

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38}



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

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

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

Program: view source print?

8/3/2019 38613220 C Pattern Codes


01#include <stdio.h> 02/* 03 * nos = Num. of spaces required in the triangle. 04* i = Counter for the num. of charcters to print in each row 05 * skip= A flagfor checking whether to 06 * 07 * 08 */ 09int triangle(int nos, int i, int skip){ 10 char prnt =

*

; 11 int s, j; 12 for (s = nos; s >= 1; s--) { 13 printf(""); 14 } 15 for (j = 1; j <= i; j++) { 16 if (skip != 0) { 17 18 19 if (i == 4 && j == 1) { continue; } skip a character in a row.

20 } 21 printf("%2c", prnt); 22 } 23 return 0; 24} 25 26int main() {

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27 int i, nos = 4; 28 for (i = 1; i <= 7; (i = i + 2)) { 29 triangle(nos, i, 0);30 nos--; 31 printf("\n"); 32 } 33 nos = 5; 34 for (i = 1; i <= 4; i++) { 35triangle(1, i, 0); //one space needed in each case of the formation 36triangle(nos,i, 1); //skip printing one star in the last row. 37nos = nos - 2; 38printf("\n"); 39} 40nos = 1; 41for (i = 3; i >= 1; i--) { 42 triangle(1, i, 0); 43 triangle(nos, i, 0); 44 nos = nos + 2; 45 printf("\n"); 46 } 47 nos = 1; 48 for (i = 7;i >= 1; (i = i - 2)) { 49 triangle(nos, i, 0); 50 nos++; 51 printf("\n"); 52 }

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53 return 0; 54}



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

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

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

Program: view source print? 01#include <stdio.h> 02 03/* 04 * nos = Num. of spaces required in the triangle. 05 * i = Counter for the num. of characters to print in each row

8/3/2019 38613220 C Pattern Codes


06 * skip= A flag for check whether to 07 * 08 * 09 */ 10 11int triangle(int nos, int i, int skip) { 12 char prnt =

*

; 13 int s, j; 14 for (s = nos; s >= 1; s--) { 15 printf(" "); 16 } 17 for (j = 1; j <= i; j++) { 18 if (skip != 0) { 19if (i == 9 && j == 1) { 20 continue; skip a character in a row.

21 } 22 } 23if (i == 1 i == 9) { 24 printf("%2c", prnt); 25} 26else if (j ==1 j == i) { 27 printf("%2c", prnt); 28 } else { 29 printf(" "); 30} } 31retur

n 0; }

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32int main() { 33int i, nos = 0, nosp = -1, nbsp = -1; 34for (i = 9; i >= 1; (i= i - 2)) { 35 triangle(nos, i, 0); 36 triangle(nosp, i, 1); 37 triangle(nbsp, i, 1); 38 printf("\n"); 39 nos++; 40 nosp = nosp + 2; 41 nbsp = nbsp + 2; 42 } 43nos = 3, nosp = 5, nbsp = 5; 44 for (i = 3; i <= 9; (i = i + 2)) { 45 triangle(nos, i, 0); 46 triangle(nosp, i, 1); 47 triangle(nbsp, i, 1); 48 printf("\n"); 49 nos--; 50 nosp = nosp - 2; 51 nbsp = nbsp - 2; 52 } 53 return 0; 54}

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Number pattern programs# Write a C program to print the following pattern:

1 01 101 0101 10101

Program: view source print? 01 02 03 04 05 06 07 08 09 10 11 12 13 14 int main(void) { int i, j; for (i = 0; i < 4; i++) { for (j = 0; j <= i; j++) { if (((i +

j) % 2) == 0) { // Decides on as to which digit to print. printf("0"); } else {printf("1"); } printf("\t"); } printf("\n"); #include <stdio.h>

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15 16 17

} return 0; }

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Explanation: This is a right angle triangle composed of 0′s and 1′s. Back to top

End of Question1 Start of Question2 # Write C program to print the following pattern:

0 11 235 8 13 21

Program: view source print? 01 02 03 04 05 06 int main(void) { int i, j, a = 0,b = 1, temp = 1; for (i = 1; i <= 4; i++) { for (j = 1; j <= i; j++) { #include<stdio.h>

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07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23

if (i == 1 && j == 1) { // Prints the

0

individually first printf("0"); continue; } printf("%d ", temp); // Prints the next digit in the series //Computes theseries temp = a + b; a = b; b = temp; if (i == 4 && j == 3) { // Skips the 4thcharacter of the base break; } } printf("\n"); } return 0; }

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Explanation: This prints the Fibonacci series in a right angle triangle formation where the base has only three characters. Back to top

End of Question2 Start of Question3 # Write C program to print the following pattern:

8/3/2019 38613220 C Pattern Codes


1 121 12321 1234321 12321 121 1

Program: view source print? 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 void sequence(int x); int main() { /* c taken for columns */ int i, x = 0, num = 7; for(i = 1; i <= num; i++) { if (i <= (num / 2) + 1) { x = i; } else { x = 8 - i; }sequence(x); puts("\n"); } #include <stdio.h>

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16 17 18 19 20 21 22 23 24 25 26 27 28

return 0; }

void sequence(int x) { int j;

for (j = 1; j < x; j++) { printf("%d", j); } for (j = x; j > 0; j--) { printf("%

d", j); } }


End of Question3 Start of Question4 # Write a C program to print the following pattern:

2 456 6 7 8 9 10 456 2

8/3/2019 38613220 C Pattern Codes


Program: view source print? 01 02 03 04 int main(void) { int prnt; #include <stdio.h>

05 int i, j, k, r, s, sp, nos = 3, nosp = 2; //nos n nosp controls the spacing factor 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 // Prints the uppertriangle for (i = 1; i <= 5; i++) { if ((i % 2) != 0) { for (s = nos; s >= 1; s--) { printf(" "); } for (j = 1; j <= i; j++) { if (i == 5 && j == 5) { //Provide

s the extra space reqd betn 9 n 10 printf(" "); } prnt = i + j; printf("%2d", prnt); } } if ((i % 2) != 0) { printf("\n"); nos--; } // as 10 is a 2 digit no.

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24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

} // Prints the lower triangle skipin its base.. for (k = 3; k >= 1; k--) { if ((k % 2) != 0) { for (sp = nosp; sp >= 1; sp--) { printf(" "); } for (r = 1; r <=k; r++) { prnt = k + r; printf("%2d", prnt); } } if ((k % 2) != 0) { printf("\n"); nosp++; } } return 0; }

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Explanation: This is a diamond formation composed of numbers. The numbers are inthe following order next_no=i+j where next_no = The next no to be printed i = index of the outer for loop j = index of the inner for loop

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Back to top


1 333 55555

1 333 55555

7777777 7777777 55555 333 1 55555 333 1

Program: view source print? 01 02 03 04 05 06 07 08 09 10 int main(void) { int i, j, k, s, p, q, sp, r, c = 1, nos = 13; for (i = 1; c <= 4; i++) { if ((i % 2)!= 0) { // Filters out the even line nos. for (j = 1; j <= i; j++) { // The upper left triangle printf("%2d", i); } for (s = nos; s >= 1; s--) { // The spacingfactor #include <stdio.h>

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11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

printf(" "); } for (k = 1; k <= i; k++) { // The upper right triangle printf("%2d", i); } printf("\n"); nos = nos - 4; // Space control ++c; } } nos = 10; // Space control re intialized c = 1; for (p = 5; (c < 4 && p != 0); p--) { if ((p %2) != 0) { // Filters out the even row nos for (q = 1; q <= p; q++) { // Lower left triangle printf("%2d", p); } for (sp = nos; sp >= 1; sp--) { // Spacing fact

or printf(" "); } for (r = 1; r <= p; r++) { // Lower right triangle printf("%2d", p); }

printf("\n"); --c;

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37 38 39 40 41 42

nos = nos + 8; // Spacing control. } }

return 0; }

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Explanation: Here we are printing only the odd row nos along with thier respective line number. This structure can divided into four identical right angle triangles which are kind of twisted and turned placed in a particular format . Back to top


0 -2-3 0 -4-3-2-1 0 -2-3 0 0

Program: view source print?

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01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

#include <stdio.h>

int main(void) { int i, j, k, r, s, sp, nos = 2, nosp = 1; for (i = 1; i <= 5; i++) { if ((i % 2) != 0) { for (s = nos; s >= 1; s--) { //for the spacing factor.printf(" "); } for (j = 1; j <= i; j++) { printf("%2d", j-i); } } if ((i % 2) !

= 0) { printf("\n"); nos--; } } for (k = 3; k >= 1; k--) { if ((k % 2) != 0) { for (sp = nosp; sp >= 1; sp--) { // for the spacing factor. printf(" "); } for (r= 1; r <= k; r++) { printf("%2d", r-k); }

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27 28 29 30 31 32 33 34

} if ((k % 2) != 0) { printf("\n"); nosp++; } } return 0; }

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Explanation:This can be seen as a diamond composed of numbers. If we use the con

ventional nested for loop for its construction the numbers can be seen to flowing the following function f(x) -> j-i where j= inner loop index i= outer loop index Back to top


77777777777 7 7 7 7

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7 7 7 7 7 7

Program: view source print? 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 intmain(void) { int i, j; for (i = 11; i >= 1; i--) { for (j = 1; j <= i; j++) { if(i == 11) { printf("7"); // Makes sure the base is printed completely continue;} else if (j == i) { // Hollows the rest printf("7"); } else { printf(" "); } }printf("\n"); #include <stdio.h>

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17 18 19

} return 0; }

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Explanation: This can be seen as a hollow right-angled triangle composed of 7′s Ba

ck to top


1 10 101 1010 10101

1 10 101 1010 10101

101010 101010 1010101010101 101010 101010 10101 1010 101 10 1 10101 1010 101 101

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Program: view source print? 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 } if ((k%2)!=0) { // Applying the condition printf(" 1"); } else { } for (k=1; k<=i; k++) { if(i==7 && k==1) // Skipping the extra 1 { continue; } for (s=nos; s>=1; s--) { // Space factor printf(" "); } int main(void) { inti,j,k,s,nos=11; for (i=1; i<=7; i++) { for (j=1; j<=i; j++) { if ((j%2)!=0) { // Applying the condition printf(" 1"); } else { printf(" 0"); #include <stdio.h>

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24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 }} } } } } } }

printf(" 0");

printf("\n"); nos=nos-2; // Space Control

nos=1; for ( i=6; i>=1; i--) { // It shares the same base for (j=1; j<=i; j++) {if (j%2!=0) { printf(" 1"); } else { printf(" 0");

for(s=nos; s>=1; s--) // Spacing factor { printf(" ");

for (k=1; k<=i; k++) { if (k%2!=0) { printf(" 1"); } else { printf(" 0");

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50 51 52 53 54 } }

printf("\n"); nos=nos+2;

return 0;



1 24 369 24 1

Program: view source print? 01 02 03 04 05 #include <stdio.h> int main(void) { int i,j; for (i=1; i<=3 ; i++) { for (j=1; j<=i; j++) {

8/3/2019 38613220 C Pattern Codes


06 07 08 09 10 11 12 13 14 15 16 17 } } } } }

printf("%2d", (i*j));

printf("\n");

for (i=2; i>=1; i--) { // As they share the same base for (j=1; j<=i; j++) { pri

ntf("%2d",i*j);

printf("\n");

return 0;

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Explanation: This can be seen as two right angle triangles sharing th same baseThe numbers are following the following function f(x) = i *j where i = Index ofthe Outer loop j = Index of the inner loop Back to top

End of Question9 Start of Question10 # Write a C program to print the followingpattern:

1

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10 100 1000 10000 100000 1000000 100000 10000 1000 100 10 1

Program: view source print? 01 02 03 04 05 06 07 08 09 10 int main(void) { int i,j; for (i=1; i<=7; i++) { for (j=1; j<=i; j++) { if (j==1) { printf(" 1"); } else { printf(" 0"); // Applying the condition #include <stdio.h>

8/3/2019 38613220 C Pattern Codes


11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 } } } } }

}

printf("\n");

for (i=6; i>=1; i--) { //As it shares the same base i=6 for (j=1; j<=i; j++) { i

f (j==1) { // Applying the condition printf(" 1"); } else { printf(" 0"); }

printf("\n");

return 0;

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Explanation: This can be seen as two right angle triangles sharing the same basewhich is composed of 0′s n 1′s. The first column is filled with 1′s and rest with 0′s

8/3/2019 38613220 C Pattern Codes


8/3/2019 38613220 C Pattern Codes


38613220 C Pattern Codes

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