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;
}
}
}
Related
Problem Statement:
While playing an RPG game, you were assigned to complete one of the hardest quests in this game. There are n monsters you'll need to defeat in this quest. Each monster i is described with two integer numbers - poweri and bonusi. To defeat this monster, you'll need at least poweri experience points. If you try fighting this monster without having enough experience points, you lose immediately.
You will also gain bonusi experience points if you defeat this monster. You can defeat monsters in any order. The quest turned out to be very hard - you try to defeat the monsters but keep losing repeatedly. Your friend told you that this quest is impossible to complete. Knowing that, you're interested, what is the maximum possible number of monsters you can defeat?
Input:
The first line contains an integer, n, denoting the number of monsters.
The next line contains an integer, e, denoting your initial experience.
Each line i of the n subsequent lines (where 0 ≤ i < n) contains an integer, poweri, which
represents power of the corresponding monster.
Each line i of the n subsequent lines (where 0 ≤ i < n) contains an integer, bonusi, which represents bonus for defeating the corresponding monster.
Sample cases:
Input 2 123 78 130 10 0
Output 2
Output description
Initial experience level is 123 points.
Defeat the first monster having power of 78 and bonus of 10. Experience level is now 123+10=133.
Defeat the second monster.
What I have tried:
public static int defeat(int [] monster,int bonus[],int n,int exp){
if(n==0)
return 0;
if(n==1 && monster[0]<=exp)return 1;
if(n==1 && monster[0]>exp) return 0;
if(monster[n-1]<=exp){
return defeat(monster,bonus,n-1,bonus[n-1]+exp )+ defeat(monster,bonus,n-1,exp);
}else{
return defeat(monster,bonus,n-1,exp);
}
}
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int n = s.nextInt();
int exp = s.nextInt();
int monst[] = new int[n];
int bonus[] = new int[n];
for (int i = 0; i < n; i++) {
monst[i] = s.nextInt();
}
for (int i = 0; i < n; i++) {
bonus[i] = s.nextInt();
}
System.out.println(defeat(monst,bonus,n,exp));
}
I am not getting correct answers with this solution.
I thought of this problem as 0/1 knapsack problem( correct me If I am wrong). Also can you provide me DP solution of this problem.
You can just sort the monsters from lowest to highest power required and defeat them in that order.
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int n = s.nextInt();
int exp = s.nextInt();
int monst[] = new int[n];
int bonus[] = new int[n];
for (int i = 0; i < n; i++) {
monst[i] = s.nextInt();
}
for (int i = 0; i < n; i++) {
bonus[i] = s.nextInt();
}
class Monster {
private final int power, bonus;
public Monster(int power, int bonus){
this.power = power;
this.bonus = bonus;
}
}
Monster[] monsters = new Monster[n];
for(int i = 0; i < n; i++) monsters[i] = new Monster(monst[i], bonus[i]);
Arrays.sort(monsters, Comparator.comparingInt(m -> m.power));
int count = 0;
for(Monster m: monsters){
if(exp < m.power) break;
exp += m.bonus;
++count;
}
System.out.println(count);
}
Perhaps I am too simplistic, but I would try it as below:
First create a class Monster like this:
public class Monster
{
final int m_Power;
final int m_Bonus;
public Monster( final int power, final int bonus )
{
m_Power = power;
m_Bonus = bonus;
}
public final int getPower() { return m_Power; }
public final int getBonus() { return m_Bonus; }
}
Next, initialise a list of Monsters like this:
public static void main( String... args )
{
final var scanner = new Scanner( System.in );
final var n = scanner.nextInt();
final var experience = scanner.nextInt();
final var power [] = new int [n];
for( var i = 0; i < n; ++i )
{
power [i] = scanner.nextInt();
}
List<Monster> monsters = new ArrayList<>( n );
for( var i = 0; i < n; ++i )
{
monsters.add( new Monster( power [i], scanner.nextInt();
}
monsters.sort( (m1,m2) ->
{
final var p = m1.getPower() - m2.getPower();
return p == 0 ? m2.getBonus() - m1.getBonus(() : p;
} ); //*1
System.out.println( defeat( monsters, experience ) );
}
*1 -> this implementation of the comparator works well only for power and bonus values that are small compared to MAX_INT.
The list monsters now contains the monsters sorted by their power in ascending order; monsters with the same power are ordered by their bonus values in descending order.
Now my implementation of defeat() would look like this:
public final int defeat( final List<Monster> m, final int initialExperience )
{
var experience = initialExperience;
var retValue = 0;
final Stack<Monster> monsters = new LinkedList<>( m );
while( !monsters.empty() )
{
var monster = monsters.pop();
if( experience > monster.getPower() )
{
experience += monster.getBonus();
++retValue;
}
else break;
}
return retValue;
}
n=int(input())
e=int(input())
P=[]
for i in range(n):
P.append(int(input()))
B=[]
for i in range(n):
B.append(int(input()))
c=0
f=True
while n>0 and f:
f=False
i=0
while i<n:
if e>=P[i]:
e+=B[i]
P.pop(i)
B.pop(i)
n-=1
c+=1
f=True
i-=1
i+=1
print(c)
import java.io.*;
import java.util.*;
class Player {
int exp;
public int getExp() {
return exp;
}
public void setExp(int exp) {
this.exp = exp;
}
}
class Monster {
int power;
int bonus;
public int getPower() {
return this.power;
}
public int getBonus() {
return this.bonus;
}
public void setBonus(int bonus) {
this.bonus = bonus;
}
public void setPower(int power) {
this.power = power;
}
}
class Calc {
public int calc() {
Scanner sc = new Scanner(System.in);
//number of monsters
System.out.println("enter the number of monsters");
int n = sc.nextInt();
int arr[] = new int[n];
System.out.println("enter the power of the player");
Player player = new Player();
player.setExp(sc.nextInt());
//declaration of array type object of monster
Monster monster[] = new Monster[n];
//value setting completed
for (int i = 0; i < n; i++) {
//allocation of object to the real
monster[i] = new Monster();
System.out.println("enter the power of monster");
monster[i].setPower(sc.nextInt());
System.out.println("enter the bonus that to be earned after killing the monster");
monster[i].setBonus(sc.nextInt());
}
//calculate win or loose
int count = 0;
int flag = 0;
//**
for (int i = 0; i < n; i++) {
if (player.getExp() >= monster[i].getPower()) {
player.setExp(player.getExp() + monster[i].getBonus());
count = count + 1;
// flag = flag + 1;
} else if (player.getExp() < monster[i].getPower()) {
for (int j = 0; j < n; j++)
arr[j] = i;
//count = count;
flag = flag + 1;
}
//**
for (int t = 0; t < flag; t++) {
if (player.getExp() >= monster[arr[t]].getPower()) {
count = count + 1;
}
}
}
return count;
}
}
public class Main {
public static void main(String[] args) {
// write your code here
System.out.println("welcome to the loosing game");
Calc c = new Calc();
System.out.println("no of monster killed" + c.calc());
}
}
#include<bits/stdc++.h>
using namespace std;
struct Monsters
{
int power;
int bonus;
};
class Solution
{
public:
//Function to find the maximum number of activities that can
//be performed by a single person.
static bool compare(Monsters a, Monsters b)
{
if(a.power < b.power) return 1;
else if(a.power > b.power) return 0;
else return 1;
}
int MonstersKill(vector<int> power, vector<int> bonus, int n, int exp)
{
Monsters arr[n];
for(int i = 0; i < n; i++)
{
arr[i].power = power[i];
arr[i].bonus = bonus[i];
}
sort(arr, arr+n, compare);
int count = 0;
for(int i = 0; i < n; i++)
{
if(arr[i].power <= exp)
{
count++;
exp += arr[i].bonus;
}
else
break;
}
return count;
}
};
int main()
{
int n, exp;
cin >> n >> exp;
vector<int> power(n), bonus(n);
//adding elements to arrays power and bonus
for(int i=0;i<n;i++)
cin>>power[i];
for(int i=0;i<n;i++)
cin>>bonus[i];
Solution ob;
cout << ob.MonstersKill(power, bonus, n, exp) << endl;
return 0;
}
Time Complexity : O(N * Log(N))
Auxilliary Space: O(N)
I need to fill a 2D Array with numbers between 2 and 6, given by the user (is just part of a bigger proyect) but when I give the number I only get another request for a number.
public static int[][] crearTablero(int tamaño)
{
int[][] tablero = new int[tamaño][tamaño];
return tablero;
}
public static void imprimeTablero(int[][] tablero)
{
for(int i = 0; i<tablero.length; i++)
{
for(int j = 0; j<tablero[i].length; j++)
{
System.out.print(tablero[i][j] + " ");
}
System.out.println();
}
}
public static void swap(int[][] tablero, int x1, int y1, int x2, int y2)
{
int temp = tablero[x1][y1];
tablero[x1][y1] = tablero[x2][y2];
tablero[x2][y2] = temp;
}
public static void rellenarTablero(int[][] tablero) {
for (int x = 0; x < tablero.length; x++) {
for (int y = 0; y < tablero[x].length; y++) {
tablero[x][y] = aleatorio(numeroColores());
}
}
}
public static void shuffleBoard(int[][] tablero)
{
Random rnd = new Random();
int randX = 0;
for(int x = 0; x<tablero.length; x++)
{
randX = rnd.nextInt(tablero.length);
int[] temp = tablero[x];
tablero[x] = tablero[randX];
tablero[randX] = temp;
}
}
public static int numeroColores(){
int colores = 0;
System.out.print("Numero de colores (entre 2 y 6): ");
Scanner scn = new Scanner(System.in);
colores = scn.nextInt();
while(colores < 2 || colores > 6)
{
System.out.println("Invalid matrix size. Re-enter ");
}
return colores;
}
public static int aleatorio(int colores) {
int l = (int) (Math.floor(Math.random()*(colores-2)) + 2);
return l;
}
I would really appreciate some help because I don't know how to continue, Thanks.
You call numeroColores() in a for-loop in a for-loop, so you are of course asked multiple times for it.
Btw. you have an endless loop if you type in 1 or smaller or 7 or bigger with constantly getting the same line printed out and not asking for new input
Try this code to generate the random value between 2 and 6
public static int aleatorio(int colores) {
int l = 0;
while(l < 2 || l > 6) {
l = (int) (Math.floor(Math.random()*(colores-2)) + 2);
}
return l;
}
I'm a absolutly beginner in java and i want to write a code with the acm.libary, which is about the fibonacci sequence.
The result is very nice for me, but i want only print the last number of the sequence. I don't know how.
If the user type n = 5, the result need to be 8.
If the user type n = 8, the result need to be 21.
In my program it is the last number, but the program also prints all the previous numbers.
I hope you can understand me :D
Thank you in advance!
int a = 1;
int b = 0;
public void run() {
int n = readInt ("n: ");
for(int i = 1; i <= n; i++) {
println (fibonacci (n));
}
}
private int fibonacci(int n) {
int c = (a) + (b);
a = b;
b = c;
return c;
}
Try this code.
import java.util.Scanner;
public class Test {
int a = 1;
int b = 0;
public int run() {
#SuppressWarnings("resource")
Scanner s = new Scanner(System.in);
int n = s.nextInt();
int value =0;
for(int i = 1; i <= n; i++) {
value = fibonacci (n);
}
return value;
}
private int fibonacci(int n) {
int c = (a) + (b);
a = b;
b = c;
return c;
}
public static void main (String arg[])
{
Test t = new Test();
System.out.println(t.run());
}
}
You can replace the body of the loop by this:
if (i == n) {
println (fibonacci (n));
} else {
fibonacci (n);
}
Basically, I would like to know if there is a shorter way of printing out the Fibonacci numbers from 0 - 100.
What I have done is probably very basic, but here is the code:
public static void main(String[] args) {
int number[] = new int[100];
number[0] = 0;
number[1] = 1;
int sum1 = number[0] + number[0];
int sum2 = sum1 + number[1];
int sum3 = sum1 + sum2;
int sum4 = sum2 + sum3;
int sum5 = sum3 + sum4;
int sum6 = sum4 + sum5;
System.out.println(sum1);
System.out.println(sum2);
System.out.println(sum3);
System.out.println(sum4);
System.out.println(sum5);
System.out.println(sum6);
}
And I would be doing this up until 100. But I'm sure there is a quicker way of doing this. How?
You can use a loop. This example uses BigInteger as the number quickly become too large for long.
BigInteger a = BigInteger.ZERO, b = BigInteger.ONE;
System.out.println(1);
for (int i = 0; i < 100000; i++) {
BigInteger c = a.add(b);
System.out.println(c);
a = b;
b = c;
}
prints after a couple of seconds finally
420269270299515438 ... many, many digits deleted ... 9669707537501
Note: you don't need to memorize all the previous values, just the last two.
Standard way to do this is by recursion:
Have a look at this simple snippet, which returns the fibonacci number at position a:
public static long fib(int a){
if (a==1||a==2) return 1;
else return fib(a-1)+fib(a-2);
}
google search ftw!... wasn't even a minute searching
int[] fibonacci = new int[25+1];
fibonacci[1] = 1;
fibonacci[2] = 1;
for ( int i = 3; i < fibonacci.length; i++ )
{
fibonacci[i] = fibonacci[i-2] + fibonacci[i-1];
}
Print number < 100 with n=11
public int getFib(int n){
if(n==0) return 0;
else if(n==1) return 1;
else{
int temp=getFib(n-1)+getFib(n-2);
return temp;
}
}
Just in case you don't want to use the recursive way..
Here is the iterative...
public class Fib2 {
public static int fib(int n, int a, int b)
{
if (n==0)
{
System.out.print("1x +");
return a;
}
else
{
System.out.print("2x +");
return fib(n-1,b,a+b);
}
}
public static void main(String arg[])
{
System.out.println(fib(0,1,1));
}
}
Well, 100 isn't one of the Fibonacci numbers, however if we include 144, then this is a nice succinct example:
class F
{
public static void main( final String[] a )
{
int a = 1, b = 1;
for ( ; b < 145; a = b + (b = a) )
System.out.println( b );
}
}
If you have to do this without arrays here is the way to do it. Just change 20 to 100. This is the way I had to do it for an assignment and it is simple to understand for those still learning.
public static void main(String[] args)
{
int number1 = 1;
int number2 = 1;
int count;
int fib;
System.out.print(number1 + " " + number2 + " ");
for(count = 3; count <= 20; count++)
{
fib = number1 + number2;
number1 = number2;
number2 = fib;
System.out.print(fib + " ");
}
}
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);
}
}