Skip to main content

BINARY TO DECIMAL

AJ Java Programs

Program 2:

Write a program to Convert Binary Number into a Decimal Equivalent
Example: 11101 = 29

Syntax:

import java.io.*;
class bintodec //binary to decimal
{
public static void main(String args[])throws IOException
{
    String binary;
    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
    System.out.println("Enter a biary Number");
    binary=br.readLine();
    int decimal=0;
    int power=0;
 
    //while(binary.length()>0)
    while(binary.length()>0)
    {
        int temp = Integer.parseInt(binary.charAt((binary.length())-1)+"");
        decimal+=temp*Math.pow(2, power++);
        binary=binary.substring(0,binary.length()-1);
    }
    System.out.println(decimal);
}
}

Tested Ok Program
If You Don't Understand anything or want any program please write your request in Comment Section
And Please Like The Post

Comments

Popular posts from this blog

Arranging according to potential.

PROGRAM: Requested by Sir  Ankur Acharya   ARRANGING WORDS IN A STRING ACCORDING TO THEIR POTENTIAL The encryption of alphabets are to be done as follows: A = 1 B = 2 C = 3 . . . Z = 26 The potential of a word is found by adding the encrypted value of the alphabets. Example: KITE Potential = 11 + 9 + 20 + 5 = 45 Accept a sentence which is terminated by either “ . ” , “ ? ” or “ ! ”. Each word of sentence is separated by single space. Decode the words according to their potential and arrange them in ascending order. Output the result in format given below: Example 1 INPUT : THE SKY IS THE LIMIT. POTENTIAL : THE = 33 SKY = 55 IS = 28 THE = 33 LIMIT = 63 OUTPUT : IS THE THE SKY LIMIT Example 2 INPUT : LOOK BEFORE YOU LEAP. POTENTIAL : LOOK = 53 BEFORE = 51 YOU = 61 LEAP = 34 OUTPUT : LEAP BEFORE LOOK YOU SYNTAX import java.io.*; class Potential {     public static void main(String args[])throws IOException     {         BufferedReader br=new BufferedReader(new Inp

SMITH NUMBER

PROGRAM: SMITH NUMBER A   Smith number  is a composite number, the sum of whose digits is the sum of the digits of its prime factors obtained as a result of prime factorization (excluding 1). The first few such numbers are 4, 22, 58, 85, 94, 121 ……………….. For Example 58=> 5+8=13 2+(2+9) of 29 =13 Since sum of digits=sum of digits of prime factors 58 is a SMITH NUMBER SYNTAX: import java.io.*; class smith_no { static int sum,sum1;     public static void main(String args[])throws IOException     {         BufferedReader br=new BufferedReader(new InputStreamReader(System.in));         int n;int c=0;         System.out.println("Enter a Number");         n=Integer.parseInt(br.readLine());         int n1=n;         while(n1!=0){         int d=n1%10;         sum1+=d;         n1=n1/10;}         for(int i=2;i<n;i++)         {             if(n%i==0)             {                 for(int j=2;j<i;j++)                 {                     if(i

MAGICAL/ MAGIC NUMBER USING RECURSION

PROGRAM :  MAGICAL NUMBER BY RECURSION This  program  checks if a  number  is a  Magic number  in  JAVA . A  number  is said to be a  Magic number  if the sum of its digits are calculated till a single digit is obtained by recursively adding the sum of its digits. If the single digit comes to be 1 then the  number  is a  magic number . Example: 10999 => 1+0+9+9+9=28=>2+8=>10=>1+0= 1                  199    =>1+9+9 =>19=> 1+9=>10=>1+0= 1 SYNTAX import java.io.*; class magic  {    static int c;     public static void main(String arsg[])throws IOException     {         BufferedReader br=new BufferedReader(new InputStreamReader(System.in));         int n;int d=0;int m;         System.out.println("ENter a NUmber");         n=Integer.parseInt(br.readLine());         m=n;         while(n!=0)         {d=n%10;             n=n/10;             c++;         }         magical(m);              }     public static void magical(int x)