#include < stdio.h >
#include < conio.h >
#define BILMAX 10
int main()
{
int num[BILMAX],i;
printf("insert 10 number and separated by space:\n");
for (i = 0;i<10;i++)
{
scanf("%d",&num[i]);}
printf("\n\nEven Number : \n");
for(i = 0; i < 10; i++)
{
if(num[i] % 2==0)
printf("%d",num[i]);
}
printf("\n\nOdd Number : \n");
for(i = 0;i < 10; i++)
{
if(num[i] % 2 !=0)
printf("%d",num[i]);
}
getch();
return 0;
}
my output is like this
insert 10 number and separated by space:
1 2 3 4 5 6 7 8 9 10
Even Number:
246810
Odd Number:
13579
i want my output like this
insert 10 number and separated by space:
1 2 3 4 5 6 7 8 9 10
Even Number:
2 4 6 8 10
Odd Number:
1 3 5 7 9
i want my output have a space.
Please help
If you want to print a space after each number, change this line:
printf("%d",num[i]);
to
printf("%d ",num[i]);
Notice the space after %d.
Space an be inserted in a printf just by leaving a simple blank between quotes: " "
Just like printing your first program "Hello World" (space between Hello and World). Whatever you type inside the code it gets printed as it is.
For printing a space after each number do this
printf("%d ",num[i]);
Space can be provided by using escape sequence \t printf("%d \t",num[i]);
Related
// printing in spiral order matrix
#include<iostream>
using namespace std;
int main(){
int n,m;
cin>>n>>m;
int arr[n][m];
for(int i=0; i<n; i++){
for(int j=0; j<m; j++){
cin>>arr[i][j];
}
}
// print
for(int i=0; i<n; i++){
for(int j=0; j<m; j++){
cout<<arr[i][j]<<" ";
}
cout<<endl;
}
// spiral print
int row_start=0,row_end=n-1,col_start=0,col_end=m-1;
while(row_start<=row_end && col_start<=col_end){
for(int j=col_start; j<=col_end; j++){
cout<<arr[row_start][j]<<" ";
}
row_start++;
for(int i=row_start; i<=row_end; i++){
cout<<arr[i][col_end]<<" ";
}
col_end--;
for(int j=col_end; j>=col_start; j--){
cout<<arr[row_end][j]<<" ";
}
row_end--;
for(int i=row_end; i>=row_start; i--){
cout<<arr[i][col_start]<<" ";
}
col_start++;
}
return 0;
}
My output is:
PS C:\Users\anmol\Desktop\c++projectwork> ./a
3 4
1 2 3 4 5 6 7 8 9 0 1 2
1 2 3 4
5 6 7 8
9 0 1 2
1 2 3 4 8 2 1 0 9 5 6 7 6
I am getting an extra '6' at the end.
which is not required however this type of problem only come when matrix is rectangular.
But code works fine for square matrix.
Please tell me where i went wrong..
Suppose you have a below matrix. Let's dry run your code on below example.
1 2 3 4
5 6 7 8
9 10 11 12
Dry Run
1st for loop: prints 1 2 3 4 row_start=1,row_end=2,col_start=0,col_end=3
2nd for loop: prints 8 12 row_start=1,row_end=2,col_start=0,col_end=2
3rd for loop: prints 11 10 9 row_start=1,row_end=1,col_start=0,col_end=2
4th for loop: prints 5 row_start=1,row_end=1,col_start=1,col_end=2
All condition of while loop are true 2nd Iteration of while loop:
1st for loop: prints 6 7 row_start=2,row_end=1,col_start=1,col_end=2
2nd for loop: won't do anything row_start=2,row_end=1,col_start=1,col_end=1
3rd for loop: prints 6
DO you see the problem?
you should run 3rd for loop only if
"row_start < row_end"
similarly you should check for
"col_start<col_end"
before running the 4th loop
I have to write a program that takes a completed sudoku board, saves only the numbers (meaning all the symbols used between the numbers to separate them such as '-', '|' etc cant be saved) into a two-dimensional array.
#include <iostream>
#include <string>
using namespace std;
int main()
{
int input[11] = { 0 };
int sudoku[9][9] = { 0 };
for (int line = 0; line <= 10; line++)
{
cin >> input[line];
}
system("PAUSE");
return 0;
}
This is the only working code I've got so far. I've tried different kinds of for loops to get this done but I can't figure why it doesn't work.
So I wanted to ask, is it even possible save all the numbers of a string into a multi-dimensional array? And if it's not, where is my approach wrong or how could I solve this task?
One example of the input would be:
.5.1.4.|.8.6.9.|.7.2.3
.8.7.2.|.3.4.5.|.6.1.9
.9.6.3.|.2.1.7.|.5.4.8
-------|-------|-------
.6.2.8.|.1.3.4.|.9.5.7
.1.9.7.|.6.5.2.|.8.3.4
.4.3.5.|.7.9.8.|.1.6.2
-------|-------|-------
.2.4.6.|.9.7.1.|.3.8.5
.7.5.1.|.4.8.3.|.2.9.6
.3.8.9.|.5.2.6.|.4.7.1
One approach is to use regular expressions. This way the formatting of the sudoku board can change but your will still be able to parse out the numbers.
The reason I broke it into two for loops was to easily ignore the row that has no numbers in it.
#include <iostream>
#include <regex>
#include <string>
#include <vector>
int main()
{
std::string line;
// this regular expression matches a single digit
std::regex exp("(\\d)");
std::smatch res;
int sudoku[9][9] = {{0}};
int row = 0;
for (int i = 0; i < 3; ++i)
{
for (int j = 0; j < 3; ++j)
{
// get a line of the board
std::getline(std::cin, line);
// search for the next digit in the line
for (int k = 0; std::regex_search(line, res, exp); ++k)
{
// convert the digit into an integer and store it in the board
sudoku[row][k] = std::stoi(res[0]);
// the rest of the line after the first match becomes the new
// line so that we can search for the next digit
line = res.suffix();
}
row += 1;
}
// ignore every third row that is used to separate the board sections
std::getline(std::cin, line);
}
for (int i = 0; i < 9; ++i)
{
for (int j = 0; j < 9; ++j)
{
std::cout << sudoku[i][j] << " ";
}
std::cout << std::endl;
}
return 0;
}
For your example board, it produces this output:
5 1 4 8 6 9 7 2 3
8 7 2 3 4 5 6 1 9
9 6 3 2 1 7 5 4 8
6 2 8 1 3 4 9 5 7
1 9 7 6 5 2 8 3 4
4 3 5 7 9 8 1 6 2
2 4 6 9 7 1 3 8 5
7 5 1 4 8 3 2 9 6
3 8 9 5 2 6 4 7 1
I am trying to create an array by taking value 'n' from the console and create an array with 'n' inits and then again take a value 'r' to work.
so far I wrote
int main(){
int n = 0;
cin >> n;
int* a = new int[n];
for(int i = 0; i< sizeof(a);i++){
cin >> a[i];
}
for(int y = 0; y < sizeof(a);y++){
cout << a[y] << " ";
}
int r = 0;
cin >> r;
rotate(a,r);
(the "cout" part is for checking the output of the array)
but no matter I try I would get an array which length doesn't equal the input 'n'. Can anyone give me some advice on it?
Here the outputs for every input from console:
(the second row is supposed to be the created array)
INPUT
6
1 2 3 4 5 6
3
OUTPUT
1 2 3 4 5 6 3 0
INPUT
10
-1 -2 3 4 5 -6 7 -8 9 0
5
OUTPUT
-1 -2 3 4 5 -6 7 -8
INPUT
1
1
1
OUTPUT
1 1 0 0 0 0 135137 0
INPUT
5
1 2 3 4 5
5
OUTPUT
1 2 3 4 5 -3 135137 0
Any ideas why those unexplainable numbers at the end?
As Algirdas said, take a close look at what SizeOf does. Also, you don't really need it. You can make it work like this:
for(int i = 0; i< n; i++){
cin >> a[i];
}
As you've got 'n' elements in your array.
Also, I know that most textbooks are really fond of arrays, but please follow Cody Gray's advice!
I'm having trouble with my C++ programming class homework.
Here is the assignment:
Write a program which reads in the number of rows and prints 'n' rows
of digits.
1
12
123
1234
12345
where the i’th row is 1234 . . . k where k = i mod 10. If a row
has more than 10 digits, the digit after 9 should start again from 0.
For instance, if the number of rows is 15, the output should be: 1
12
123
1234
12345
123456
1234567
12345678
123456789
1234567890
12345678901
123456789012
1234567890123
12345678901234
123456789012345
This is what I have so far:
#include <iostream>
using namespace std;
int main()
{
int rows(0);
int i(0);
int j(0);
cout << "Enter number of rows: ";
cin >> rows;
int k=rows;
i=1;
while (i <= rows)
{
j=1;
while(j <= i)
{
cout << j;
j++;
}
cout << endl ;
i++;
}
return (0);
}
This works perfectly until I get to the 10th row. I'm not sure how to get the counter to reset back to 0 and go 1-9 again. I'm guessing an if statement, but I don't know how to correctly implement it. Any help would be greatly appreciated.
Using cout << j%10 will always print the ones digit of whatever j is equal to. So when j = 10, cout << j%10 will print 0, etc.
To expand on the other answer a bit:
% is referred to as the remainder operator. It is also sometimes called "modulo" or "modulus". It returns the remainder of the first argument divided by the second argument. So for this situation, since you want the number to be between 0 and 9, consider what happens when you have n % 10.
8 / 10 == 0 remainder 8
9 / 10 == 0 remainder 9
10 / 10 == 1 remainder 0
11 / 10 == 1 remainder 1
12 / 10 == 1 remainder 2
Etc.
In C++, as in most programming languages, you typically start counting at 0 and you exclude the final number in your range. So if you had n % 56, your output starting at 0 would go up to 55, and then reset to 0 again. The classical argument for this was written by Dijkstra:
http://www.cs.utexas.edu/users/EWD/transcriptions/EWD08xx/EWD831.html
Please help me answer this question , void Function with parameters. I do not understand it very well
I want to write a program so that the user can input both characters that form the pattern
I need to define two (characters) variables which are (star) and (next), the statement must include three parameters – one that will be an (int) for the pattern
and two that will be (characters) for the character to display the pattern.
I have to change the function (drawpattern) to have three values parameters
an (int) parameter indicating the size of the pattern and two (characters) indicating the character that will be used to display the pattern .
input (4) for size, (Y) for the character to be used to start the pattern and (+) for the second and every alternate group
This is what I have done so far and I do not now if is okay:
#include <iostream>
using namespace std;
void drawPattern(int size, char start, char next)
{
for (int i = 0; i <= size; i++)
for (int j = 0; j <= size; j++)
{
if ((i / size) % 3 == 0)
if ((j / size3) % 3 == 0)
cout << '4';
else
cout << 'Y';
else
if ((j / size) % 3 == 0)
cout << '+';
else
cout << '4';
}
}
int main ()
{
int size;
char start, next;
cout << "Please enter number ( 4 ) for the size of the pattern : ";
cin >> size;
cout << " Now enter leter ( Y ) to start the pattern: ";
cin >> start;
cout << "Lastly enter the ( + ) for the other pattern: ";
cin >> size;
cout << " This is the output pattern: " << endl;
drawPattern(size, start, next);
return 0;
}
It looks like you are hard coding your output in the drawPattern function and ignoring the actual input from the user. I think you should probably replace the 'Y' and '+' with the corresponding argument passed to the function, since I'm pretty sure the professor would not be happy about hard coded values.
At line 11, you have a typo. It says size3 where it should say size.
You are also making a logical mistake.
This code
if ((i / size) % 3 == 0)
is wrong in concept. At least as the code is written now. Normally you would write like this:
if (i % 3 == 0)
This if statement will be true every third row. The % (modulu) calculates the remainder of the integer (whole number) division firstnumber / secondnumber. If you have a sequence of i going from 0 to 10, this is what i % 3 outputs
i: 0 1 2 3 4 5 6 7 8 9 10
i%3: 0 1 2 0 1 2 0 1 2 0 1
As you can see, i % 3 == 0 is true when i is divisible by 3.
Your code does something different. Let's say size = 10. Then you calculate the (integer) division i / size. Lastly you calculate (i / size) % 3. However i is always less than size, except at the last turn of the loop. Let's look at the values again:
i: 0 1 2 3 4 5 6 7 8 9 10
size: 10 10 10 10 10 10 10 10 10 10 10
i/size: 0 0 0 0 0 0 0 0 0 0 1
Since the value of i / size only changes once, the calculation (i / size) % 3 is meaningless.