Esta sección te permite ver todos los mensajes escritos por este usuario. Ten en cuenta que sólo puedes ver los mensajes escritos en zonas a las que tienes acceso en este momento.
import java.io.*;
public class UVa11152_ColourfulFlowers {
public static void main(String[] args) throws IOException {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
String line;
StringTokenizer stk;
while ((line = in.readLine()) != null) {
stk = new StringTokenizer(line);
double a = Double.parseDouble(stk.nextToken());
double b = Double.parseDouble(stk.nextToken());
double c = Double.parseDouble(stk.nextToken());
double s = (a + b + c) / 2.0;
double At = Math.sqrt(s * (s - a) * (s - b) * (s - c));
double rc = (a * b * c) / (4.0 * At);
double ri = 2.0 * At / (a + b + c);
double Ac = Math.PI * rc * rc;
double Ai = Math.PI * ri * ri;
double sun = Ac - At;
double violets = At - Ai;
double roses = Ai;
System.out.printf(Locale.ENGLISH, "%.4f %.4f %.4f%n", sun, violets, roses);
}
in.close();
Citardouble s = (a + b + c) / 2.0;
double At = Math.sqrt(s * (s - a) * (s - b) * (s - c));
Citardouble rc = (a * b * c) / (4.0 * At);
double ri = 2.0 * At / (a + b + c);
import java.util.*;
import java.io.*;
public class UVa10991_Region {
public static void main(String[] args) throws IOException {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
int T = Integer.parseInt(in.readLine());
while (T-- > 0) {
StringTokenizer stk = new StringTokenizer(in.readLine());
double R1 = Double.parseDouble(stk.nextToken());
double R2 = Double.parseDouble(stk.nextToken());
double R3 = Double.parseDouble(stk.nextToken());
double a = R2 + R3;
double b = R1 + R3;
double c = R1 + R2;
double alpha = Math.acos((b * b + c * c - a * a) / (2.0 * b * c));
double beta = Math.acos((a * a + c * c - b * b) / (2.0 * a * c));
double gamma = Math.acos((a * a + b * b - c * c) / (2.0 * a * b));
double s = (a + b + c) / 2.0;
double A = Math.sqrt(s * (s - a) * (s - b) * (s - c));
double slice1 = alpha * R1 * R1 / 2.0;
double slice2 = beta * R2 * R2 / 2.0;
double slice3 = gamma * R3 * R3 / 2.0;
double res = A - slice1 - slice2 - slice3;
System.out.printf(Locale.ENGLISH, "%.6f%n", res);
}
in.close();
System.exit(0);
}
}
CitarThe Tower of Babylon
Perhaps you have heard of the legend of the Tower of Babylon. Nowadays many details of this tale have been forgotten. So now, in line with the educational nature of this contest, we will tell you the whole story:
The babylonians had n types of blocks, and an unlimited supply of blocks of each type. Each type-i block was a rectangular solid with linear dimensions tex2html_wrap_inline32 . A block could be reoriented so that any two of its three dimensions determined the dimensions of the base and the other dimension was the height. They wanted to construct the tallest tower possible by stacking blocks. The problem was that, in building a tower, one block could only be placed on top of another block as long as the two base dimensions of the upper block were both strictly smaller than the corresponding base dimensions of the lower block. This meant, for example, that blocks oriented to have equal-sized bases couldn't be stacked.
Your job is to write a program that determines the height of the tallest tower the babylonians can build with a given set of blocks.
Input and Output
The input file will contain one or more test cases. The first line of each test case contains an integer n, representing the number of different blocks in the following data set. The maximum value for n is 30. Each of the next n lines contains three integers representing the values tex2html_wrap_inline40 , tex2html_wrap_inline42 and tex2html_wrap_inline44 .
Input is terminated by a value of zero (0) for n.
For each test case, print one line containing the case number (they are numbered sequentially starting from 1) and the height of the tallest possible tower in the format "Case case: maximum height = height"
Sample Input
1
10 20 30
2
6 8 10
5 5 5
7
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
5
31 41 59
26 53 58
97 93 23
84 62 64
33 83 27
0
Sample Output
Case 1: maximum height = 40
Case 2: maximum height = 21
Case 3: maximum height = 28
Case 4: maximum height = 342
Citar
You have devised a new encryption technique which encodes a message by inserting between its characters
randomly generated strings in a clever way. Because of pending patent issues we will not discuss in
detail how the strings are generated and inserted into the original message. To validate your method,
however, it is necessary to write a program that checks if the message is really encoded in the final
string.
Given two strings s and t, you have to decide whether s is a subsequence of t, i.e. if you can remove
characters from t such that the concatenation of the remaining characters is s.
Input
The input contains several testcases. Each is specified by two strings s, t of alphanumeric ASCII
characters separated by whitespace. Input is terminated by EOF.
Output
For each test case output, if s is a subsequence of t.
Sample Input
sequence subsequence
person compression
VERDI vivaVittorioEmanueleReDiItalia
caseDoesMatter CaseDoesMatter
Sample Output
Yes
No
Yes
No
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
/**
* a->97 z->122 A->65 A->90
*
* @author qwery.azc
*/
public class Main {
public String pertenecer(String cadenita, String cadenota) {
int piv = 0;
int cajaIndex=0;
String res="No";
for (int i = 0; i < cadenita.length(); i++) {
char caracter=cadenita.charAt(i);
for (int j = 0; j < cadenota.length() ; j++) {
if (caracter == cadenota.charAt(j)) {
piv++;
i++;
}
}
}
if(piv==cadenita.length()){
res="Yes";
}
System.out.println(res);
return res;
}
public static void main(String argumentos[]) throws IOException {
Main m = new Main();
StringTokenizer stk;
String line;
String cad1,cad2;
BufferedReader scanner = new BufferedReader(new InputStreamReader(System.in));
while ((line=scanner.readLine())!=null) {
stk=new StringTokenizer(line," ");
cad1=stk.nextToken();
cad2=stk.nextToken();
m.pertenecer(cad1, cad2);
}
scanner.close();
}
}
cuando uso n>>2 es igual n/2
Is Bigger Smarter?
Some people think that the bigger an elephant is, the smarter it is. To disprove this, you want to take
the data on a collection of elephants and put as large a subset of this data as possible into a sequence
so that the weights are increasing, but the IQ's are decreasing.
Input
The input will consist of data for a bunch of elephants, one elephant per line, terminated by the endof-file.
The data for a particular elephant will consist of a pair of integers: the first representing its size
in kilograms and the second representing its IQ in hundredths of IQ points. Both integers are between
1 and 10000. The data will contain information for at most 1000 elephants. Two elephants may have
the same weight, the same IQ, or even the same weight and IQ.
Output
Say that the numbers on the i-th data line are W[i] and S[i]. Your program should output a sequence
of lines of data; the first line should contain a number n; the remaining n lines should each contain a
single positive integer (each one representing an elephant). If these n integers are a[1], a[2],..., a[n] then
it must be the case that
W[a[1]] < W[a[2]] < ... < W[a[n]]
and
S[a[1]] > S[a[2]] > ... > S[a[n]]
In order for the answer to be correct, n should be as large as possible. All inequalities are strict:
weights must be strictly increasing, and IQs must be strictly decreasing.
There may be many correct outputs for a given input, your program only needs to find one.
Sample Input
6008 1300
6000 2100
500 2000
1000 4000
1100 3000
6000 2000
8000 1400
6000 1200
2000 1900
Sample Output
4
4
5
9
7
Quirksome Squares
The number 3025 has a remarkable quirk: if you split its decimal representation in two strings of equal length (30 and 25) and square the sum of the numbers so obtained, you obtain the original number:
displaymath26
The problem is to determine all numbers with this property having a given even number of digits.
For example, 4-digit numbers run from 0000 to 9999. Note that leading zeroes should be taken into account. This means that 0001 which is equal to tex2html_wrap_inline28 is a quirksome number of 4 digits. The number of digits may be 2,4,6 or 8. Although maxint is only 32767 and numbers of eight digits are asked for, a well-versed programmer can keep his numbers in the range of the integers. However efficiency should be given a thought.
Input
The input of your program is a textflle containing numbers of digits (taken from 2,4,6,8), each number on a line of its own.
Output
The output is a textfile consisting of lines containing the quirksome numbers (ordered according to the input numbers and for each input number in increasing order).
Warning: Please note that the number of digits in the output is equal to the number in the corresponding input line : leading zeroes may not be suppressed.
Sample Input
2
2
Sample Output
00
01
81
00
01
81
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.Map;
/**
*
* @author qwery,azc
*/
public class RadixSort {
public ArrayList<Integer> radixSort(ArrayList<Integer> entryData) {
int max = Collections.max(entryData);
max = cantidadDigitos(max);
Map<Integer, ArrayList> regid = new HashMap();
int dataAux = 0;
int digMenorSig = 0;
int porciento = 10;
ArrayList<Integer> lista = new ArrayList<>();
ArrayList<Integer> copiaSort = new ArrayList<>();
for (int i = 0; i < max; i++) {
porciento *= 10;
for (int j = 0; j < entryData.size(); j++) {
dataAux = entryData.get(j);
digMenorSig = dataAux % 10;
lista.add(dataAux);
regid.put(digMenorSig, lista);
dataAux = dataAux / porciento;
}
for (Map.Entry<Integer, ArrayList> entry : regid.entrySet()) {
Integer key = entry.getKey();
ArrayList<Integer> value = entry.getValue();
for (int j = 0; j < value.size(); j++) {
copiaSort.add(value.get(j));
}
}
entryData=copiaSort;
}
for (int i = 0; i < entryData.size(); i++) {
System.err.println(entryData.get(i));
}
//System.out.println("max " + max);
return null;
}
public int cantidadDigitos(int data) {
int res = 0;
while (data > 0) {
data = data / 10;
res++;
}
return res;
}
public static void main(String args[]) {
RadixSort r = new RadixSort();
ArrayList<Integer> a = new ArrayList();
a.add(34);
a.add(27);
a.add(38);
r.radixSort(a);
}
}
<html>
<head>
<title>Geolocalizacion</title>
<meta name="viewport" content="initial-scale=1.0, user-scaleble=no">
<meta charset="utf-8">
<style >
#map-canvas{
height: 380px;
width: 300px;
margin: :0px;
padding: 0px;
background-color: #000;
}
</style>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js"></script>
<script src="fancywebsocket.js"></script>
<script language="javascript" src="https://code.jquery.com/jquery-1.12.4.js"></script>
<script src="https://maps.googleapis.com/maps/api/js?v=3.exp&signed_in=true"></script>
<script type="text/javascript" src="http://maps.google.com/maps/api/js?sensor=false"></script>
<script>
alert("lslslsl");
var map;
var marker;
function initialize(){
var mapOptions={
zoom:15;
};
map=new google.maps.Map(document.getElementById('map-canvas'),mapOptions);
if (navigator.geolocation) {
navigator.geolocation.getCurrentPosition(function(position){
var pos=new google.maps.LatLng(position.coords.latitude,position.coords.longitude)
var goldStar={
path:google.maps.SymbolPath.CIRCLE,
strokeColor:'#276ED0',
fillColor:'#276ED0',
fillOpacity:.9,
strokeWeight:1,
scale:6,
};
var marker=new google.maps.Marker({
position:pos,
icon:goldStar,
draggable:true,
animation:google.maps.Animation.DROP,
map:map,
});
map.setCenter(pos);
},function(){
handleNoGeolocation(true);
});
}else{
handleNoGeolocation(false);
}
}
function animar(){
navigator.geolocation.getCurrentPosition(function(position)
{
var pos =new google.maps.LatLng(position.coords.latitude, position.coords.longitude);
map.panTo(pos);
var goldStar={
path:google.maps.SymbolPath.CIRCLE,
strokeColor:'#FF4E51',
fillColor:'#FF4E51',
fillOpacity:.9,
strokeWeight:1,
scale:5,
};
var marker=new google.maps.Marker({
position:pos,
icon:goldStar,
draggable:true,
map:map,
});
var option={
map:map,
position:pos,
};
send(position.coords.latitude+","+position.coords.longitude);
map.setCenter(options.position);
});
}
function handleNoGeolocation(errorFlag){
if (errorFlag) {
var content='Error:the service';
}else{
var content="tu buscador";
}
var options={
map=map,
position:new google.maps.LatLng(60,105),
content:content,
};
var infowindow=new google.maps.infoWindow(options);
map.setCenter(options.position);
}
google.maps.event.addDomListener(window,'load',initialize);
setTimeout(function(){animar()},5000);
function pocision(pos){
}
</script>
</head>
<body>
<div id="map-canvas">
</div>
</body>
</html>Citar
Big Mod
Calculate
displaymath25
for large values of B, P, and M using an efficient algorithm. (That's right, this problem has a time dependency !!!.)
Input
Three integer values (in the order B, P, M) will be read one number per line. B and P are integers in the range 0 to 2147483647 inclusive. M is an integer in the range 1 to 46340 inclusive.
Output
The result of the computation. A single integer.
Sample Input
3
18132
17
17
1765
3
2374859
3029382
36123
Sample Output
13
2
13195