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Closed 10 years ago.
public void mystery1(int n) {
if (n <= 1) {
System.out.print(n);
} else {
mystery1(n / 2);
System.out.print(", " + n);
}
}
What gives this code for odd numbers. Becuase when we divide it it will not be an integer.
There is not mystery, because result of the integer division in Java is integer.
In Java or most other programming languages, when you divide an integer by an integer, the result will be an integer. If a decimal number occurs, say for example:
5/2=2.5
then, the number before the decimal point will be treated as the integer and 2 will be chosen.
In case you want to explicitly convert the integer into float or double, you can use any of the following conversions:
(float) 3/2;
(double) n/2;
The above explicitly converts it to a decimal.
n / 2, this is an integer division, where the fraction part will be ignored.
System.out.println(3/2); // prints 1
System.out.println(3.0/2); // prints 1.5
System.out.println(3/2.0); // prints 1.5
System.out.println(3.0/2.0); // prints 1.5
Param will rounded to int, for example if param will be 5, the next call the function will be with param 2
Related
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Closed 9 years ago.
I have a set =(1, 2, 3), and I need to get all possible subsets of set with different sequences (with repeating elements).The out put look:
1
2
3
1,2
1,3
2,1
2,3
3,1
3,2
1,2,3
1,3,2
2,1,3
2,3,1
3,1,2
3,2,1
please can someone help me with that? thxx
If you want subsets, then Google Guava has a method for you:
http://docs.guava-libraries.googlecode.com/git/javadoc/com/google/common/collect/Sets.html#powerSet(java.util.Set)
But in your example you have some duplicate sets (remember, sets are unordered). So you might want to get all the possible permutations of each subset as well:
http://docs.guava-libraries.googlecode.com/git/javadoc/com/google/common/collect/Collections2.html#permutations(java.util.Collection)
The problem can be solved by finding all combinations using bitwise operations.
The idea is: Generate all the subsets of a given array (set), this set is known as a power set .For each of the subset(Combination), find its permutation too.
Refer the following tutorials, to learn, how can you find all combinations using Bit Wise Operations. http://www.codechef.com/wiki/tutorial-bitwise-operations
void PERMUTE()
{
/*PERMUTATION FUNCTION THAT PERMUTES THE SUBSET ARRAY*/
}
public static void main(String[] args)
{
// TODO code application logic here
int Set[]={1,2,3};
int n=Set.length;
for (int i = 0; i <= (1 << n); ++i)
{
System.out.print("[");
int subsetSz=0;
int A[]=new int[100];
for (int j = 0; j < n; ++j)
{
if ((i & 1 << j)!=0)
{
System.out.print(Set[j]+",");
A[subsetSz++]=Set[j];
}
}
System.out.println("]");
/*Permute the subset*/
if(subsetSz>1)
{
PERMUTE(A);
}
}
}
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Closed 10 years ago.
hey guys this is my problem..
I want to generate random letters in an specific length of a word, but the Starting Letter should corresponds to value of the variable I declare.
Example:
A3 should generate AER
A5 should generate AJIEH
B2 should generate BJ
Working with the variable names will be tedious (although it is possible I suppose through reflection). You could, however, try something like this:
public static String genString(char first, int len) {
String s = "";
for (int i = 1 ; i < len ; i++)
s += (char)(Math.random() * ('Z' - 'A' + 1) + 'A');
return first + s;
}
For example:
System.out.println(genString('A', 4));
Output (one of many possible):
AVGH
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Closed 10 years ago.
Please explain how recursion works, in the simplest way you can.
As been pointed out, recursion is when a function calls itself. Here illustrated using factorial, where factorial(5) mathematically evalutates to values 5 * 4 * 3 * 2 * 1.
public int factorial(int x) {
if (x == 1) return 1;
else return x * factorial (x - 1);
}
// Try routine
factorial(3);
Which evaluates as
factorial(1) = 1 = 1
factorial(2) = 2 * factoral(1) = 2 * 1
factorial(3) = 3 * (2 * factorial(2) * (factorial(1)) = 3 * 2 * 1
...
Usually, it's a function that calculates one result itself, and calls itself to get the rest of the results.
For example, to get all positive numbers less or equal to 3, the function says, "One result is 3, and the rest of the results are all positive numbers less or equal to 2. Let's call myself with 2 and add that result to the one I calculated."
Of course, recursive functions need to be careful to have an "end condition" or they will never return a result. In the case of this example, the end condition would be when the function is called with 0, it should just return 0.
Here's a simple example of recursive method: -
public int recur(int count) {
if (count < 10) {
return count + recur(count++);
}
return count;
}
System.out.println(recur(0)); // Invoke first time
Recursion is a method which calls itself from within itself.
Example:
public void recur(int x)
recur(10);
end
Recursion is when a function calls itself:
int factorial(int integer)
{
if (integer == 1)
return 1;
else
return (integer*(factorial(integer-1)));
}
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Closed 10 years ago.
I need to print the first k digits and the last k digits in
n^n (n to the power of n, where n is an integer)
For example :
Input Output
n k First k digits Last k digits
4 2 --> 25 56
9 3 --> 387 489
I sense that it requires some clever mathematics, however I am unable to think of anything as such. Please suggest a way to approach the problem.
The last k digits are easy, you just need to calculate it modulo 10^k. To do this, after every multiplication, just apply the modulo, ie. intermediate_result %= 10^k.
Of course, you will need to calculate 10^k using some other method, because ^ does not mean power of in C or Java.
To find the first k digits, see first n digits of an exponentiation.
Thanks everyone for their help. My final code is
#include <stdio.h>
#include <math.h>
long int lastKdigits(long long n,int k)
{
long long i,res=1,div=pow(10,k);
for(i=1;i<=n;i++)
{
res=(res*n)%div;
}
return res;
}
long int firstKdigits(long long n,int k)
{
long double x, y;
x = n*log10(n);
y = floor(pow(10,x-floor(x) +k-1));
return ((int)y);
}
int main()
{
long long n;
int k;
scanf("%lld %d",&n,&k);
printf("%ld\t",firstKdigits(n,k));
printf("%ld\n",lastKdigits(n,k));
}
return 0;
}
For last k digits it's pretty easy you just need to calculate n^n (mod 10^k) but I don't know any solution for the other k diggits!
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Closed 10 years ago.
What do you think Float.MIN_VALUE equals to?
The next code explains where my last 5 hours went to, trying solving a bug.
public static void main(String[] args) {
compareToZero(Float.MIN_VALUE); // Out = true false false
compareToZero(Float.MAX_VALUE); // Out = true false false
System.out.println("Float minimum " + Float.MIN_VALUE); // Out = 1.4E-45
System.out.println("Float maximum " + Float.MAX_VALUE); // Out = 3.4028235E38
}
private static void compareToZero(float value1) {
System.out.print((value1 > 0) + " ");
System.out.print((value1 < 0) + " ");
System.out.print((value1 == 0) + "\n");
}
I didn't imagined that minimum value of float will be a positive value... Can't find any use for it.
Per the documentation for Float.MIN_VALUE:
A constant holding the smallest positive nonzero value of type float, 2-149. It is equal to the hexadecimal floating-point literal 0x0.000002P-126f and also equal to Float.intBitsToFloat(0x1).
While the name is debatable as the "true minimum value" of a float is -Float.MAX_VALUE, I suspect MIN_VALUE was chosen for consistency with the other numeric types. Using the names MIN_RANGE_VALUE and MAX_RANGE_VALUE (or similar) might have made the difference more clear.
To understand why this is the "minimum value" requires understanding a little bit about how Java (or IEEE-754) floating point values work. With this insight, after reading the documentation, it is clear that Float.MIN_VALUE is the minimum non-zero value representable by the mantissa and exponent components of a float. Or, the smallest positive value a float can represent.
The "true minimum value" is -Float.MAX_VALUE because Float.MAX_VALUE represents the maximum value that the mantissa and exponent components of a float can represent. Since the sign for a float is stored as a discrete bit, this range limit is the same for both positive and negative numbers.
This differs from how integers work in Java (and on most CPUs): they are encoded using two's complement. (Some computer systems used a discrete sign bit, which is called "one's complement", which then has two integer values of zero: 0 and -0!)
Happy researching!