Hello coders, today we are going to solve Byte to Bit Codechef Solution.
Problem
In the magical land of Byteland, there are three kinds of citizens:
- a Bit – 2ms2ms after a Bit appears, it grows up and becomes a Nibble (i.e. it disappears and a Nibble appears)
- a Nibble – 8ms8ms after a Nibble appears, it grows up and becomes a Byte
- a Byte – 16ms16ms after a Byte appears, it grows up, splits into two Bits and disappears
Chef wants to know the answer to the following question: what would the population of Byteland be immediately before the time NmsNms if only 1 Bit appeared at time 0ms0ms?
Help him and find the population (number of citizens) of each type.
Input
- The first line of the input contains a single integer TT denoting the number of test cases. The description of TT test cases follows.
- The first and only line of each test case contains a single integer NN.
Output
For each test case, print a single line containing three space-separated integers — the number of Bits, Nibbles and Bytes.
Constraints
- 1≤T≤1041≤T≤104
- 1≤N≤1,6001≤N≤1,600
Subtasks
Subtask #1 (25 points): 1≤N≤1401≤N≤140
Subtask #2 (75 points): original constraints
Sample Input 1
2 2 3
Sample Output 1
1 0 0 0 1 0
Explanation
Immediately before the time 2ms2ms, there is only one Bit. At 2ms2ms, this Bit grows up, so immediately before 3ms3ms, there is only one Nibble in Byteland.
Byte to Bit CodeChef Solution in JAVA
import java.util.Scanner;
class Byte_to_Bit {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int t = input.nextInt();
while(t-- > 0) {
int n = input.nextInt();
long ans = 1;
while(n > 26) {
ans *= 2;
n -= 26;
}
if (n<=2) {
System.out.println(ans + " 0" + " 0");
} else if (n>2 && n<= 10) {
System.out.println("0 " + ans + " 0");
} else {
System.out.println("0 " + "0 " + ans);
}
}
input.close();
}
}Byte to Bit CodeChef Solution in CPP
#include <iostream>
using namespace std;
#define ll long int
int main() {
// your code goes here
int t;
cin>>t;
while(t--)
{
ll n;
cin>>n;
ll x=1,y=0,z=0;
ll count=0;
while(1)
{
count+=2;
if(count>=n)
{
break;
}
else
{
y=x;
x=0;
count+=8;
if(count>=n)
{
break;
}
else
{
z=y;
y=0;
count+=16;
if(count>=n)
{
break;
}
else
{
x=2*z;
z=0;
}
}
}
}
cout<<x<<" "<<y<<" "<<z<<endl;
}
return 0;
}Byte to Bit CodeChef Solution in Python
for _ in range(int(input())):
n=int(input())
if(n%26==0):
a=n//26-1
else:
a=n//26
a=2**a
b=n%26
if b in range(1,3):
print(a,0,0)
elif b in range(3,11):
print(0,a,0)
else:
print(0,0,a)Disclaimer: The above Problem (Byte to Bit) is generated by CodeChef but the solution is provided by Chase2learn.This tutorial is only for Educational and Learning purpose.