I am getting wrong results resolving below task:
Generalized harmonic numbers. Write a program GeneralizedHarmonic.java that takes two integer command-line arguments n and r and uses a for loop to compute the nth generalized harmonic number of order r, which is defined by the following formula:
formula
public class GeneralizedHarmonic {
public static void main(String[] args) {
int a = Integer.parseInt(args[0]);
int b = Integer.parseInt(args[1]);
int i;
double sum = 0;
for (i = 0; i <= a; i++) {
sum += 1 / Math.pow(i, b);
}
System.out.println(sum);
}
}
This is my code but I could not get the correct test output.The output result is always Infinity.
test outputs
You have initliazed int i = 0 in for-loop for (i = 0; i <= a; i++) and so the first element of your harmonic number isn't , but .
The code that works:
public class GeneralizedHarmonic {
public static void main(String[] args) {
int a = Integer.parseInt(args[0]);
int b = Integer.parseInt(args[1]);
double sum = 0;
for (int i = 1; i <= a; i++) {
sum += 1 / Math.pow(i, b);
}
System.out.println(sum);
}
}
I have tried this program many times I didn't get the proper output till now please help me to solve this type of program.
input:n=3
output: 001 to 999
input:n=4
output:0001 to 9999
input:n=2
output:01 to 99
public static void main(String[] args)
{
Scanner scan = new Scanner(System.in);
int number = scan.nextInt();
int sum=1,result=0;
while(number!=0)
{
result=result+(9*sum);
sum=sum*10;
number--;
}
System.out.println(result);
for(int i=1;i<=result;i++)
{
System.out.printf("%02d ",i);//here i manually mentioned the %02d but i want to take user input
}
}
You can use this code
int number = 3;
String mask = "%0" + (number) + "d%n";
int max = (int)Math.pow(10, number)-1;
for (int x = 1; x <= max; x++)
System.out.printf(mask, x);
thanks to #RalfRenz
Can you try below code ?
class Main
{
public static void main(String[] args)
{
Scanner scan = new Scanner(System.in);
int num = scan.nextInt();
String masked = "%0" + (num) + "d%n";
int max = (int)Math.pow(10, num)-1;
for (int k = 1; k <= max; k++)
System.out.printf(masked, k);
}}
Here's My Code; and I can find an array with this and I would like to calculate the mean of the values (overall) after this I would like to calculate standard deviation of this but I couldn't understand the question exactly so I dont have a method for now. Here's the question for standard deviation (Write a method that takes two parameters --a set of int values in an array and a double value representing their mean-- and computes and returns the standard deviation of the values using the given mean.)
import java.util.*;
public class Test
{
final static int N = 100;
static int limit = 0;
static int[] list;
static int i, j;
static int sum = 0;
static Scanner scan = new Scanner (System.in);
public static int[] generateArray ()
{
System.out.print ("Enter your array limit: ");
limit = scan.nextInt();
list = new int[limit];
for(i = 0; i < limit; i++)
{
list[i] = (int) (Math.random() * 2 * N - N);
}
return list;
}
public static void printArray()
{
for(j = 0; j < limit; j++)
System.out.print (list[j] + "\t");
}
public static void meanArray()
{
sum = sum + list[j]; //PROBLEM HERE
System.out.println (sum);
}
public static void main(String[] args)
{
generateArray();
printArray();
meanArray(); //PROBLEM HERE
}
}
To generate the mean value, add up all values in your list and devide them by the number of values:
public static void meanArray() {
double result = 0;
for(int i : list) {
result += i;
}
result /= list.length;
System.out.println(result);
}
I tried making a Java program executing the Fibonacci sequence.
Here's my code:
import java.io.*;
public class Fibonacci{
public static void main(String[]args){
BufferedReader Data=new BufferedReader (new InputStreamReader(System.in));
int ctr1=0;
int ctr2=0;
int num1=0;
int num2=0;
int num3=0;
try{
System.out.println("How many numbers would you want to see?");
ctr2=Integer.parseInt(Data.readLine());
for(int ans=0; ctr1==ctr2; ctr1++){
num1++;
System.out.println(num2 + "\n" + num1);
ans=num1+num2;
System.out.println(ans);
ans=num3;
}
}catch(IOException err){
System.out.println("Error!" + err);
}catch(NumberFormatException err){
System.out.println("Invald Input!");
}
}
}
Obviously, I'm a beginner in Java and I don't know how to properly use the for statement. Would somebody be kind enough to make my code work? Or maybe make a way shorter code that works. I'm a beginner so be cool. Thanks :)
Fibonacci series in java is actually quite simple and can be done with just one single for-loop!!!!
import java.io.*;
class fibonacci{
public static void main() throws NumberFormatException, IOException{
BufferedReader Data=new BufferedReader (new InputStreamReader(System.in));
int a,b,c,d;
System.out.println("Upto How many numbers do you want to see?");
d=Integer.parseInt(Data.readLine());
for (a=0,b=1,c=a;a<d;c=a,a+=b,b=c){
System.out.println(a);
}
}
}
This has been done using buffered reader........ If you are said to use only bufferedreader go for this else you can use Scanner class which is much simple and easy to use because you don't have to catch or throw any exceptions.....
Scanner program:-
import java.util.*;
class fibonacci{
public static void main(){
Scanner sc = new Scanner(System.in);
int a,b,c;
System.out.println("Upto How many numbers do you want to see?");
d=sc.nextInt();
for (a=0,b=1,c=a;a<d;c=a,a+=b,b=c){
System.out.println(a);
}
}
}
Now as I said in one loop you can do it.... Here is another method where you do the swapping inside the body of the loop and not in the arguments of it...
And this is much simplier to understand for beginners as u don't have to pass multiple variables inside the arguments and yeah its a bit longer
import java.util.*;
class fibonacci{
public static void main(){
Scanner sc = new Scanner(System.in);
int a = 0,b = 1,c,d;
System.out.println("Upto How many numbers do you want to see?");
d=sc.nextInt();
System.out.println(a +"\n" +b);//\n is used to go to next line....
for (c=0;c<d;c++){
c = a + b;//Doing and printing the fibonacci...
System.out.println(c);
a = b;
b = c;//Swapping the values...
}
}
}
So here i have given you three methods that should give the same output(Most probably) choose whichever is convenient for you..
Look at this code snippet which is much easier than yours to understand. Solution tip is simple, you keep 2 pointers for the first 2 fibonacci numbers and update them appropriately in the loop. In the example below, the loop executes 10 times, you can modify it as desired.
static void fibonacci() {
int ptr1 = 1, ptr2 = 1;
int temp = 0;
System.out.print(ptr1 + " " + ptr2 + " ");
for (int i = 0; i < 10; i++) {
System.out.print(ptr1 + ptr2 + " ");
temp = ptr1;
ptr1 = ptr2;
ptr2 = temp + ptr2;
}
}
Output:
1 1 2 3 5 8 13 21 34 55 89 144
Expanding on the answers, if you want to look really cool use recursion.
public class Fibonacci {
public static long fib(int n) {
if (n <= 1) return n;
else return fib(n-1) + fib(n-2);
}
public static void main(String[] args) {
int N = 300; // how many numbers you want to generate
for (int i = 1; i <= N; i++)
System.out.println(i + ": " + fib(i));
}
}
Here is Google search of what it is, hope those resources help: http://bit.ly/1cWxhUS
I'm a beginner in java as well however I've found an easy way to create a Fibonacci number using an array. The basic principle of a Fibonacci number is the addition of the current number and the number that came before.
Here is my code:
//Creation of array
int [ ] fib = new int[size];
//Assigning values to the first and second indexes of array named "fib"
fib [0] = 0;
fib [1] = 1;
//Creating variable "a" to use in for loop
int a = 1
//For loop which creates a Fibonacci number
for( int i = 2; i < size ; i++)
{
fib[i] = a;
a = fib[i] + fib[i-1];
}
This is another algorithm which I found online and I kind of simplified the code from it.
public static BigInteger fib(BigInteger x) {
if (x.intValue() < 0){return x.intValue() % 2 == 0 ?fib(x.multiply(BigInteger.valueOf(-1))).multiply(BigInteger.valueOf(-1)) : fib(x.multiply(BigInteger.valueOf(-1)));}
int n = Integer.valueOf(x.toString());
BigInteger a = BigInteger.ZERO,b = BigInteger.ONE;
for (int bit = Integer.highestOneBit(n); bit != 0; bit >>>= 1) {
BigInteger d = a.multiply(b.shiftLeft(1).subtract(a));
BigInteger e = a.multiply(a).add(b.multiply(b));
a = d;
b = e;
if ((n & bit) != 0) {
BigInteger c = a.add(b);
a = b;
b = c;
}
}
return a;
}
I know there is a chance that you wont understand how to use BigInteger, so I am giving you this link, just trying to be helpful.
Here we get Fibonacci Series up to n.
public static void fibSequence(int n) {
int sum = 0;
for (int x = 0, y = 1; sum < n; x = y, y = sum, sum = x + y) {
System.out.print(sum + " ");
}
}
Example:
Input: n = 20
Output: 0 1 1 2 3 5 8 13
more simple way
public static void main(String[] args) {
int first = 1;
int second = 2;
for (int i = 0; i < 20; i++) {
if (i == 0)
System.out.print(first);
System.out.print("," + second);
int temp = second;
second = first + second;
first = temp;
}
}```
program output :: 1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946
import java.util.*;
public class sequence1
{
public static void main(String[] args)
{
sequence1 fs=new sequence1();
fs.fibonacci();
}
public void fibonacci()
{
int numb1 = 1;
int numb2 = 1;
int temp = 0;
#SuppressWarnings("resource")
Scanner input=new Scanner(System.in);
System.out.println("How Many Terms? (Up To 45)");
int x=input.nextInt();
x=x-2;
System.out.println(numb1);
System.out.println(numb2);
for (int i = 0; i < x; i++)
{
System.out.println(numb1 + numb2 + " ");
temp = numb1;
numb1 = numb2;
numb2 = temp + numb2;
}
}
}
This function return the fibonacci series
/**
* #param startElement
* #param secondElent
* #param length :length of fibonacci series
* #return fibonacciseries : contain the series of fibonacci series
*/
public int[] createFibonacciSeries(int startElement, int secondElent,
int length) {
int fibonacciSeries[] = new int[length];
fibonacciSeries[0] = startElement;
fibonacciSeries[1] = secondElent;
for (int i = 2; i < length; i++) {
fibonacciSeries[i] = fibonacciSeries[i - 1]
+ fibonacciSeries[i - 2];
}
return fibonacciSeries;
}
import java.util.*;
class MyFibonacci {
public static void main(String a[]){
int febCount = 15;
int[] feb = new int[febCount];
feb[0] = 0;
feb[1] = 1;
for(int i=2; i < febCount; i++){
feb[i] = feb[i-1] + feb[i-2];
}
for(int i=0; i< febCount; i++){
System.out.print(feb[i] + " ");
}
}
}
public class FibonacciExercitiu {
public static void main(String[] args) {
int result = fib(6); //here we test the code. Scanner can be implemented.
System.out.println(result);
}
public static int fib(int n) {
int x = 1;
int y = 1;
int z = 1; //this line is only for declaring z as a variable. the real assignment for z is in the for loop.
for (int i = 0; i < n - 2; i++) {
z = x + y;
x = y;
y = z;
}
return z;
}
/*
1. F(0) = 1 (x)
2. F(1) = 1.(y) =>Becomes x for point4
3.(z)F(2) = 2 (z) =>Becomes Y for point4 // becomes X for point 5
4.(z)F(3) = 3 // becomes y for point 5
5.(z)F(4) = 5 ..and so on
*/
}
public static int[] fibonachiSeq(int n)
{
if (n < 0)
return null;
int[] F = new int[n+1];
F[0] = 0;
if (n == 0)
return F;
F[1] = 1;
for (int i = 2; i <= n; i++)
{
F[i] = F[i-1] + F[i-2];
}
return F;
}
Using while loop
class Feb
{
static void Main(string[] args)
{
int fn = 0;
int sn = 1;
int tn = 1;
Console.WriteLine(fn);
Console.WriteLine(sn);
while (true)
{
tn = fn + sn;
if (tn >10)
{
break;
}
Console.WriteLine(tn);
fn = sn;
sn = tn;
}
Console.Read();
}
}
public class Febonacci {
public static void main(String[] args) {
int first =0;
int secend =1;
System.out.print(first+","+secend);
for (int k=1;k<7;k++){
System.out.print(","+(first+secend ));
if(k%2!=0)
first+=secend;
else
secend+=first;
}
}
}
public class FibonacciSeries {
public static void main(String[] args) {
int a=0, c=0, b=1;
for(int i=0; i<10; i++) {
System.out.print(c+" ");
a = c + b;
c = b;
b = a;
}
}
}
I tried to find the factorial of a large number e.g. 8785856 in a typical way using for-loop and double data type.
But it is displaying infinity as the result, may be because it is exceeding its limit.
So please guide me the way to find the factorial of a very large number.
My code:
class abc
{
public static void main (String[]args)
{
double fact=1;
for(int i=1;i<=8785856;i++)
{
fact=fact*i;
}
System.out.println(fact);
}
}
Output:-
Infinity
I am new to Java but have learned some concepts of IO-handling and all.
public static void main(String[] args) {
BigInteger fact = BigInteger.valueOf(1);
for (int i = 1; i <= 8785856; i++)
fact = fact.multiply(BigInteger.valueOf(i));
System.out.println(fact);
}
You might want to reconsider calculating this huge value. Wolfram Alpha's Approximation suggests it will most certainly not fit in your main memory to be displayed.
This code should work fine :-
public class BigMath {
public static String factorial(int n) {
return factorial(n, 300);
}
private static String factorial(int n, int maxSize) {
int res[] = new int[maxSize];
res[0] = 1; // Initialize result
int res_size = 1;
// Apply simple factorial formula n! = 1 * 2 * 3 * 4... * n
for (int x = 2; x <= n; x++) {
res_size = multiply(x, res, res_size);
}
StringBuffer buff = new StringBuffer();
for (int i = res_size - 1; i >= 0; i--) {
buff.append(res[i]);
}
return buff.toString();
}
/**
* This function multiplies x with the number represented by res[]. res_size
* is size of res[] or number of digits in the number represented by res[].
* This function uses simple school mathematics for multiplication.
*
* This function may value of res_size and returns the new value of res_size.
*/
private static int multiply(int x, int res[], int res_size) {
int carry = 0; // Initialize carry.
// One by one multiply n with individual digits of res[].
for (int i = 0; i < res_size; i++) {
int prod = res[i] * x + carry;
res[i] = prod % 10; // Store last digit of 'prod' in res[]
carry = prod / 10; // Put rest in carry
}
// Put carry in res and increase result size.
while (carry != 0) {
res[res_size] = carry % 10;
carry = carry / 10;
res_size++;
}
return res_size;
}
/** Driver method. */
public static void main(String[] args) {
int n = 100;
System.out.printf("Factorial %d = %s%n", n, factorial(n));
}
}
Hint: Use the BigInteger class, and be prepared to give the JVM a lot of memory. The value of 8785856! is a really big number.
Use the class BigInteger. ( I am not sure if that will even work for such huge integers )
Infinity is a special reserved value in the Double class used when you have exceed the maximum number the a double can hold.
If you want your code to work, use the BigDecimal class, but given the input number, don't expect your program to finish execution any time soon.
The above solutions for your problem (8785856!) using BigInteger would take literally hours of CPU time if not days. Do you need the exact result or would an approximation suffice?
There is a mathematical approach called "Sterling's Approximation
" which can be computed simply and fast, and the following is Gosper's improvement:
import java.util.*;
import java.math.*;
class main
{
public static void main(String args[])
{
Scanner sc= new Scanner(System.in);
int i;
int n=sc.nextInt();
BigInteger fact = BigInteger.valueOf(1);
for ( i = 1; i <= n; i++)
{
fact = fact.multiply(BigInteger.valueOf(i));
}
System.out.println(fact);
}
}
Try this:
import java.math.BigInteger;
public class LargeFactorial
{
public static void main(String[] args)
{
int n = 50;
}
public static BigInteger factorial(int n)
{
BigInteger result = BigInteger.ONE;
for (int i = 1; i <= n; i++)
result = result.multiply(new BigInteger(i + ""));
return result;
}
}
Scanner r = new Scanner(System.in);
System.out.print("Input Number : ");
int num = r.nextInt();
int ans = 1;
if (num <= 0) {
ans = 0;
}
while (num > 0) {
System.out.println(num + " x ");
ans *= num--;
}
System.out.println("\b\b=" + ans);
public static void main (String[] args) throws java.lang.Exception
{
BigInteger fact= BigInteger.ONE;
int factorialNo = 8785856 ;
for (int i = 2; i <= factorialNo; i++) {
fact = fact.multiply(new BigInteger(String.valueOf(i)));
}
System.out.println("Factorial of the given number is = " + fact);
}
import java.util.Scanner;
public class factorial {
public static void main(String[] args) {
System.out.println("Enter the number : ");
Scanner s=new Scanner(System.in);
int n=s.nextInt();
factorial f=new factorial();
int result=f.fact(n);
System.out.println("factorial of "+n+" is "+result);
}
int fact(int a)
{
if(a==1)
return 1;
else
return a*fact(a-1);
}
}