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Exponential notation of a decimal number

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  • Difficulty Level : Medium
  • Last Updated : 02 Jun, 2022

Given a positive decimal number, find the simple exponential notation (x = a·10^b) of the given number. Examples:

Input : 100.0
Output : 1E2
Explanation:
The exponential notation of 100.0 is 1E2.

Input :19
Output :1.9E1
Explanation:
The exponential notation of 16 is 1.6E1.

Approach: The simplest way is to find the position of the first non zero digit and the position of the dot. The difference between that positions is the value of b (if the value is positive you should also decrease it by one). Below is the implementation of the above approach: 

C++




// C++ code to find the exponential notation
#include <bits/stdc++.h>
using namespace std;
 
// function to calculate the exponential
// notation
void FindExponent(char s[], int n)
{
    int i, j, b, c;
    for (i = 0; s[i] == '0' || s[i] == '.'; i++)
        ;
    for (j = n - 1; s[j] == '0' || s[j] == '.'; j--)
        ;
 
    c = find(s, s + n, '.') - s;
    putchar(s[i]);
 
    if (i != j)
        putchar('.');
 
    for (int k = i + 1; k <= j; k++)
        if (s[k] != '.')
            putchar(s[k]);
    if (i < c)
        b = c - i - 1;
    else
        b = c - i;
    if (b)
        printf("E%d", b);
}
 
// main function
int main()
{
    char s[] = "100";
    int n = strlen(s);
    FindExponent(s, n);
}

Javascript




// JavaScript code to find the exponential notation
 
// function to calculate the exponential
// notation
function FindExponent(s, n){
    // Storing the result in an array
    const res = new Array();
     
    let i, j, b, c;
    for (i = 0; s[i] == '0' || s[i] == '.'; i++);
    for (j = n - 1; s[j] == '0' || s[j] == '.'; j--);
     
    // Finding the first index in s which is equal to '.'
    c = s.findIndex(function (element) {
        return element == '.';
    });
 
    // if '.' not found then put c = n.
    if(c == -1){
        c = n;
    }
     
    res.push(s[i]);
     
    if (i != j){
        res.push('.');
    }
         
    for (let k = i + 1; k <= j; k++){
        if (s[k] != '.'){
            res.push(s[k]);
        }
    }
 
    if (i < c){
        b = c - i - 1;
    }     
    else{
       b = c - i;
    
     
    if (b){
        res.push("E");
        res.push(b);
    
    console.log(res.join(''));
}
 
// main function
{
    let s = "100";
    let n = s.length;
    FindExponent(s.split(''), n);
}
 
// The code is contributed by Gautam goel (gautamgoel962)

Output:

1E2


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