# Int Spot

uva.onlinejudge.org www.lightoj.com

## Wednesday, June 29, 2016

## Saturday, June 25, 2016

### DOWNLOAD AND INSTALL WINDOWS MOVIE MAKER

# This is the link to download:

## Before checking link SEE THIS:

## Friday, June 24, 2016

## Wednesday, June 22, 2016

### BOUNCING STAR!!! (COMPLETED!!)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 | #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)

body, html { margin:0; padding:0; color:#333; background:#fff; font-family:Verdana,Helvetica,Arial,sans-serif; } input { margin:.5em 0; } #container { width:750px; margin:1em auto; } #container > div { margin:auto 1em; } #code { float:left; width:50%; } #html { float:right; width:50%; } #code2 { margin-right:1em; } #html2 { margin-left:1em; } #main textarea { width:100%; height:10em; } #html2 textarea { float:right; } #options { clear:both; } #divstyles { width:50%; } #preview { padding-bottom: 3em; } #footer { border-top:1px dotted #000; } #footer p { text-align:center; font-size:75%; }

**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 3***n* + 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. inputn2. print

n3. if

n= 1 then STOP4. if

nis odd then

n=3*n+15. else

n=n/26. 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; }