## Algorithm

Problem Name: beecrowd | 1178

# Array Fill III

Adapted by Neilor Tonin, URI Brazil

Timelimit: 1

Read a number X. Put this X at the first position of an array N [100]. In each subsequent position (1 up to 99) put half of the number inserted at the previous position, according to the example below. Print all the vector N.

## Input

The input contains a double precision number with four decimal places.

## Output

For each position of the array N print "N[i] = Y", where i is the array position and Y is the number stored in that position. Each number of N[...] must be printed with 4 digits after the decimal point.

 Input Sample Output Sample 200.0000 N[0] = 200.0000 N[1] = 100.0000 N[2] = 50.0000 N[3] = 25.0000 N[4] = 12.5000 ...

## Code Examples

### #1 Code Example with C++ Programming

```Code - C++ Programming```

``````
#include <iostream>
#include <iomanip>
using namespace std;
int main(){
double n;
cin >> n;
for(int i = 0; i  <  100; i++){
cout << fixed << setprecision(4);
cout << "N[" << i << "] = " << n << "\n";
n = n / 2;
}
}
``````
Copy The Code &

Input

cmd
200.0000

Output

cmd
N[0] = 200.0000 N[1] = 100.0000 N[2] = 50.0000 N[3] = 25.0000 N[4] = 12.5000 ...

### #2 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const lines = input.split('\n');

let n = parseFloat(lines.shift());
for(var i = 0; i  <  100; i++){
console.log(`N[\${i}] = \${(((n).toLocaleString('en-US',{ minimumFractionDigits: 4 })).toString()).replace(/,/g, "")}`);
n = n / 2.0;
}
``````
Copy The Code &

Input

cmd
200.0000

Output

cmd
N[0] = 200.0000 N[1] = 100.0000 N[2] = 50.0000 N[3] = 25.0000 N[4] = 12.5000 ...

### #3 Code Example with Python Programming

```Code - Python Programming```

``````
n = float(input())
for i in range(0, 100):
print(f'N[{i}] = {n:.4f}')
n = n / 2
``````
Copy The Code &

Input

cmd
200.0000

Output

cmd
N[0] = 200.0000 N[1] = 100.0000 N[2] = 50.0000 N[3] = 25.0000 N[4] = 12.5000 ...