print a matrix in a specific sequence 
hello all ,
i want to print a matrix like this 
a=[5
 4
 3
 2
 1
 0
 5
 4
 3
 2
 1
 0 
 .
 .];

i want numbers from 1 to 5 to be repeated in this form several times
you can use the repmat function here
if you have to print the patter n (here 10) times, use the following code:
a = [5 4 3 2 1 0]';
n = 10;
b =repmat(a, n,1)
Hi Valdeth. You should use input and output file streams if you want to write and read from a file. Basically you open the file for writing and once you are done close the file. Later you can open the file for reading. The example below reads numbers from the file and just prints them out. If you want to read the file line by line you should look at the getline() function. I have just used your initial array to come up with a result per array which gives the product of all negative numbers in the array. Note that there **HAS** to be at least 1 negative number per array.



```
const int m=4, n=5;
	int i,j;
	int num;
	int result;
	int A[m][n]={{5,3,-4,8,2},
				 {1,5,2,7,-8},
				 {-3,9,5,2,4},
				 {4,-1,3,6,8}};

	ofstream outfile;
	ifstream infile; 

	outfile.open("c:/matrix/matrix.txt");

	for (i=0;i<m;i++)
	{
		result = 1;

		for (j=0;j<n;j++)
		{
			outfile << setw(3)<< A[i][j];

                            // Multiply negative numbers in each array (has to be at least 1 negative number in each array)
			if (A[i][j] < 0)              				                      result *= A[i][j];  // result = result x A[i][j]
		}
		cout << "Result of array " << i << " is " << result << endl;
		outfile << "\n";
	}

	outfile.close(); // Close for writing

	infile.open("c:/matrix/matrix.txt"); // opens the file
    
	infile >> num;
	while ( !infile.eof() ) 
	{ // keep reading until end-of-file
		cout << "The next number is " << num << endl;
		infile >> num; // sets EOF flag if no value found
	}
	infile.close();  //Close for reading
``` ```
[b]#include[/b] <stdio.h> 

[b]void[/b] [b]foo[/b]([b]int[/b] n)
{
   [b]if[/b] ( n >= 0 )
   {
	  [b]printf[/b]("%d ", n);
	  [b]foo[/b](n - 1);
   }
}

[b]void bar[/b]([b]int[/b] n)
{
   [b]if[/b] ( n >= 0 )
   {
	  [b]bar[/b](n - 1);
	  [b]printf[/b]("%d ", n);
   }
}

[b]int main[/b]() 
{
   [b]int[/b] value = 5;
   [b]foo[/b](value); [b]putchar[/b]('\n');
   [b]bar[/b](value); [b]putchar[/b]('\n');
   [b]return[/b] 0; 
}

/* my output
5 4 3 2 1 0 
0 1 2 3 4 5 
*/
``` c = input()  

t = len(c)  

temp = ""  

for i in range(0,len(c)):  

    if t<3:  

        break  

    else:  

        if i in (0,1,t-2,t-1):  

            temp = temp+c[i]  

        else:  

            continue  

print(temp) finding the correct answer here is important, but perhaps equally important is finding it using the method of matrix reduction, since those are the instructions.the augmented matrix is4 -2 -5 -514 -1 -2 -364 -2 4 -24you'll have to forgive the poor formatting. this answer box doesn't have a proper math symbol editor.using elementary row operations, the reduced row echelon form of the augmented matrix is1 0 0 -60 1 0 60 0 1 3from which we conclude that the system has a unique solution (-6,6,3). >printf("%c", len\_char); //<this does not print anything too!!  

What did you expect it to print? You realize that the %c format modifier takes the integer value you supply and prints the character representation of that integer, right? In other words, it's not going to print '3'...ever. As long as the lengths are single digits, you can say len\_char + '0' and get the correct representation, but once the length exceeds 9, you're SOL with this method. 1 a c d
2 a
3 c
4 b
5 c
i got a 100% [@cssc](/u/cssc), another option would be to take the regex way.

first, it finds the index of the 5th pipe.

then, it takes the substring until that index.

give this a try:

inputstr.substring(0, regex.matches(inputstr, "\|")(4).index + 1) 10.1 sequences polynomials: easy to evaluate, integrate, diﬀerentiate general functions: -a sequence is a list of #s -sequences are written as: -many series correspond to functions formally, the sequence converges to l if for every positive number , there corresponds an integer n such that for all I'm not going to go through all of your code because I'm in a hurry, but this should simplify it quite a bit. At a minimum it will make it easier to read.



```
alphabet = ['A','B','C','D'....etc]
count = 0
infile = open('infile.txt','r')
for character in infile.readlines():
    if character in alphabet:
        count += 1

print count
``` direction: a word and number arrangement machine when given an input line of word and numbers rearranges them following a particular rule in each step. the following is an illustration of input and rearrangement. input: politics 61 policy 116 partner 93 role 45 clear 27 38 point step i: role politics policy 116 partner 93 45 clear 27 38 point 61 step ii: role clear politics policy partner 93 45 27 38 point 116 61 step iii: role clear partner politics policy 93 45 38 point 27 116 61 step iv: role clear partner politics policy 93 38 point 45 27 116 61 step v: role clear partner politics point policy 93 38 45 27 116 61 step v, is the last step of the above arrangement as the intended arrangement is obtained. as per the rule followed in the given steps find out the appropriate steps for the given input. input: energy 54 78 place 23 issue 211 145 flat rent taxi 17in step iii, what is the difference between fifth element from left and second element from right?