Wednesday, June 29, 2016

Cyberwort

CYBERWORT

Private Forms:

Click here

-------------------------------------

Public Docs:

Wordbook: Click here

Important links:  Click here

Bookshelf: Click here

Code: Click here

Wednesday, June 22, 2016

BOUNCING STAR!!! (COMPLETED!!)


SCREENCASTING:

Code:

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#include <stdio.h>
#include <windows.h>
#include <math.h>

void h_newl(int h){
	while(h--){
		printf("\n");
	}
}
void stars(){
	printf("*\n");
}


int main(){
    int c;
    printf("Number of drops: ");
    scanf("%d",&c);
    while(c--){
	int x;
	float t;
	/* drop down */
	for ( x=1; x<=11; ++x){
		 t= 700*(sqrt (2*x/9.8) - sqrt (2*(x-1)/9.8));
		Sleep((int)t);
		system("cls");
		h_newl(x-1);
		stars();
	}
	/* drop up */
	for ( x=11; x>=1; --x){
		t= 700*(sqrt (2*x/9.8) - sqrt (2*(x-1)/9.8));
		if(x!=11)Sleep(t);
		system("cls");
		h_newl(x-1);
		stars();
	}
    }
	return 0;
}

Monday, May 2, 2016

UVa problem-12577 (Solved)


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Labayk Allahuma Labayk. Labayk La shareeka laka Labayk. Innal hamda wannimata laka wal mulk. La shareeka Lak
(Here I am at your service, oh Lord, here I am - here I am. No partner do you have. Here I am. Truly, the praise and the favor are yours, and the dominion. No partner do you have.)
These are the words chanted by some two million people from world heading, as if pulled by a magnet, to one single spot on Earth. As has happened every year for 14 centuries, Muslim pilgrims gather in Makkah to perform rituals based on those conducted by the Prophet Muhammad during his last visit to the city.
Performing these rituals, known as the Hajj, is the fth pillar of Islam and the most signi cant manifestation of Is- lamic faith and unity. Undertaking the Hajj at least once is a duty for Muslims who are physically and nancially able to make the journey to Makkah. The emphasis on nancial ability is meant to ensure that a Muslim takes care of his family rst. The requirement that a Muslim be healthy and physically capable of undertaking the pilgrimage is intended
to exempt those who cannot endure the rigors of extended travel.
The pilgrimage is the religious high point of a Muslim's life and an event that every Muslim dreams of undertaking. Umrah, the lesser pilgrimage, can be undertaken at any time of the year; Hajj, however, is performed during a ve-day period from the ninth through the thirteenth of Dhu Al-Hijjah, the twelfth month of the Muslim lunar calendar.
It is generally presumed that the Hajj performed on Friday is called `Hajj-e-Akbar' and it is a superior kind of Hajj as compared with the Hajj performed on other days of the week.
But, the correct meaning of the term, as explained by a large number of the commentators of the Holy Quran is that the Umrah, which can be performed at any time throughout the year, was generally called `Hajj-e-Asghar' (the minor Hajj). In order to distinguish hajj from Umrah the former was named `Hajj-e-Akbar' (the greater hajj). Therefore, each and every hajj is Hajj-e-Akbar, no matter whether it is performed on Friday or on any other day. The word `Akbar' (greater) is used only to distinguish it from Umrah which is a minor Hajj.
Input
There will be several lines in the input terminated with a line containing a single `*'. This last line should not be processed. Each of the lines will contain either Hajj or Umrah.
Output
For each line of the input, output either `Hajj-e-Akbar' or `Hajj-e-Asghar' in separate lines without quotations. For exact format refer to the sample.
Sample Input
Hajj Umrah Hajj Umrah
*
Sample Output
Case 1: Hajj-e-Akbar
Case 2: Hajj-e-Asghar
Case 3: Hajj-e-Akbar
Case 4: Hajj-e-Asghar

Code:

#include <stdio.h>
#include <string.h>
int main(){
    int i=1;
    char inp[100];
    while(scanf("%s", inp)!=EOF){
    if(inp[0]!='*'){
        if (strcmp(inp,"Hajj")==0){
            printf("Case %d: Hajj-e-Akbar\n",i);
        }else if (strcmp(inp,"Umrah")==0){
            printf("Case %d: Hajj-e-Asghar\n",i);
        }
            i++;
    }else{
    return 0;
 }
    }
    return 0;
}

Tuesday, April 26, 2016

UVa problem-100(Solved)

The 3n + 1 problem


 The 3n + 1 problem 

Background

Problems in Computer Science are often classified as belonging to a certain class of problems (e.g., NP, Unsolvable, Recursive). In this problem you will be analyzing a property of an algorithm whose classification is not known for all possible inputs.

The Problem

Consider the following algorithm:

 
  1.    input n

2. print n

3. if n = 1 then STOP

4. if n is odd then
n=3*n+1

5. else
n=n/2

6. GOTO 2

Given the input 22, the following sequence of numbers will be printed 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1

It is conjectured that the algorithm above will terminate (when a 1 is printed) for any integral input value. Despite the simplicity of the algorithm, it is unknown whether this conjecture is true. It has been verified, however, for all integers n such that 0 < n < 1,000,000 (and, in fact, for many more numbers than this.)

Given an input n, it is possible to determine the number of numbers printed (including the 1). For a given n this is called the cycle-length of n. In the example above, the cycle length of 22 is 16.

For any two numbers i and j you are to determine the maximum cycle length over all numbers between i and j.

The Input

The input will consist of a series of pairs of integers i and j, one pair of integers per line. All integers will be less than 1,000,000 and greater than 0.

You should process all pairs of integers and for each pair determine the maximum cycle length over all integers between and including i and j.

You can assume that no operation overflows a 32-bit integer.

The Output

For each pair of input integers i and j you should output i, j, and the maximum cycle length for integers between and including i and j. These three numbers should be separated by at least one space with all three numbers on one line and with one line of output for each line of input. The integers i and j must appear in the output in the same order in which they appeared in the input and should be followed by the maximum cycle length (on the same line).

Sample Input

1 10
100 200
201 210
900 1000

Sample Output

1 10 20
100 200 125
201 210 89
900 1000 174

My Code:


#include <stdio.h>

typedef long long int lint;

lint m,n,i,c=0, maxi=1,a,b,d;

lint cyf(lint x,lint y);

int main()
{
    while(scanf("%lld %lld", &a,&b)!=EOF)
    {maxi=1;
        m= (a > b) ? a : b;
    n= (a > b) ? b : a;

    for (d=n; d<=m; d++)
    {
        i=d;
        c=0;
        here:
            c=c+1;
            if(i==1){

                goto there;}
                else
                {
                    if (i%2!=0)
                        i=3*i+1;
                    else
                        i=i/2;

                    goto here;
                }
                there:
        maxi=(maxi>=c) ? maxi : c;

    }


        printf("%lld %lld %lld\n", a,b,maxi);
    }
    return 0;
}