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Real Pacquiao-Marquez Live stream 2011

Saturday, November 12, 2011 Posted by Glenn Von Posadas 0 comments

Main Stream:

If neither of the streams are working,
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Other Live Stream:










Above is a free live streaming of the fight between Manny Pacquiao and Juan Manuel Marquez. No copyright violations intended. The video is owned by a third party video sharing website.

View in full screen to avoid advertisements from popping-out. If video pauses, hit F5 or click refresh in your browser or try other channels located in the sidebar.


Watch the Pacquiao vs Marquez free live streaming by simply sharing this website on your Facebook, Twitter and/or Google+. The live stream here is intended to show the boxing fans who will win between Pacquiao and Mayweather. It is rumored that the winner of the Pacquiao vs Marquez fight will fight Manny Pacquiao next.

Share first before watching Pacquiao vs Marquez 3 live streaming.

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Convert Password input into Asterisk in C, C++

Posted by Glenn Von Posadas 0 comments

Here is my source code about converting the inputted password into asterisk * in C and C++




#include "iostream.h"
#include "conio.h"
#include "cstring.h" // for strcmp

using namespace std;
int main(void)
{
int c;
int i;
char pw[80];
start:

printf("\n1Enter Password:");

for ( i = 0; i < 79 && (c = getch()) != '\r'; ++i ) {
pw[i] = c;
putch('*');
}

pw[i] = '\0';


if ( strcmp(pw, "CORRECT") == 0 ) {
printf("\nPASSWORD CORRECT");
} else {
goto start;
}
getch();

return 0;
}

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Customized Shoes by Mcki Took

Monday, November 7, 2011 Posted by Glenn Von Posadas 0 comments

Customized Shoes by Mcki Took
Link
Price Range: P1.5k to 3k (depends on design and color scheme)

Contact Mcki for more info

Facebook: click here
Contact #: +63915-3060879



-------------SAMPLES------------
(note: click the photo to enlarge)








































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Compute Pi using Assembly Language 8086

Posted by Glenn Von Posadas 0 comments

Source Codes:

compiled using any 8086 Assembler/Emulator
.radix 10
.model tiny
.code

; This program uses almost all of the 64K code segment. There might
; be rare situations where the last 4 digits of the printout are
; wrong, but this is less likely than in the previous version.
; The last 4 digits should be 1224. (9,304 digits of pi)

; Program assumes there is 64K of memory available for the program,
; without checking. In the rare case where there isn't 64K
; available, it may crash the computer.
; i.e. don't LOADHIGH this program.

org 100h
start:
mov cx,2326*14 ;32564 ; this MUST be a multiple of 14
; might be possible to increase
; to 2327*14, but that last 4 digits
; is much more likely to be corrupted
; and/or corrupt the rest of the array
; since that ends the scratch area
; at FFF1h.
mov di,Offset Buffer
mov bx,cx
mov ax,2000 ; initial value of the array
mov bp,10000 ; keeping 10000 constant in extra register
rep stosw ; wipes the buffer used.
; offset 0172h through offset FFD9

; FFDA-FFFE is stackspace for 18 words.
; the workspace shrinks by an additional 14 words
; before calling an INT. This program is smaller
; than previous version by 12 bytes, meaning that
; stack is larger by 6 words, making it less likely
; that the stack will overrun the workspace.

; cx = 0000, which is the original value of the remainder.

; bx is decremented from 32564 by 14's. Lesser value would
; not have correct values toward the low end of the output.
; outer-counter is "i", inner-counter is "j"
; because counter is decremented at the end, array is accessed
; by subtracting 2 from the counter.

OuterLoop:
xor ax,ax
push bx
cwd ; dx:ax initialized as 0000:0000

InnerLoop:
push bx
mov si,dx ; save dx thru the multiply
mul bx ; bx * ax -> dx:ax
xchg si,ax ; get the original dx to multiply it
push dx
mul bx
pop dx
add ax,dx
xchg di,ax ; dx:ax * bx -> di:si

mov ax,bp ; ax = 10,000
add bx,bx ; align for 2-byte array elements
mul word ptr [bx+offset Buffer-2]
; because bx is decremented at the end,
; must subract 2 to point at correct item
add ax,si
adc dx,di ; dx:ax = dx:ax + di:si
dec bx ; bx = 2*j -1
xor di,di
xchg di,ax
xchg dx,ax
div bx ; dx = 0, ax = high part of prev di:si
; dx:ax / bx -> ax, dx=remainder
xchg di,ax
mov si,dx
xor dx,dx
div bx ; dx = 0, ax = low part of prev di:si
xchg si,ax
xchg dx,ax
div bx
mov [bx+offset buffer-1],dx ; because bx was decremented
; above, just subtract 1 to
; access correct element.
xor dx,dx
add ax,si ; si:di + ax -> dx:ax
adc dx,di
pop bx
dec bx
jnz InnerLoop

div bp ; div by 10,000 - resulting mod has 4 decimal digits

push dx ; save remainder for use in next loop.
add ax,cx
mov bl,10 ; dividing by 10 yields decimal digits.
mov cx,4

Num1:
cwd ; "xor dx,dx" not needed, value is always < 7fff
div bx ; divide by 10
push dx
loop Num1

mov cl,4

Num2:
; pop dx
; add dl,"0"
; mov ah,2
; int 21H

pop ax ; these 3 lines save 3 bytes compared to
add al,"0" ; the previous 4 lines, but the output
int 29H ; can't be redirected to a file

loop Num2

pop cx ; restore remainder pushed in DX above
pop bx ; restore outer-loop's value
sub bx,+14 ; decrement by 14.

; jnz OuterLoop
jnbe OuterLoop ; changed from JNZ in case someone
; didn't make BX a multiple of 14.

ret ; .COM always has 0000 on the stack.
; instruction at offset 0000 is always
; an INT 20H, which ends the program.

ProgramEnd:
Buffer=$ ;+1 ;align even without adding a byte

end start
; This is offset 016Eh. Next 32564 words through FFD5h
; are used as workspace.


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How to put loading design in C++

Posted by Glenn Von Posadas 0 comments

Wonder how to put a loading design in C++? The ASCII Table will be necessary in your design.

Here is my example:


and Here is my code:



/*I got this idea from Engr. Billy Jo from MAPUA. :"> */

cout<<"\n\n\n\n\n\t\t\t Loading...Please Wait...";
cout<<"\t\t\t";
gotoxy(3,9);
printf("\t\t\tÉÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍ»");
gotoxy(3,10);
printf("\t\t\tº°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°º");
gotoxy(3,11);
printf("\t\t\tÈÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍͼ");

for(int k=25;k<58;k++)
{
Sleep(90);
gotoxy(k,10);
printf("±");
}

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String Initials in C++

Posted by Glenn Von Posadas 0 comments

#include "iostream.h"
#include "conio.h"

using namespace std;
int main(void)
{
string f,m,l;
//string testing..getting the initials
cout<<"ENTER YOUR FULL NAME (first middle and lastname)";
cin>>f>>m>>l;
string initals = f.substr(0,1)
+ m.substr(0,1) + l.substr(0,1);
cout<<"Your initials are: "<

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