Printing A - Revision history

Arjun Suresh at 14:25, 23 July 2014

2014-07-23T14:25:13Z

Arjun Suresh at 16:51, 28 June 2014

2014-06-28T16:51:59Z

Arjun Suresh at 16:48, 28 June 2014

2014-06-28T16:48:50Z

Arjun Suresh at 16:45, 28 June 2014

2014-06-28T16:45:03Z

Arjun Suresh at 16:44, 28 June 2014

2014-06-28T16:44:37Z

Arjun Suresh at 16:43, 28 June 2014

2014-06-28T16:43:21Z

Arjun Suresh at 16:31, 28 June 2014

2014-06-28T16:31:40Z

Arjun Suresh at 16:31, 28 June 2014

2014-06-28T16:31:23Z

Arjun Suresh at 10:40, 28 June 2014

2014-06-28T10:40:51Z

Arjun Suresh at 20:15, 26 June 2014

2014-06-26T20:15:21Z

← Older revision		Revision as of 14:25, 23 July 2014
Line 186:		Line 186:
	{{Template:FBD}}		{{Template:FBD}}

−	~~[[Category:Arrays]]~~
	[[Category:Placement Coding Questions from Arrays]]		[[Category:Placement Coding Questions from Arrays]]

@@ Line 24: / Line 24: @@
 for 1<= n <= 6
-The recurrence relation is drawn from the fact that, the optimal solution at position $n$ will be obtained by CTRL-A, CTRL-C, CTRL-V sequence starting at either $(n-3)^{th}$  position or $(n-4)^{th}$ position or $(n-5)^{th}$ position. We needn't consider this sequence starting at $(n-6)^{th}$ position because that will mean the sequence repeating again at $(n-3)^{th}$ position which we have considered. Each CTRL-V puts the same number of characters as was during which the previous CTRL-A was typed. So, if we take $sol[n-3]$, that would mean CTRL-A starting at $n-2$, CTRL-C at $n-1$ and CTRL-V at $n$. So, $sol[n-3]$ is doubled at $n$. If we take $sol[n-4]$, there will be CTRL-V at positions $n-1$ and $n$, and so $sol[n]$ becomes $3$ times $sol[n-4]$. Similarly, if we take $sol[n-5]$, there will be 3 CTRL-V at $n-2$, $n-1$ and $n$ and $sol[n]$ becomes 4 times $sol[n-5]$.
+The recurrence relation is drawn from the fact that, the optimal solution at position $n$ will be obtained by CTRL-A, CTRL-C, CTRL-V sequence starting at either $(n-3)^{th}$  position or $(n-4)^{th}$ position or $(n-5)^{th}$ position. We needn't consider solution at position $(n-6)$ because that is already considered while calculating the solution at position $(n-3)$ which we are considering. Each CTRL-V prints the same number of characters as was on the screen during which the previous CTRL-A was typed. So, if we take $sol[n-3]$, that would mean CTRL-A starting at $n-2$, CTRL-C at $n-1$ and CTRL-V at $n$. So, $sol[n-3]$ is doubled at $n$. If we take $sol[n-4]$, there will be CTRL-V at positions $n-1$ and $n$, and so $sol[n]$ becomes $3$ times $sol[n-4]$. Similarly, if we take $sol[n-5]$, there will be 3 CTRL-V at $n-2$, $n-1$ and $n$ and $sol[n]$ becomes 4 times $sol[n-5]$.
 <syntaxhighlight lang="c" name="printa">

@@ Line 35: / Line 35: @@
+}
-//Returns 1 for a > (b,c), 2 for b > (a,c) and 3 for c > (a,b)
+//Returns 1 if a = max(a,b,c), 2 if b = max(a,b,c) and 3 if c = max(a,b,c)
 int maxp(int a, int b, int c)
+{
@@ Line 111: / Line 111: @@
 #include <stdio.h>
-//Returns max of a,b,c
+//Returns max of a,b,c,d
 int max(int a, int b, int c, int d)
+{
@@ Line 118: / Line 118: @@
-//Returns 1 for a > (b,c), 2 for b > (a,c) and 3 for c > (a,b)
+//Returns 1 if a = max(a,b,c,d), 2 if b = max(a,b,c,d), 3 if c = max(a,b,c,d) and 4 if d = max(a,b,c,d)
 int maxp(int a, int b, int c, int d)
+{

@@ Line 174: / Line 174: @@
 			printf("CTRL-A\nCTRL-C\nCTRL-V\nCTRL-V\nCTRL-V\nCTRL-V\n");
                          break;
-        case 4:
+                case 4:
      		        print(prin, n-7);
 			printf("CTRL-A\nCTRL-C\nCTRL-V\nCTRL-V\nCTRL-V\nCTRL-V\nCTRL-V\n");

@@ Line 173: / Line 173: @@
 			print(prin, n-6);
 			printf("CTRL-A\nCTRL-C\nCTRL-V\nCTRL-V\nCTRL-V\nCTRL-V\n");
-            break;
+                        break;
          case 4:
-    		print(prin, n-7);
+    		        print(prin, n-7);
 			printf("CTRL-A\nCTRL-C\nCTRL-V\nCTRL-V\nCTRL-V\nCTRL-V\nCTRL-V\n");
+	}

← Older revision		Revision as of 16:43, 28 June 2014
Line 99:		Line 99:
	</syntaxhighlight>		</syntaxhighlight>

		+	===Alternate Solution===

	The above solution assumes that after CTRL-C, we can do a mouse click as otherwise CTRL-V will override whatever was selected on the screen. If mouse click is not allowed, recurrence relation should be modified as:		The above solution assumes that after CTRL-C, we can do a mouse click as otherwise CTRL-V will override whatever was selected on the screen. If mouse click is not allowed, recurrence relation should be modified as:

@@ Line 20: / Line 20: @@
 A key to the solution is that we can continue pressing CTRL-V multiple times and the characters in the buffer will get printed continuously. The recurrence relation for the solution is
-  sol[n] = max(2*sol[n-3], 3*sol[n-4], 4*sol[n-5]) | n > 6
+  sol[n] = max(2 * sol[n-3], 3 * sol[n-4], 4 * sol[n-5]) | n > 6
 and
 for 1<= n <= 6

@@ Line 96: / Line 96: @@
+	}
+}
 </syntaxhighlight>