I'm creating a program that calculates the multiples of seven from 1 to 100. However my code prints one multiple above 100, 105. I'm not sure how to fix this. I've tried doing num <= 100, num < 100, num < 99, num = 0, but nothing works. Thank you!
Code:
int main()
{
//variables
int i = 1, num;
printf("Multiples of Seven from 1 to 100: \n");
while(num < 100) {
num = i * 7;
printf(" %d ", num);
i++;
} //while end
} //main end
Make while loop condtion as,
while(i * 7 <= 100)
or change the lines from this:
num = i * 7;
printf(" %d ", num);
to this:
printf(" %d ", num);
num = i * 7;
if you choose to go this way you should initialize num..
this is the output
Multiples of Seven from 1 to 100:
0 7 14 21 28 35 42 49 56 63 70 77 84 91 98
Related
How can i run a command only one time inside of for loop?
I couldn't find how to do it so i wrote code like that. But the problem on this code you can see when the program goes to else command it doesn't work.
I just wanted to see odd and even numbers with using for loop but I don't want to get result like;
Even numbers :12
Even numbers :28
Even numbers :46
I just want to see only one time print Even numbers and then print numbers like;
Even numbers:
12
28
46
etc.
I hope I could explain clearly.
My alternative but wrong code is:
#include <stdio.h>
int main()
{
int num [] = {12, 14, 16, 33, 65, 98, 45, 25, 87, 18, 20};
printf("even numbers:");
printf("\t\t\t\t Odd numbers:");
for (int i = 0; i < 11; i++) {
if (num[i] % 2 == 0) {
printf("\n%d", num[i]);
}
else {
printf("\t\t\n%d");
}
}
return 0;
}
You want two loops:
#include <stdio.h>
int main(void)
{
int num[] = {12,14,16,33,65,98,45,25,87,18,20};
size_t n = sizeof num / sizeof *num;
printf ("Even numbers:\t\t\t\tOdd numbers:\n");
for (size_t i = 0; i < n; i++) {
if (num[i] % 2 == 0) {
printf ("%d ", num[i]);
}
}
printf("\r\t\t\t\t\t");
for (size_t i = 0; i < n; i++) {
if (num[i] % 2 != 0) {
printf ("%d ", num[i]);
}
}
printf("\n");
return 0;
}
Output:
Even numbers: Odd numbers:
12 14 16 98 18 20 33 65 45 25 87
Notice the \r to go to the beginning of the line
Here a different format:
#include <stdio.h>
int main(void) {
unsigned num []={12,14,16,33,65,98,45,25,87,18,20};
printf("even numbers"
"\todd numbers\n");
const char *prefix[] = {"", "\t\t"};
for(unsigned i = 0; i < sizeof(num) / sizeof(*num); i++) {
printf("%s%u\n", prefix[num[i] % 2], num[i]);
}
}
and the output is:
even numbers odd numbers
12
14
16
33
65
98
45
25
87
18
20
Based on the assumption that the least changes to your shown code, with explanation, are most helpful, here is my explained solution:
#include <stdio.h>
int main()
{
int num [] = {12, 14, 16, 33, 65, 98, 45, 25, 87, 18, 20};
printf("even numbers:");
printf("\tOdd numbers:\n"); // newline after the output, one tab, no blanks
for (int i = 0; i < 11; i++) {
if (num[i] % 2 == 0) {
printf("%d\n", num[i]); // newline after output
}
else {
printf("\t\t%d\n", num[i]); // tabs, then output, then newline
}
}
return 0;
}
this gets you an output of:
even numbers: Odd numbers:
12
14
16
33
65
98
45
25
87
18
20
Your problem was only caused by missapplied whitespaces.
(Apart from accidentally dropping the parameter for the odd output....)
Doing newlines after output is a good practice, but that is a matter of taste.
Important is to not output tabulators followed by a newline; because the newline spoils the effect of the tabulators.
Here is the version with newlines before output (for really minimal changes), but I recommend against it.
#include <stdio.h>
int main()
{
int num [] = {12, 14, 16, 33, 65, 98, 45, 25, 87, 18, 20};
printf("even numbers:");
printf("\t\t\t\t Odd numbers:");
for (int i = 0; i < 11; i++) {
if (num[i] % 2 == 0) {
printf("\n%d", num[i]);
}
else {
printf("\n\t\t\t\t\t %d", num[i]); // newline, tabs, blank, output
}
}
return 0;
}
This gets you an output of:
even numbers: Odd numbers:
12
14
16
33
65
98
45
25
87
18
20
Wider, because I left the multiple tabulators and the unneeded blank in.
EDIT: OP has been edited to show desired output. I had add '\n' to two lines of this to effect the change.
"Factoring out" common processing into a function is always a good idea.
#include <stdio.h>
void show( int num[], int nItems, char *title, int rem ) {
printf( "%s\n", title );
for( int i = 0; i < nItems; i++ )
if( num[i]%2 == rem )
printf( "%d\n", num[ i ] );
printf( "\n" );
}
int main () {
int num [] = { 12, 14, 16, 33, 65, 98, 45, 25, 87, 18, 20 };
show( num, sizeof num/sizeof num[0], "even numbers: ", 0 );
show( num, sizeof num/sizeof num[0], "odd numbers: ", 1 );
return 0;
}
Output:
even numbers:
12
14
16
98
18
20
odd numbers:
33
65
45
25
87
I have created a program to search for prime numbers. It works without problems until the entered number is smaller than 52, when it is bigger output prints out some blank (0) numbers and I don't know why. Also other numbers have blank output.
My code is:
#include <stdio.h> //Prime numbers
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <unistd.h>
int c[100], n, a[50], d, e, b = 1;
void sort() {
for (int i = 1; i < n; i++) {
if (c[i] > 1) {
a[b] = c[i];
printf("%d %d %d\n", a[1], b, i);
b++;
e = 2;
d = 0;
while (d <= n) {
d = c[i] * e;
c[d - 1] = 0;
e++;
}
}
}
}
int main() {
printf("Enter number as an limit:\n");
scanf("%d", &n);
for (int i = 0; i < n; i++) {
c[i] = i + 1;
}
sort();
printf("Prime numbers between 1 and %d are:\n", n);
for (int i = 1; i < b; i++) {
printf("%d ", a[i]);
}
return 0;
}
Here is output for 25:
Enter number as an limit:
25
2 1 1
2 2 2
2 3 4
2 4 6
2 5 10
2 6 12
2 7 16
2 8 18
2 9 22
Prime numbers between 1 and 25 are:
2 3 5 7 11 13 17 19 23
But for 83 is:
Enter number as an limit:
83
2 1 1
2 2 2
2 3 4
2 4 6
2 5 10
2 6 12
2 7 16
2 8 18
2 9 22
2 10 28
2 11 30
2 12 36
2 13 40
2 14 42
2 15 46
2 16 52
0 17 58
0 18 60
0 19 66
0 20 70
0 21 72
0 22 78
0 23 82
Prime numbers between 1 and 83 are:
0 3 5 7 11 0 17 19 23 29 31 37 0 43 47 53 0 61 67 71 73 79 83
Blank spots always spots after 17th prime number. And always the blank numbers are the same. Can you help me please what is the problem?
The loop setting entries in c for multiples of c[i] runs too far: you should compute the next d before comparing against n:
for (d = c[i] * 2; d <= n; d += c[i]) {
c[d - 1] = 0;
}
As a matter of fact you could start at d = c[i] * c[i] because all lower multiples have already been seen during the previous iterations of the outer loop.
Also note that it is confusing to store i + 1 into c[i]: the code would be simpler with an array of booleans holding 1 for prime numbers and 0 for composite.
Here is a modified version:
#include <stdio.h>
int main() {
unsigned char c[101];
int a[50];
int n, b = 0;
printf("Enter number as a limit:\n");
if (scanf("%d", &n) != 1 || n < 0 || n > 100) {
printf("invalid input\n");
return 1;
}
for (int i = 0; i < n; i++) {
c[i] = 1;
}
for (int i = 2; i < n; i++) {
if (c[i] != 0) {
a[b] = i;
//printf("%d %d %d\n", a[0], b, i);
b++;
for (int d = i * i; d <= n; d += i) {
c[d] = 0;
}
}
}
printf("Prime numbers between 1 and %d are:\n", n);
for (int i = 0; i < b; i++) {
printf("%d ", a[i]);
}
printf("\n");
return 0;
}
Output:
chqrlie$ ./sieve4780
Enter number as a limit:
25
Prime numbers between 1 and 25 are:
2 3 5 7 11 13 17 19 23
chqrlie$ ./sieve4780
Enter number as a limit:
83
Prime numbers between 1 and 83 are:
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79
Your problem seems to be caused by the fact that you have declared an array with size 50, but in fact it goes further than that: imagine you want to use Eratosthenes' procedure to find the first 10,000 prime numbers. Does this mean that you need to declare an array of size 10,000 first (or even bigger), risking to blow up your memory?
No: best thing to do is to work with collections where you don't need to set the maximum size at declaration time, like a linked list, a vector, ..., like that you can make your list grow as much as you like during runtime.
So this is my code for printing pascal triangle using 2d arrays but its not giving me the desired output and I cannot determine what's wrong with the logic/code.
#include <stdio.h>
int main()
{
int num, rows, col, k;
printf("Enter the number of rows of pascal triangle you want:");
scanf("%d", &num);
long a[100][100];
for (rows = 0; rows < num; rows++)
{
for (col = 0; col < (num - rows - 1); col++)
printf(" ");
for (k = 0; k <= rows; k++)
{
if (k == 0 || k == rows)
{
a[rows][k] = 1;
printf("%ld", a[rows][k]);
}
else
a[rows][k] = (a[rows - 1][k - 1]) + (a[rows - 1][k]);
printf("%ld", a[rows][k]);
}
printf("\n");
}
return 0;
}
You don't have curly braces around the statements after the else, so it looks like you'll double-printf() when the condition of the if-statement is true.
I copied the source into codechef.com/ide and changed the io for num to be just assigned to 6 which produced the following output:
Enter the number of rows of pascal triangle you want:
11
1111
11211
113311
1146411
1151010511
It looks like your close, but you want 1, 11, 121, 1331 etc right?
Wraping the else case produced the following output:
if (k == 0 || k == rows)
{
a[rows][k] = 1;
printf("(%ld)", a[rows][k]);
}
else{// START OF BLOCK HERE
a[rows][k] = (a[rows - 1][k - 1]) + (a[rows - 1][k]);
printf("(%ld)", a[rows][k]);
}//END OF BLOCK HERE, NOTE THAT IT INCLUDES THE PRINT IN THE ELSE CASE NOW
OUTPUT:
Enter the number of rows of pascal triangle you want:
(1)
(1)(1)
(1)(2)(1)
(1)(3)(3)(1)
(1)(4)(6)(4)(1)
(1)(5)(10)(10)(5)(1)
But i added () to make it clearer to me. I also added a "/n" to the end of the first printf that asks for the value of num, so that the first line is on a new line.
printf("Enter the number of rows of pascal triangle you want:\n");
You can do that without using any arrays:
#include <stdlib.h>
#include <stdio.h>
int num_digits(int number)
{
int digits = 0;
while (number) {
number /= 10;
++digits;
}
return digits;
}
unsigned max_pascal_value(int row)
{
int result = 1;
for (int num = row, denom = 1; num > denom; --num, ++denom)
result = (int)(result * (double)num / denom );
return result;
}
int main()
{
printf("Enter the number of rows of pascals triangle you want: ");
int rows;
if (scanf("%d", &rows) != 1) {
fputs("Input error. Expected an integer :(\n\n", stderr);
return EXIT_FAILURE;
}
int max_digits = num_digits(max_pascal_value(rows));
for (int i = 0; i <= rows; ++i) {
for (int k = 0; k < (rows - i) * max_digits / 2; ++k)
putchar(' ');
int previous = 1;
printf("%*i ", max_digits, previous);
for (int num = i, denom = 1; num; --num, ++denom) {
previous = (int)(previous * (double)num / denom );
printf("%*i ", max_digits, previous);
}
putchar('\n');
}
}
Output:
Enter the number of rows of pascals triangle you want: 15
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1
1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1
1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1
int main (void)
{
int range, i, j;
printf("Input size of multiplication table: ");
scanf("%i", &range);
int output[range][range];
for (i = 0; i<=range; ++i)
{
for (j=0; j<=range; ++j)
{
output[i][j] = i * j;
if(j!=range && output[i][j] != 0)
{
printf("%3i ", output[i][j]);
}
else if (j==range)
{
printf("%3i", output[i][j]);
}
else if (output[i][j] == 0)
{
printf("%3i "), i+2;
}
else
{
printf("%3i", j + i - range);
}
}
printf("\n");
}
return 0;
}
I am having it output:
0 1 2 3 4 0
5 1 2 3 4 5
10 2 4 6 8 10
15 3 6 9 12 15
20 4 8 12 16 20
25 5 10 15 20 25
and I need the 0 on the end to be a five and the first column to be 0,1,2,3,4,5 instead of 0,5,10,15,20,25.
If anyone could help I would appreciate it.
You have two primary problems, (1) you fail to validate your user input, and (2) your loop bounds are incorrect, e.g.
Any time you take user input, you must validate that you actually received what you expected and that any conversion required, was completed successfully. Failure to validate will lead to Undefined Behavior it invalid (or no) input is provide. (e.g. What if the user enters foo instead of 10?) When using scanf, you must validate the return which provides the count of the number of conversions that successfully took place, e.g.
printf ("Input size of multiplication table: ");
if (scanf("%i", &range) != 1) { /* VALIDATE ALL USER INPUT */
fprintf (stderr, "error: invalid input.\n");
return 1;
}
That is bare minimum. You can also check if the return is EOF to indicate the user canceled input with a [Ctrl+D] (or [Ctrl+Z] on windoze -- must be explicitly enabled on Win10).
Next, your loop bound are for (i = 0; i < range; i++) not i <= range, that invokes Undefined Behavior by attempting to access memory outside your array bounds. Simply fix the loop condition, e.g.
for (i = 0; i< range; i++) { /* fill multiplication table */
for (j = 0; j< range; j++) {
output[i][j] = (i + 1) * (j + 1); /* i+1 * j+1 */
}
}
Putting it altogether, you could do something similar to:
#include <stdio.h>
int main (void) {
int range, i, j;
printf ("Input size of multiplication table: ");
if (scanf("%i", &range) != 1) { /* VALIDATE ALL USER INPUT */
fprintf (stderr, "error: invalid input.\n");
return 1;
}
int output[range][range]; /* variable length array */
for (i = 0; i< range; i++) { /* fill multiplication table */
for (j = 0; j< range; j++) {
output[i][j] = (i + 1) * (j + 1); /* i+1 * j+1 */
}
}
for (i = 0; i< range; i++) { /* output table */
for (j = 0; j< range; j++)
printf (" %3d", output[i][j]);
putchar ('\n');
}
return 0;
}
note: the trivial parts of the table is omitted (e.g. 0 * anything), and duplicated rows of 1 * anything are also not shown. If you need to additional rows, you can add them back.
Example Use/Output
$ ./bin/multable
Input size of multiplication table: 10
1 2 3 4 5 6 7 8 9 10
2 4 6 8 10 12 14 16 18 20
3 6 9 12 15 18 21 24 27 30
4 8 12 16 20 24 28 32 36 40
5 10 15 20 25 30 35 40 45 50
6 12 18 24 30 36 42 48 54 60
7 14 21 28 35 42 49 56 63 70
8 16 24 32 40 48 56 64 72 80
9 18 27 36 45 54 63 72 81 90
10 20 30 40 50 60 70 80 90 100
Look things over and let me know if you have further questions.
To Show the 1X Rows
You simply update your print routine as follows:
for (i = 0; i< range; i++) { /* output table */
if (!i) {
printf (" ");
for (j = 0; j< range; j++)
printf (" %3d", output[i][j]);
putchar ('\n');
}
printf (" %3d", i + 1);
for (j = 0; j< range; j++)
printf (" %3d", output[i][j]);
putchar ('\n');
}
Example Use/Output
$ ./bin/multable1
Input size of multiplication table: 10
1 2 3 4 5 6 7 8 9 10
1 1 2 3 4 5 6 7 8 9 10
2 2 4 6 8 10 12 14 16 18 20
3 3 6 9 12 15 18 21 24 27 30
4 4 8 12 16 20 24 28 32 36 40
5 5 10 15 20 25 30 35 40 45 50
6 6 12 18 24 30 36 42 48 54 60
7 7 14 21 28 35 42 49 56 63 70
8 8 16 24 32 40 48 56 64 72 80
9 9 18 27 36 45 54 63 72 81 90
10 10 20 30 40 50 60 70 80 90 100
Check the solution below. It's all about properly managing iterators (note the less-equal sign in the ranges and the subtraction of indexes by 1 in the assignment). You can do it very concisely by assigning to output inside the printf and using a ternary-if. (I also used dynamic allocation in order to comply with ISO standards.)
#include <stdio.h>
#include <stdlib.h>
int main()
{
int range, i, j;
printf("Enter the size of the multiplication table:\n");
scanf("%d", &range);
int ** output = (int **) malloc(sizeof(int *) * (long unsigned int) range);
for(i = 0; i <= range; ++i)
{
output[i] = (int *) malloc(sizeof(int) * (long unsigned int) range);
for(j = 0; j <= range; ++j)
printf("%3d ", !i ? j : !j ? i : (output[i - 1][j - 1] = i * j));
printf("\n");
}
return 0;
}
I have a program that outputs a huge array of integers to stdout, each integer in a line. Ex:
103
104
105
107
I need to write another program that reads in that array and fill up the spaces where the number isn't an increment of 1 of the previous number. The only different between numbers is going to be 2 (105,107), which makes it easier.
This is my code to do that logic:
printf("d",num1);
if ((num2-num1) != 1)
numbetween = num1 + 1;
printf("%d", numbetween);
printf("%d", num2);
else(
printf("%d",num2);
)
So the output of this program will now be:
103
104
105
106
107
My issue is reading the numbers. I know I can do while (scanf("%hd", &num) != EOF) to read all the lines one at a time. But to do the logic that I want, I'm going to need to read two lines at a time and do computation with them, and I don't know how.
You could always just read the first and last numbers from the file, and then print everything in between.
int main( void )
{
// get the first value in the file
int start;
if ( scanf( "%d", &start ) != 1 )
exit( 1 );
// get the last value in the file
int end = start;
while ( scanf( "%d", &end ) == 1 )
;
// print the list of numbers
for ( int i = start; i <= end; i++ )
printf( "%d\n", i );
}
Read first num then add missing if needed when you read next int
#include <stdio.h>
#include <stdlib.h>
int main()
{
int previous = 0;
int num;
scanf("%hd", &previous);
while (scanf("%hd", &num) != EOF) {
for (int i = previous; i < num; i++) {
printf("%d\n" , i);
}
previous = num;
}
printf("%d\n" , previous);
return 0;
}
this input
100
102
103
105
107
110
returns this output
100
101
102
103
104
105
106
107
108
109
110
While you can read the first and last, to fill the range, what you are really doing is finding the min and max and printing all values between them inclusively. Below the names are left first and last, but they represent min and max and will cover your range regardless whether the values are entered in order. Taking that into consideration, another approach insuring you cover the limits of the range of int would be:
#include <stdio.h>
int main (void) {
int num = 0;
int first = (1U << 31) - 1; /* INT_MAX */
int last = (-first - 1); /* INT_MIN */
/* read all values saving only first (min) and last (max) */
while (scanf (" %d", &num) != EOF) {
first = num < first ? num : first;
last = num > last ? num : last;
}
/* print all values first -> last */
for (num = first; num <= last; num++)
printf ("%d\n", num);
return 0;
}
Input
$ cat dat/firstlast.txt
21
25
29
33
37
41
45
49
53
57
61
65
69
73
77
81
85
89
93
97
101
Output
$ ./bin/firstlast < dat/firstlast.txt
21
22
23
24
25
26
27
28
29
<snip>
94
95
96
97
98
99
100
101
Note: you can change the types to conform to your expected range of data.