1.
Write a C program to print the
following pattern:
2. *
3. *
*
4. *
* *
5. * *
* *
6.
Write a C program to print the
following pattern:
7. * *
8. *
* * * * *
9. *
* * * * * * * * *
10.* * * * *
* * * * * *
11. Write a C program to print the following pattern:
12.* *
13. * *
14.* * * *
15. *
* * *
16.* *
* * *
17. *
* * *
18.* * * *
19. * *
20.* *
21. Write a C program to print the following pattern:
22.* *
* * *
23. *
* * *
24. *
* *
25. *
*
26. *
27. *
*
28. *
* *
29. *
* * *
30.* *
* * *
31. Write a C program to print the following pattern:
32.*
33. * * *
34. * * * * *
35. * * * * * * *
36. * * * * * * * * *
37. * * * * * * *
38. * * * * *
39. * * *
40. *
41. * * *
42. * * * * *
43. Write a C program to print the following pattern:
44.*
45. * *
46. * * *
47. * * * *
48. * * *
49. * *
50. *
51. Write a C program to print the following pattern:
52.* * * * *
* * * *
53.* * *
* * * * *
54.* * * * * *
55.* * * *
56.* *
57.* * * *
58.* * * *
* *
59.* * *
* * * * *
60.* * * * *
* * * *
61. Write a C program to print the following pattern:
62.* * * * *
* * * * * * * * * * * *
63. * * * * * * * * * * * * * *
64. * * * * * * * * * *
65. * * * * * *
66. * * * * * * * * *
67. * * * * * * *
68. * * * * *
69. * * *
70. *
71. Write a C program to print the following pattern:
72.*
73. * * *
74. * * * * *
75.* * * * *
* *
76.* *
77.* * * *
78.* * * * * *
79.* * * * *
* *
80.* * * * * *
81.* * * *
82.* *
83.* * * * *
* *
84. * * * * *
85. * * *
86. *
87. Write a C program to print the following pattern:
88.* * * * *
* * * * * * * * * * * * * * * * * * * *
89. *
* * *
* *
90. *
* * *
* *
91. *
* * *
* *
92. * * *
93. *
* * *
* *
94. *
* * *
* *
95. *
* * *
* *
96.* * * * *
* * * * * * * * * * * * * * * * * * * *
*
*
*
*
* *
* *
* *
Program:
/*
This is a simple mirror-image of a right angle triangle */
#include
int
main() {
char prnt = '*';
int i, j, nos = 4, s;
for (i = 1; i <= 5; i++) {
for
(s = nos; s >= 1; s--) { // Spacing
factor
printf(" ");
}
for (j = 1; j <= i; j++) {
printf("%2c", prnt);
}
printf("\n");
--nos;
// Controls the spacing factor
}
return 0;
}
- Write C program to print the following pattern:
3. * *
4. *
* * * * *
5. *
* * * * * * * * *
*
* * * * * * * * * *
Program:
#include
int
main() {
char prnt = '*';
int i, j, k, s, c = 1, nos = 9;
for (i = 1; c <= 4; i++) {
// As we want to print the columns in odd
sequence viz. 1,3,5,.etc
if ((i % 2) != 0) {
for (j = 1; j <= i; j++) {
printf("%2c", prnt);
}
for
(s = nos; s >= 1; s--) { //The spacing factor
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
++c;
}
}
return 0;
}
- Write C program to print the following pattern:
7. * *
8. * *
9. * * * *
10. *
* * *
11.* *
* * *
12. *
* * *
13.* * * *
14. * *
* *
Program:
#include
int
main() {
char prnt = '*';
int i, j, k, s, p, r, nos = 7;
for (i = 1; i <= 5; i++) {
for (j = 1; j <= i; j++) {
if ((i % 2) != 0 && (j % 2) != 0) {
printf("%3c",
prnt);
}
else
if ((i % 2) == 0 && (j % 2) == 0) {
printf("%3c",
prnt);
}
else
{
printf(" ");
}
}
for
(s = nos; s >= 1; s--) { // for the
spacing factor
printf(" ");
}
for (k = 1; k <= i; k++) { //Joining
seperate figures
if
(i == 5 && k == 1) {
continue;
}
if
((k % 2) != 0) {
printf("%3c",
prnt);
}
else
{
printf(" ");
}
}
printf("\n");
nos
= nos - 2; // space control
} nos = 1;
// remaining half..
for
(p = 4; p >= 1; p--) {
for (r = 1; r <= p; r++) {
if
((p % 2) != 0 && (r % 2) != 0) {
printf("%3c",
prnt);
}
else
if ((p % 2) == 0 && (r % 2) == 0) {
printf("%3c",
prnt);
}
else
{
printf(" ");
}
}
for
(s = nos; s >= 1; s--) {
printf(" ");
}
for (k = 1; k <= p; k++) {
if ((k % 2) != 0) {
printf("%3c", prnt);
}
else
{
printf(" ");
}
}
nos = nos + 2; // space control
printf("\n");
}
return 0;
}
Download Code Explanation: This can be seen as an inverted diamond
composed of stars. It can be 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 ‘*’, else a 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 triagles aligned as per the
requirement.
- Write a C program to print the following pattern:
16.* *
* * *
17. *
* * *
18. *
* *
19. *
*
20. *
21. *
*
22. *
* *
23. *
* * *
* *
* * *
Program:
#include
int
main() {
char prnt = '*';
int i, j, s, nos = 0;
for (i = 9; i >= 1; (i = i - 2)) {
for (s = nos; s >= 1; s--) {
printf(" ");
}
for (j = 1; j <= i; j++) {
if ((i % 2) != 0 && (j % 2) != 0) {
printf("%2c", prnt);
} else {
printf(" ");
}
}
printf("\n");
nos++;
}
nos = 3;
for (i = 3; i <= 9; (i = i + 2)) {
for
(s = nos; s >= 1; s--) {
printf(" ");
}
for (j = 1; j <= i; j++) {
if ((i % 2) != 0 && (j % 2) != 0) {
printf("%2c", prnt);
} else {
printf(" ");
}
}
nos--;
printf("\n");
}
return 0;
}
- Write a C program to print the following pattern:
25.*
26. * * *
27. * * * * *
28. * * * * * * *
29. * * * * * * * * *
30. * * * * * * *
31. * * * * *
32. * * *
33. *
34. * * *
* * * * *
Program:
#include
int
main() {
char prnt = '*';
int i, j, k, s, nos = 4;
for (i = 1; i <= 5; i++) {
for
(s = nos; s >= 1; s--) {
printf(" ");
}
for (j = 1; j <= i; j++) {
printf("%2c", prnt);
}
for (k = 1; k <= (i - 1); k++) {
if
(i == 1) { continue;
}
printf("%2c",
prnt);
}
printf("\n"); nos--;
}
nos = 1;
for
(i = 4; i >= 1; i--) {
for (s = nos; s >= 1; s--) {
printf(" ");
}
for (j = 1; j <= i; j++) {
printf("%2c", prnt);
}
for (k = 1; k <= (i - 1); k++) {
printf("%2c", prnt);
}
nos++;
printf("\n");
}
nos = 3;
for (i = 2; i <= 5; i++) {
if
((i % 2) != 0) {
for
(s = nos; s >= 1; s--) {
printf(" ");
}
for (j = 1; j <= i; j++) {
printf("%2c", prnt);
}
}
if ((i % 2) != 0) {
printf("\n");
nos--;
}
}
return 0;
}
- Write a C program to print the following pattern:
36.*
37. * *
38. * * *
39. * * * *
40. * * *
41. * *
*
Program:
/*
This can be seen as two right angle
triangles sharing the same base
which is modified by adding few extra
shifting spaces
*/
#include
//
This function controls the inner loop and the spacing
//
factor guided by the outer loop index and the spacing index.
int
triangle(int nos, int i) {
char prnt = '*';
int s, j;
for (s = nos; s >= 1; s--) { // Spacing factor
printf("
");
}
for (j = 1; j <= i; j++) { //The inner loop
printf("%2c", prnt);
}
return 0;
}
int
main() {
int i, nos = 5;
//draws the upper triangle
for (i = 1; i <= 4; i++) {
triangle(nos, i); //Inner loop construction
nos++; // Increments the spacing factor
printf("\n"); }
nos
= 7; //Draws the lower triangle skipping
its base.
for
(i = 3; i >= 1; i--) {
int j = 1;
triangle(nos, i); // Inner loop construction
nos = nos - j; // Spacing factor
printf("\n");
}
return 0;
}
- Write a C program to print the following pattern:
43.* * * * *
* * * *
44.* * *
* * * * *
45.* * * * * *
46.* * * *
47.* *
48.* * * *
49.* * * * * *
50.* * *
* * * * *
*
* * * * * * * *
Program:
#include
int
main() {
char prnt = '*';
int i, j, k, s, nos = -1;
for (i = 5; i >= 1; i--) {
for (j = 1; j <= i; j++) {
printf("%2c", prnt);
}
for
(s = nos; s >= 1; s--) {
printf(" ");
}
for (k = 1; k <= i; k++) {
if (i == 5 && k == 5) {
continue;
}
printf("%2c", prnt);
}
nos = nos + 2;
printf("\n");
}
nos = 5;
for (i = 2; i <= 5; i++) {
for (j = 1; j <= i; j++) {
printf("%2c", prnt);
}
for
(s = nos; s >= 1; s--) {
printf(" ");
}
for (k = 1; k <= i; k++) {
if (i == 5 && k == 5) {
break;
}
printf("%2c", prnt);
}
nos = nos - 2;
printf("\n");
}
return 0;
}
- Write a C program to print the following pattern:
52.* * * * *
* * * * * * * * * * * *
53. * * * * * * * * * * * * * *
54. * * * * * * * * * *
55. * * * * * *
56. * * * * * * * * *
57. * * * * * * *
58. * * * * *
59. * * *
*
Program:
#include
int
main() {
char prnt = '*';
int i, j, k, s, sp, nos = 0, nosp = -1;
for (i = 9; i >= 3; (i = i - 2)) {
for (s = nos; s >= 1; s--) {
printf(" ");
}
for (j = 1; j <= i; j++) {
printf("%2c",
prnt);
}
for
(sp = nosp; sp >= 1; sp--) {
printf(" ");
}
for (k = 1; k <= i; k++) {
if (i == 9 && k == 1) {
continue;
}
printf("%2c",
prnt);
}
nos++;
nosp
= nosp + 2;
printf("\n");
}
nos
= 4;
for
(i = 9; i >= 1; (i = i - 2)) {
for (s = nos; s >= 1; s--) {
printf(" ");
}
for (j = 1; j <= i; j++) {
printf("%2c", prnt);
}
nos++;
printf("\n");
}
return 0;
}
- Write a C program to print the following pattern:
61.*
62. * * *
63. * * * * *
64.* * * * *
* *
65.* *
66.* * * *
67.* * * * * *
68.* * * * *
* *
69.* * * * * *
70.* * * *
71.* *
72.* * * * *
* *
73. * * * * *
74. * * *
*
Program:
#include
/*
* nos = Num. of spaces required in the
triangle.
* i =
Counter for the num. of charcters to print in each row
* skip= A flag for checking whether to
*
skip a character in a row.
*
*/
int
triangle(int nos, int i, int skip) {
char prnt = '*';
int s, j;
for (s = nos; s >= 1; s--) {
printf("
");
}
for (j = 1; j <= i; j++) {
if (skip != 0) {
if (i == 4 && j == 1) {
continue;
}
}
printf("%2c", prnt);
}
return 0;
}
int
main() {
int i, nos = 4;
for (i = 1; i <= 7; (i = i + 2)) {
triangle(nos, i, 0);
nos--;
printf("\n");
}
nos = 5;
for (i = 1; i <= 4; i++) {
triangle(1,
i, 0); //one space needed in each case of the formation
triangle(nos,
i, 1); //skip printing one star in the last row.
nos
= nos - 2;
printf("\n");
}
nos
= 1;
for
(i = 3; i >= 1; i--) {
triangle(1, i, 0);
triangle(nos, i, 0);
nos = nos + 2;
printf("\n");
}
nos = 1;
for (i = 7; i >= 1; (i = i - 2)) {
triangle(nos, i, 0);
nos++;
printf("\n");
}
return 0;
}
- Write a C program to print the following pattern:
76.* * * * *
* * * * * * * * * * * * * * * * * * * *
77. *
* * *
* *
78. *
* * *
* *
79. *
* * *
* *
80. * * *
81. *
* * *
* *
82. *
* * *
* *
83. *
* * *
* *
*
* * * * * * * * * * * * * * * * * * * * * * * *
Program:
#include
/*
* nos = Num. of spaces required in the
triangle.
* i =
Counter for the num. of characters to print in each row
* skip= A flag for check whether to
*
skip a character in a row.
*
*/
int
triangle(int nos, int i, int skip) {
char prnt = '*';
int s, j;
for (s = nos; s >= 1; s--) {
printf("
");
}
for (j = 1; j <= i; j++) {
if (skip != 0) {
if (i == 9 && j == 1) {
continue;
}
}
if
(i == 1 || i == 9) {
printf("%2c", prnt);
}
else
if (j == 1 || j == i) {
printf("%2c", prnt);
} else {
printf("
");
} }
return
0; }
int
main() {
int
i, nos = 0, nosp = -1, nbsp = -1;
for
(i = 9; i >= 1; (i = i - 2)) {
triangle(nos, i, 0);
triangle(nosp, i, 1);
triangle(nbsp, i, 1);
printf("\n");
nos++;
nosp = nosp + 2;
nbsp = nbsp + 2;
}
nos = 3, nosp = 5, nbsp = 5;
for (i = 3; i <= 9; (i = i + 2)) {
triangle(nos, i, 0);
triangle(nosp, i, 1);
triangle(nbsp, i, 1);
printf("\n");
nos--;
nosp = nosp - 2;
nbsp = nbsp - 2;
}
return 0;
}
Read more: http://cmagical.blogspot.com/2010/01/10-challenging-star-pattern-programs-in.html#ixzz1jhAznHcm
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