someone plz help me..

[rcplguy15]

Write a C++ program that takes 2 characters (c and d), and a positive integer (n) and outputs the following
drawing of size n n made of the characters c and d, as shown below:
Enter two characters and an integer: # _ 6
#_#_#_
_#_#_#
#_#_#_
_#_#_#
#_#_#_
_#_#_#
Press any key to continue . . .

this is the question..
I did till here but I don't know how to continue
#include <iostream>
using namespace std;

int main()
{
int n;
char c, d;
cout << "Enter two characters and an integer: " << flush;
cin >> c >> d >> n;

someone plz help me..thanks...:S....

[sfuo]

I came up with a way to output the characters and semi commented what is going on.

#include <iostream>

using namespace std;

int main()
{
    int n;
    char in[2]; //change to an array (see output)
    cout << "Enter two characters and an integer: " << flush;
    cin >> in[0] >> in[1] >> n;

    for( int i = 0; i < n; i++ )
    {
    	for( int c = 0; c < n; c++ )
    		cout << in[((c%2)+(i%2))%2]; //output the character based on the row and col and then mod by 2 so it does not leave the array's range
    	cout << endl;
    }
    return 0;
}

[burgercho]

This should do it:

bool toggle = 1;
for(int i=0;i<n;i++) {
    for(int y=0;y<n;y++) {
            if(toggle) cout << c;
            else cout << d;
            toggle = !toggle;
    }
    cout << endl;
}

[sundip]

void print(char c,char d,int n)
{
    for(int i = 0;i<n;i++)
    {
       for(int j = 0;j<n;j++)
       {
           printf("%c%c",c,d);
       }
       char temp = c;
       c = d;
       d = temp;
       printf("\n");
    } 
}

[mrnutty]

Well since everyone is posting their solution, might as well post mines :

void printPattern(char ch1, char ch2, const int MAX){	
	for(int r = 0; r != MAX; ++r){
		for(int c = 0; c != MAX; ++c){
			cout << (c % 2 == 0 ? ch1 : ch2 ) ;
		}
		cout << endl;
	}
}

[sfuo]

Well since everyone is posting their solution, might as well post mines :

void printPattern(char ch1, char ch2, const int MAX){	
	for(int r = 0; r != MAX; ++r){
		for(int c = 0; c != MAX; ++c){
			cout << (c % 2 == 0 ? ch1 : ch2 ) ;
		}
		cout << endl;
	}
}

Problem here is that it does not change based on the rows. For even rows it should output the 1st char and odd ones should output 2nd char.