Write a C program to print the following pattern: 1 0 1 1 0 1 0 1 0 1 1 0 1 0 1 Program: view source print ? 01 #include <stdio.h> 02 03 int main(void) { 04 int i, j; 05 for (i = 0; i < 4; i++) { 06 for (j = 0; j <= i; j++) { 07 if (((i + j) % 2) == 0) { // Decides on as to which digit to print. 08 printf("0"); 09 } else { 10 printf("1"); 11 } 12 printf("\t"); 13 } 14 printf("\n"); 15 } 16 return 0; 17 } Download Code Explanation: This is a right angle triangle composed of 0′s and 1′s. Back to top
Transcript
Write a C program to print the following pattern: 10 11 0 10 1 0 11 0 1 0 1
Program: view sourceprint ? 01 #include <stdio.h>
02
03 int main(void) {
04 int i, j;
05 for (i = 0; i < 4; i++) {
06 for (j = 0; j <= i; j++) {07 if (((i + j) % 2) == 0) { // Decides on as to which digit to print.
08 printf("0");
09 } else {
10 printf("1");
11 }
12 printf("\t");
13 }
14 printf("\n");
15 }
16 return 0;17 }
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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: 01 12 3 58 13 21
Program:
view sourceprint ? 01 #include <stdio.h>
02
03 int main(void) {
04 int i, j, a = 0, b = 1, temp = 1;
05 for (i = 1; i <= 4; i++) {
06 for (j = 1; j <= i; j++) {07 if (i == 1 && j == 1) { // Prints the '0' individually first
08 printf("0");
09 continue;
10 }
11 printf("%d ", temp); // Prints the next digit in the series
12 //Computes the series
13 temp = a + b;
14 a = b;
15 b = temp;
16 if (i == 4 && j == 3) { // Skips the 4th character of the base17 break;
18 }
19 }
20 printf("\n");
21 }
22 return 0;23 }
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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: 112112321
1234321123211211
Program:
view sourceprint ? 01 #include <stdio.h>
02
03 void sequence(int x);
04 int main() {
05 /* c taken for columns */
06 int i, x = 0, num = 7;
07 for (i = 1; i <= num; i++) {
08 if (i <= (num / 2) + 1) {
09 x = i;
10 } else {11 x = 8 - i;
12 }
13 sequence(x);
14 puts("\n");
15 }
16 return 0;17 }
18
19 void sequence(int x) {
20 int j;
21
22 for (j = 1; j < x; j++) {23 printf("%d", j);
24 }
25 for (j = x; j > 0; j--) {
26 printf("%d", j);
27 }
28 }
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End of Question3 Start of Question4
Write a C program to print the following pattern: 2 4 5 6 6 7 8 9 10 4 5 6 2
Program:
view sourceprint ? 01 #include <stdio.h>
02
03 int main(void) {
04 int prnt;
05 int i, j, k, r, s, sp, nos = 3, nosp = 2; //nos n nosp controls the spacing factor
06 // Prints the upper triangle
07 for (i = 1; i <= 5; i++) {
08 if ((i % 2) != 0) {
09 for (s = nos; s >= 1; s--) {
10 printf(" ");
11 }
12 for (j = 1; j <= i; j++) {13 if (i == 5 && j == 5) { //Provides the extra space reqd betn 9 n 10
Explanation: This is a diamond formation composed of numbers. The numbers are in the following order next_no=i+j wherenext_no = The next no to be printedi = index of the outer for loopj = index of the inner for loopBack to top
End of Question4 Start of Question5
Write a C program to print the following pattern: 1 1 3 3 3 3 3 3
31 for (r = 1; r <= p; r++) { // Lower right triangle
32 printf("%2d", p);
33 }
34
35 printf("\n");
36 --c;
37 nos = nos + 8; // Spacing control.
38 }
39 }
40
41 return 0;
42 }
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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
End of Question5 Start of Question6
Write a C program to print the following pattern: 0 -2-3 0-4-3-2-1 0 -2-3 0 0
Program:
view sourceprint ? 01 #include <stdio.h>
02
03 int main(void) {
04 int i, j, k, r, s, sp, nos = 2, nosp = 1;05 for (i = 1; i <= 5; i++) {
06 if ((i % 2) != 0) {
07 for (s = nos; s >= 1; s--) { //for the spacing factor.
21 for (sp = nosp; sp >= 1; sp--) { // for the spacing factor.
22 printf(" ");
23 }
24 for (r = 1; r <= k; r++) {25 printf("%2d", r-k);
26 }
27 }
28 if ((k % 2) != 0) {29 printf("\n");
30 nosp++;
31 }
32 }
33 return 0;
34 }
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Explanation:This can be seen as a diamond composed of numbers. If we use the conventional nested for loop for its construction the numbers can be seen to flowing the following function f(x) -> j-iwherej= inner loop indexi= outer loop indexBack to top
End of Question6 Start of Question7
Write a C program to print the following pattern: 77777777777 7 7 7 7 7 7 7 7 77
Program:
view sourceprint ? 01 #include <stdio.h>
02
03 int main(void) {
04 int i, j;
05 for (i = 11; i >= 1; i--) {06 for (j = 1; j <= i; j++) {
07 if (i == 11) {
08 printf("7"); // Makes sure the base is printed completely
09 continue;
10 } else if (j == i) { // Hollows the rest11 printf("7");
12 } else {
13 printf(" ");
14 }
15 }
16 printf("\n");
17 }
18 return 0;19 }
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Explanation: This can be seen as a hollow right-angled triangle composed of 7′sBack to top
06 for (j=1; j<=i; j++) {07 if ((j%2)!=0) { // Applying the condition
08 printf(" 1");
09 } else {
10 printf(" 0");11 }
12 }
13 for (s=nos; s>=1; s--) { // Space factor
14 printf(" ");
15 }
16 for (k=1; k<=i; k++) {17 if(i==7 && k==1) // Skipping the extra 1
18 {
19 continue;
20 }
21 if ((k%2)!=0) { // Applying the condition
22 printf(" 1");
23 } else {
24 printf(" 0");25 }
26 }
27 printf("\n");
28 nos=nos-2; // Space Control
29 }
30 nos=1;31 for ( i=6; i>=1; i--) { // It shares the same base
32 for (j=1; j<=i; j++) {
33 if (j%2!=0) {
34 printf(" 1");
35 } else {
36 printf(" 0");37 }
38 }
39 for(s=nos; s>=1; s--) // Spacing factor
40 {
41 printf(" ");
42 }
43 for (k=1; k<=i; k++) {
44 if (k%2!=0) {
45 printf(" 1");
46 } else {
47 printf(" 0");
48 }
49 }
50 printf("\n");51 nos=nos+2;
52 }
53 return 0;
54 }
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End of Question8 Start of Question9
Write a C program to print the following pattern: 12 43 6 92 41
Program:
view sourceprint ? 01 #include <stdio.h>
02 int main(void) {
03 int i,j;
04 for (i=1; i<=3 ; i++) {
05 for (j=1; j<=i; j++) {
06 printf("%2d", (i*j));
07 }
08 printf("\n");
09 }
10 for (i=2; i>=1; i--) { // As they share the same base
11 for (j=1; j<=i; j++) {
12 printf("%2d",i*j);
13 }
14 printf("\n");
15 }
16 return 0;17 }
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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 *jwherei = Index of the Outer loopj = Index of the inner loopBack to top
End of Question9 Start of Question10
Write a C program to print the following pattern: 1 1 0 1 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 1 0 1
Program:
view sourceprint ? 01 #include <stdio.h>
02
03 int main(void) {
04 int i,j;
05 for (i=1; i<=7; i++) {
06 for (j=1; j<=i; j++) {
07 if (j==1) { // Applying the condition
08 printf(" 1");
09 } else {
10 printf(" 0");11 }
12 }
13 printf("\n");
14 }
15 for (i=6; i>=1; i--) { //As it shares the same base i=6
16 for (j=1; j<=i; j++) {
17 if (j==1) { // Applying the condition
18 printf(" 1");
19 } else {
20 printf(" 0");21 }
22 }
23 printf("\n");
24 }
25 return 0;
26 }
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Explanation: This can be seen as two right angle triangles sharing the same base which is composed of 0′s n 1′s. The first column is filled with 1′s and rest with 0′s
Write a program to display the multiplication table of a given number.
Enter any Number: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
Explanation: We need to multiply the given number (i.e. the number for which we want the multiplication table) with value of ‘i’ which increments from 1 to 10.
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 decremented for numbers to be displayed in alternate columns.
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Write C program to print the following pattern:
1 1 2 1 1 2 3 2 1 1 2 3 4 3 2 1 1 2 3 4 5 4 3 2 1
Program:
view source
print ?
01 #include<stdio.h>
02 int main() {
03 /* c taken for columns */
04 int i, j, c = 9, m, k;
05 for (i = 1; i <= 5; i++) {
06 /* k is used for spaces */
07 for (k = 1; k <= c; k++) {
08 printf(" ");
09 }
10 for (j = 1; j <= i; j++) { printf("%2d", j); } for (m = j - 2; m > 0; m--) {
Explanation: Here ‘i’ loop is used for printing numbers in rows and ‘k’ loop is used for providing 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:
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:
Explanation: In the above program we have taken two variables ‘num’ and ‘sum’. ‘num’ is used to check the condition and to display the numbers up to 5. ‘sum’ is used to add the numbers 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.
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