"""Any base to decimal
>>> CHECK THE SOURCE!<<<
Conversion of number from any base to decimal!
Answer by paw88789 on Nov 2,2017 on Math StackExchange:+------------------------------------------------------------------------+
A similar thing occurs if someone reads you, digit-by-digit,......left-to-right, a base ten numeral:
E.g. The actual numeral is 3702, but you only get one digit at a time:[3] Current value is 3.[7] Current the value is 3⋅10+7=37.[0] Current the value is 37⋅10+0=370.[2] Current the value is 370×10+2=3702.
End of number. The last current value is the actual value.
At each step, you multiply the previous total by the base......and add the next digit.+------------------------------------------------------------------------+"""
//* How to convert a decimal number into a binary number#include<bits/stdc++.h>usingnamespace std;classNumberConversion{int n =0, count =0, ans =0;public:voiddecimalToBinary(int n){this->n = n;while(this->n !=0){int rem =this->n %2;this->n /=2;
ans += rem *pow(10, count);
count++;}}voiddisplay(){
cout << ans << endl;}};intmain(){int number =0;
cin >> number;
NumberConversion num1;
num1.decimalToBinary(number);
num1.display();return0;}
#include<stdio.h>intmain(){int num, binary_val, decimal_val =0, base =1, rem;printf("Insert a binary num(1s and0s)
");scanf("%d",&num);
binary_val = num;while(num >0){
rem = num %10;
decimal_val = decimal_val + rem * base;
num = num /10;//these are the correct lines
base = base *2;//these are the correct lines}printf("%d
", binary_val);printf("%d
", decimal_val);return0;}
importjava.util.Scanner;publicclassBinaryToDecimal{publicstaticvoidmain(String[] args){// TODO Auto-generated method stub
Scanner sc =newScanner(System.in);
String binary = sc.next();// This is one line solution for binary to decimal in java
System.out.println(Integer.parseInt(binary,2));//This solution is based on formula for binary to decimal conversionint n=0,dec=0;for(int i=binary.length()-1;i>=0;i--){
dec = dec + Integer.parseInt(String.valueOf(binary.charAt(i)))*(int)Math.pow(2,n);
n++;}
System.out.println(dec);}}