I want to print out the following to the console:
+++++
++++*
+++**
++***
+****
*****
I am a new learner of programming, so encountering some difficulties. Can anyone help me, please? I have tried this, but is incorrect. What do I need to change?
#include<stdio.h>
int main(){
int i, j, k;
for(i=0; i<5; i++){
for(j=i; j<5; j++){
for(k=0; k<j; k++){
printf("*");
}
printf("+");
}
printf("\n");
}
return 0;
}
You have the right idea: Use three for loops.
#include <stdio.h>
int main() {
for (int i = 0; i < 6; i++) {
for (int k = i; k < 5; k++) {
printf("+");
}
for (int j = 0; j < i; j++) {
printf("*");
}
printf("\n");
}
return 0;
}
Test
+++++
++++*
+++**
++***
+****
*****
Online demo
First, generalise it and wrap it in a function. You want a square with a diagonal. It has to be an even number of characters to look right. But + and * could be any character, and the size could be 6 or all the way up to screen maximum width.
so
/* print a square with a diagonal
N - the size of the sides of the square
cha - character a (eg '+')
chb - character b (eg '*')
*/
void printdiagsquare(int N, char cha, char chb);
That's our prototype, and that's half the battle.
Now we need to check N is even and positive, then write the loops.
Let's get the test away first.
if(N < 2 || (N % 2) == 1)
printf(N must be even\n");
Now the main loop for each line
for(i=0;i<N;i++)
{
//printline code here
printf("\n");
}
Now test it. Is it printing N blank lines?
main(void)
{
printdiagsquare(6, '+', '*');
}
Now to get the lines printed.
to print N-1 '+'s is easy. We need j as the counter since i is the outer
for(j=0;j<N-1;j++)
printf("%c", cha);
But we need to generalise, we need to print 6,, 5, 4, 3 and so on as i increases.
So
for(j=0;j<N-i-1;j++)
printf("%c", cha);
I'll leave the last little bit for you to do. No point just typing ina function blindly.
You could try more optimized code for m-rows and n-columns
in 2 for loop only :-
#include <stdio.h>
int main(void) {
int m = 6; // Rows
int n = 5; // Cols
int i,j,k;
for (i = 0; i < m; i++) {
k = i;
for (j = n; j >= 0; j--) {
if(k>=j)
printf("*");
else
printf("+");
}
printf("\n");
}
return 0;
}
Related
I am new to C programming....we have 2D arrays for integers...but how to declare and take the input for 2D string arrays...I tried it for taking single character input at a time similar to integer array...but I want to take whole string as input at a time into 2D array..
code:
#include <stdio.h>
int main()
{
char str[20][20];
int i, j;
for (i = 0; i < 20; i++)
{
for (j = 0; j < 20; j++)
{
scanf("%c", &str[i][j]);
}
}
}
can anyone resolve this problem?
The declaration of a 2D string array can be described as:
char string[m][n];
where m is the row size and n is the column size.
If you want to take m strings as input with one whole string at a time...it is as follows
#include<stdio.h>
int main()
{
char str[20][20];
int i,j;
for(i=0;i<m;i++)
{
gets(str[i]);
}
}
here 'i' is the index of the string....
Hope this answer helps...
A few issues with your code. Using scanf to reach characters, you're going to read newlines. If I create a more minimal version of your code with an extra few lines to print the input, we can see this:
#include <stdio.h>
int main() {
char str[3][3];
int i, j;
for (i = 0; i < 3; i++) {
for (j = 0; j < 3; j++) {
scanf("%c", &str[i][j]);
}
}
for (i = 0; i < 3; i++) {
for (j = 0; j < 3; j++) {
printf("%c ", str[i][j]);
}
printf("\n");
}
}
And running it:
$ ./a.out
gud
ghu
ert
g u d
g h
u
e
$
We can test the input to circumvent this. If the character input is a newline character ('\n') then we'll decrement j so effectively we've sent the loop back a step. We could easily extend this boolean condition to ignore other whitespace characters like ' ' or '\t'.
#include <stdio.h>
int main() {
char str[3][3];
int i, j;
for (i = 0; i < 3; i++) {
for (j = 0; j < 3; j++) {
char temp = 0;
scanf("%c", &temp);
if (temp == '\n' || temp == ' ' || temp == '\t') {
j--;
}
else {
str[i][j] = temp;
}
}
}
for (i = 0; i < 3; i++) {
for (j = 0; j < 3; j++) {
printf("%c ", str[i][j]);
}
printf("\n");
}
}
Now,, when we run this:
$ ./a.out
gud
ghu
ert
g u d
g h u
e r t
$
Of course, one other thing you should do is to check the return value of scanf, which is an int representing the number of items read. In this case, if it returned 0, we'd know it hadn't read anything. In that case, within the inner loop, we'd probably also want to decrement j so the loop continues.
#include<stdio.h>
main()
{
char name[5][25];
int i;
//Input String
for(i=0;i<5;i++)
{
printf("Enter a string %d: ",i+1);
}
//Displaying strings
printf("String in Array:\n");
for(int i=0;i<5;i++)
puts(name[i]);
}
Simple code that accepts String in an array.
Can someone please help me i am facing runtime error while solving this problem.
I have first defined the integers and then used scanf to take the input.
Then i check whether the 2 consecutive elements of array are equal are not.
if they are equal i equate j variable to i+1 and so that it can traverse and find if same duplicate elements are side by side (eg- 15 15 15).
I increment the j element till a[j] is equal to a[i].
Then using i try to print the number with the number of occurences of it which is j-i and then assign i with vakue of j-1.
#include <stdio.h>
int main()
{
int n,j=0,i;
scanf("%d",&n);
int a[n];
for (i = 0; i < n; ++i) {
scanf("%d",&a[i]);
}
for (i = 0; i < n - 1; ++i)
{
if(a[i]==a[i+1])
{
j=i+1;
while(j<n && a[i]==a[j])
{
j++;
}
printf("%d is appearing %d times\n",a[i],j-i);
}
i=j-1;
}
return 0;
}
The input array needs to be sorted first to count duplicated, the loop logic needs to be fixed to reassign the index i.
A fixed code might like this:
#include <stdio.h>
#include <stdlib.h>
static int cmp_intp(const void *p1, const void *p2) {
return *(const int *)p1 > *(const int *)p2;
}
int main() {
int n, j = 0, i;
scanf("%d", &n);
int a[n];
for (i = 0; i < n; ++i) {
scanf("%d", &a[i]);
}
qsort(a, n, sizeof(a[0]), cmp_intp);
for (i = 0; i < n;) {
if (i < n - 1 && a[i] == a[i + 1]) {
j = i + 1;
while (j < n && a[i] == a[j]) {
j++;
}
printf("%d is appearing %d times\n", a[i], j - i);
i = j;
} else {
++i;
}
}
return 0;
}
The problem is created by the line,
i=j-1;
in the case when two consecutive elements are not equal.
move it within the if condition.
I want to print this pattern like right angled triangle
0
909
89098
7890987
678909876
56789098765
4567890987654
345678909876543
23456789098765432
1234567890987654321
I wrote the following code:
#include <stdio.h>
#include <conio.h>
void main()
{
clrscr();
int i,j,x,z,k,f=1;
for ( i=10;i>=1;i--,f++)
{
for(j=1;j<=f;j++,k--)
{
k=i;
if(k!=10)
{
printf("%d",k);
}
if(k==10)
{
printf("0");
}
}
for(x=1;x<f;x++,z--)
{
z=9;
printf("%d",z);
}
printf("%d/n");
}
getch();
}
What is wrong with this code? When I check manually it seems correct but when compiled gives different pattern
Fairly simple: use two loops, one for counting up and one for counting down. Print literal "0" between the two.
#include <stdio.h>
int main()
{
for (int i = 0; i < 10; i++) {
for (int j = 10 - i; j < 10; j++)
printf("%d", j);
printf("0");
for (int j = 9; j >= 10 - i; j--)
printf("%d", j);
printf("\n");
}
return 0;
}
Like H2CO3's, but since we're only printing single digits why not use putchar():
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
int i, j;
for(i = 0; i < 10; ++i)
{
// Left half.
for(j = 0; j < i; ++j)
putchar('9' - i + j + 1);
// Center zero.
putchar('0');
// Right half.
for(j = 0; j < i; ++j)
putchar('9' - i + j + 1);
putchar('\n');
}
return EXIT_SUCCESS;
}
Modified Code:
Check your errors:
# include<stdio.h>
# include<conio.h>
int main()
{
// clrscr();
int i,j,x,z,k,f=1;
for ( i=10;i>=1;i--,f++)
{
k=i; // K=i should be outside of loop.
for(j=1;j<=f;j++,k++)
{
if(k!=10)
{
printf("%d",k);
}
if(k==10)
{
printf("0");
}
}
z=9; //z=9 should be outside loop.
for(x=1;x<f;x++,z--)
{
printf("%d",z);
}
printf("\n");
}
//getch();
return 0;
}
You are defining k=i inside the for loop(loop which has j) so every time k gets value of i and thus it always get value of i and prints that value and your another condition(if(k==10)) will never be true because every time k takes value of i and i is less than 10 after first iteration of loop and z=9 inside loop so every time loop is executed it is taking value z=9 so it is printing wrong value.
Here's a C# version:
static void DrawNumberTriangle()
{
for (int line = 10; line >=1; line--)
{
for (int number = line; number < 10; number++)
{
System.Console.Write(number);
}
System.Console.Write("0");
for (int number = 9; number > line - 1; number--)
{
System.Console.Write(number);
}
System.Console.WriteLine();
}
}
I'd suggest renaming your i,j,x,z,k,f variables to ones that have meaning like the one's I used. This helps making your code easier to follow.
Rather than output the mid 0 using printf, why not print it using the loops itself.
The following short and simple code can be used:
int main()
{
int m = 10, n, p;
while(m >= 1)
{
for(n = m; n <= 10; n++)
printf("%d", n % 10);
for(p = n - 2; p >= m; p--)
printf("%d", p );
printf("\n");
m--;
}
return 1;
}
For high throughput (though of questionable merit in terms of clarity):
#include <stdio.h>
int main() {
char const digits[] = "1234567890";
char const rdigits[] = "9876543210";
for (int i = 0; i < 30; ++i) {
int k = i % 10;
fputs(digits + 9 - k, stdout);
for (int j = 9; j < i; j += 10) fputs(digits, stdout);
for (int j = 9; j < i; j += 10) fputs(rdigits, stdout);
fwrite(rdigits, 1, k, stdout);
fputs("\n", stdout);
}
}
#include <stdio.h>
void print(int i){
if(i == 10){
putchar('0');
return ;
} else {
printf("%d", i);
print(i+1);
printf("%d", i);
}
}
int main(void){
int i;
for(i = 10; i>0; --i){
print(i);
putchar('\n');
}
return 0;
}
I want to print this pattern like right angled triangle
0
909
89098
7890987
678909876
56789098765
4567890987654
345678909876543
23456789098765432
1234567890987654321
I wrote the following code:
#include <stdio.h>
#include <conio.h>
void main()
{
clrscr();
int i,j,x,z,k,f=1;
for ( i=10;i>=1;i--,f++)
{
for(j=1;j<=f;j++,k--)
{
k=i;
if(k!=10)
{
printf("%d",k);
}
if(k==10)
{
printf("0");
}
}
for(x=1;x<f;x++,z--)
{
z=9;
printf("%d",z);
}
printf("%d/n");
}
getch();
}
What is wrong with this code? When I check manually it seems correct but when compiled gives different pattern
Fairly simple: use two loops, one for counting up and one for counting down. Print literal "0" between the two.
#include <stdio.h>
int main()
{
for (int i = 0; i < 10; i++) {
for (int j = 10 - i; j < 10; j++)
printf("%d", j);
printf("0");
for (int j = 9; j >= 10 - i; j--)
printf("%d", j);
printf("\n");
}
return 0;
}
Like H2CO3's, but since we're only printing single digits why not use putchar():
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
int i, j;
for(i = 0; i < 10; ++i)
{
// Left half.
for(j = 0; j < i; ++j)
putchar('9' - i + j + 1);
// Center zero.
putchar('0');
// Right half.
for(j = 0; j < i; ++j)
putchar('9' - i + j + 1);
putchar('\n');
}
return EXIT_SUCCESS;
}
Modified Code:
Check your errors:
# include<stdio.h>
# include<conio.h>
int main()
{
// clrscr();
int i,j,x,z,k,f=1;
for ( i=10;i>=1;i--,f++)
{
k=i; // K=i should be outside of loop.
for(j=1;j<=f;j++,k++)
{
if(k!=10)
{
printf("%d",k);
}
if(k==10)
{
printf("0");
}
}
z=9; //z=9 should be outside loop.
for(x=1;x<f;x++,z--)
{
printf("%d",z);
}
printf("\n");
}
//getch();
return 0;
}
You are defining k=i inside the for loop(loop which has j) so every time k gets value of i and thus it always get value of i and prints that value and your another condition(if(k==10)) will never be true because every time k takes value of i and i is less than 10 after first iteration of loop and z=9 inside loop so every time loop is executed it is taking value z=9 so it is printing wrong value.
Here's a C# version:
static void DrawNumberTriangle()
{
for (int line = 10; line >=1; line--)
{
for (int number = line; number < 10; number++)
{
System.Console.Write(number);
}
System.Console.Write("0");
for (int number = 9; number > line - 1; number--)
{
System.Console.Write(number);
}
System.Console.WriteLine();
}
}
I'd suggest renaming your i,j,x,z,k,f variables to ones that have meaning like the one's I used. This helps making your code easier to follow.
Rather than output the mid 0 using printf, why not print it using the loops itself.
The following short and simple code can be used:
int main()
{
int m = 10, n, p;
while(m >= 1)
{
for(n = m; n <= 10; n++)
printf("%d", n % 10);
for(p = n - 2; p >= m; p--)
printf("%d", p );
printf("\n");
m--;
}
return 1;
}
For high throughput (though of questionable merit in terms of clarity):
#include <stdio.h>
int main() {
char const digits[] = "1234567890";
char const rdigits[] = "9876543210";
for (int i = 0; i < 30; ++i) {
int k = i % 10;
fputs(digits + 9 - k, stdout);
for (int j = 9; j < i; j += 10) fputs(digits, stdout);
for (int j = 9; j < i; j += 10) fputs(rdigits, stdout);
fwrite(rdigits, 1, k, stdout);
fputs("\n", stdout);
}
}
#include <stdio.h>
void print(int i){
if(i == 10){
putchar('0');
return ;
} else {
printf("%d", i);
print(i+1);
printf("%d", i);
}
}
int main(void){
int i;
for(i = 10; i>0; --i){
print(i);
putchar('\n');
}
return 0;
}
I am trying to write a C code that will print a pyramid structure on screen, something like this.
The corresponding code I've written is something like this.
#include <stdio.h>
#include <stdlib.h>
void printArrayFunc(char arr[9][5]) {
int i, j;
printf("=========================================\nprinting the values\n");
for (i = 0; i < 5; i++) {
for (j = 0; j < 9; j++) {
//printf("arr[%d][%d] = %d\n", i,j, arr[i][j]);
if (arr[i][j] == 1)
printf("*");
else
printf(" ");
}
printf("\n");
}
}
int main() {
int i, j;
char arr[9][5] = {
0
};
printf("============================\nfilling the values\n");
for (i = 0; i < 5; i++) {
for (j = 4 - i; j <= 4 + i; j++) {
arr[i][j] = 1;
// printf("arr[%d][%d]= %d\n",i,j,arr[i][j]);
}
//printf("\n");
}
printArrayFunc(arr);
return 0;
}
It is giving an output like
I know I'm doing some silly mistake but at this moment, I'm not able to find what is going wrong. Let me hear your comments on this.
In the function argument:
char arr[9][5]
In the loop:
for (i = 0; i<5; i++) {
for (j = 0; j<9;j++) {
if (arr[i][j] == 1)
You flipped the position of i and j. i should go from 0 to 9, j from 0 to 5.
if (arr[i][j] == 1)
printf("*");
else
printf(" ");
This statement is giving the garbage value in this statement if if condition is true then it print else statement and when else comes true it prints the garbage value.