This is a continuation to the series of challenging pattern C programs in Interview Mantra. This set of 10 puzzling programs are of type number patterns.
Also read Print Pattern Programs
- Write a C program to print the following pattern:
1 0 1 1 0 1 0 1 0 1 1 0 1 0 1
- Write a C program to print the following pattern:
0 1 1 2 3 5 8 13 21
- Write a C program to print the following pattern:
1 121 12321 1234321 12321 121 1
- Write a C program to print the following pattern:
2 4 5 6 6 7 8 9 10 4 5 6 2
- Write a C program to print the following pattern:
1 1 3 3 3 3 3 3 5 5 5 5 5 5 5 5 5 5 7 7 7 7 7 7 7 7 7 7 7 7 7 7 5 5 5 5 5 5 5 5 5 5 3 3 3 3 3 3 1 1
- Write a C program to print the following pattern:
0 -2-3 0 -4-3-2-1 0 -2-3 0 0
- Write a C program to print the following pattern:
77777777777 7 7 7 7 7 7 7 7 7 7
- Write a C program to print the following pattern:
1 1 1 0 1 0 1 0 1 1 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 1 0 1 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1 0 1 0 1 1 0 1 0 1 0 1 0 1 0 1 1 0 1 1 0 1 0 1 1
- Write a C program to print the following pattern:
1 2 4 3 6 9 2 4 1
- 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
- Write a C program to print the following pattern:
- Write C program to print the following pattern:
0 1 1 2 3 5 8 13 21
- Write C program to print the following pattern:
1 121 12321 1234321 12321 121 1
- Write a C program to print the following pattern:
2 4 5 6 6 7 8 9 10 4 5 6 2 - Write a C program to print the following pattern:
1 1 3 3 3 3 3 3 5 5 5 5 5 5 5 5 5 5 7 7 7 7 7 7 7 7 7 7 7 7 7 7 5 5 5 5 5 5 5 5 5 5 3 3 3 3 3 3 1 1
- Write a C program to print the following pattern:
0 -2-3 0 -4-3-2-1 0 -2-3 0 0 - Write a C program to print the following pattern:
77777777777 7 7 7 7 7 7 7 7 7 7 - Write a C program to print the following pattern:
1 1 1 0 1 0 1 0 1 1 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 1 0 1 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1 0 1 0 1 1 0 1 0 1 0 1 0 1 0 1 1 0 1 1 0 1 0 1 1
- Write a C program to print the following pattern:
1 2 4 3 6 9 2 4 1
- 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
1 0 1 1 0 1 0 1 0 1 1 0 1 0 1Program:
#include <stdio.h>
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");
}
return 0;
}
Explanation: This is a right angle triangle composed of 0′s and 1′s.
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End of Question1 Start of Question2
Program:
#include <stdio.h>
int main(void) {
int i, j, a = 0, b = 1, temp = 1;
for (i = 1; i <= 4; i++) {
for (j = 1; j <= i; j++) {
if (i == 1 && j == 1) { // Prints the '0' individually first
printf("0");
continue;
}
printf("%d ", temp); // Prints the next digit in the series
//Computes the series
temp = a + b;
a = b;
b = temp;
if (i == 4 && j == 3) { // Skips the 4th character of the base
break;
}
}
printf("\n");
}
return 0;
}
Explanation: This prints the Fibonacci series in a right angle triangle formation where the base has only three characters.
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End of Question2 Start of Question3
Program:
#include <stdio.h>
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");
}
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
Program:
#include <stdio.h>
int main(void) {
int prnt;
int i, j, k, r, s, sp, nos = 3, nosp = 2; //nos n nosp controls the spacing factor
// Prints the upper triangle
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) { //Provides the extra space reqd betn 9 n 10
printf(" "); // as 10 is a 2 digit no.
}
prnt = i + j;
printf("%2d", prnt);
}
}
if ((i % 2) != 0) {
printf("\n");
nos--;
}
}
// 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;
}
Explanation: This is a diamond formation composed of numbers. The numbers are in the 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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End of Question4 Start of Question5
Program:
#include <stdio.h>
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 spacing factor
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 factor
printf(" ");
}
for (r = 1; r <= p; r++) { // Lower right triangle
printf("%2d", p);
}
printf("\n");
--c;
nos = nos + 8; // Spacing control.
}
}
return 0;
}
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 .
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End of Question5 Start of Question6
Program:
#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);
}
}
if ((k % 2) != 0) {
printf("\n");
nosp++;
}
}
return 0;
}
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-i
where
j= inner loop index
i= outer loop index
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End of Question6 Start of Question7
Program:
#include <stdio.h>
int main(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");
}
return 0;
}
Explanation: This can be seen as a hollow right-angled triangle composed of 7′s
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End of Question7 Start of Question8
Program:
#include <stdio.h>
int main(void) {
int i,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");
}
}
for (s=nos; s>=1; s--) { // Space factor
printf(" ");
}
for (k=1; k<=i; k++) {
if(i==7 && k==1) // Skipping the extra 1
{
continue;
}
if ((k%2)!=0) { // Applying the condition
printf(" 1");
} else {
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");
}
}
printf("\n");
nos=nos+2;
}
return 0;
}
End of Question8 Start of Question9
Program:
#include <stdio.h>
int main(void) {
int i,j;
for (i=1; i<=3 ; i++) {
for (j=1; j<=i; j++) {
printf("%2d", (i*j));
}
printf("\n");
}
for (i=2; i>=1; i--) { // As they share the same base
for (j=1; j<=i; j++) {
printf("%2d",i*j);
}
printf("\n");
}
return 0;
}
Explanation: This can be seen as two right angle triangles sharing th same base
The numbers are following the following function f(x) = i *j
where
i = Index of the Outer loop
j = Index of the inner loop
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End of Question9 Start of Question10
Program:
#include <stdio.h>
int main(void) {
int i,j;
for (i=1; i<=7; i++) {
for (j=1; j<=i; j++) {
if (j==1) { // Applying the condition
printf(" 1");
} else {
printf(" 0");
}
}
printf("\n");
}
for (i=6; i>=1; i--) { //As it shares the same base i=6
for (j=1; j<=i; j++) {
if (j==1) { // Applying the condition
printf(" 1");
} else {
printf(" 0");
}
}
printf("\n");
}
return 0;
}
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
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About the Author: This post was written by Utsav Banerjee. You can reach Utsav on email at fugitiveland@gmail.com