# Bear and Candies 123 Codechef Solution Hello Programmers In this post, you will know how to solve the Bear and Candies 123 Codechef Solution.

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Codechef by yourself. If you find any difficulty after trying several times, then you can look for solutions.

### Problem

Bears love candies and games involving eating them. Limak and Bob play the following game. Limak eats 1 candy, then Bob eats 2 candies, then Limak eats 3 candies, then Bob eats 4 candies, and so on. Once someone can’t eat what he is supposed to eat, he loses.

Limak can eat at most A candies in total (otherwise he would become sick), while Bob can eat at most B candies in total. Who will win the game? Print “Limak” or “Bob” accordingly.

### Input

The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows.

The only line of each test case contains two integers A and B denoting the maximum possible number of candies Limak can eat and the maximum possible number of candies Bob can eat respectively.

### Output

For each test case, output a single line containing one string — the name of the winner (“Limak” or “Bob” without the quotes).

### Constraints

• 1 ≤ T ≤ 1000
• 1 ≤ A, B ≤ 1000

```10
3 2
4 2
1 1
1 2
1 3
9 3
9 11
9 12
9 1000
8 11
```

```Bob
Limak
Limak
Bob
Bob
Limak
Limak
Bob
Bob
Bob
```

### Explanation

Test case 1. We have A = 3 and B = 2. Limak eats 1 candy first, and then Bob eats 2 candies. Then Limak is supposed to eat 3 candies but that would mean 1 + 3 = 4 candies in total. It’s impossible because he can eat at most A candies, so he loses. Bob wins, and so we print “Bob”.

Test case 2. Now we have A = 4 and B = 2. Limak eats 1 candy first, and then Bob eats 2 candies, then Limak eats 3 candies (he has 1 + 3 = 4 candies in total, which is allowed because it doesn’t exceed A). Now Bob should eat 4 candies but he can’t eat even a single one (he already ate 2 candies). Bob loses and Limak is the winner.

Test case 8. We have A = 9 and B = 12. The game looks as follows:

• Limak eats 1 candy.
• Bob eats 2 candies.
• Limak eats 3 candies (4 in total).
• Bob eats 4 candies (6 in total).
• Limak eats 5 candies (9 in total).
• Bob eats 6 candies (12 in total).
• Limak is supposed to eat 7 candies but he can’t — that would exceed A. Bob wins.

### Bear and Candies 123 CodeChef Solution in JAVA

```import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
class practice
{
public static void main(String args[])
{
Scanner sc=new Scanner(System.in);
int t= sc.nextInt();
while(t -- > 0)
{
int a = sc.nextInt();
int b = sc.nextInt();
for(int i=0; i<1000; i++)
{
if(i % 2 == 1)
{
a=a-i;
if(a < 0)
{
System.out.println("Bob");
break;
}
}
else
{
b=b-i;
if(b < 0)
{
System.out.println("Limak");
break;
}
}
}
}
}
}```

### Bear and Candies 123 CodeChef Solution in CPP

```#include <iostream>
using namespace std;
int main() {
int t;
cin>>t;
while(t--){
int a,k;
int b;
cin>>a>>b;
int i=1;
int j=0;
int counter1=0;
int counter2=0;
int t1=1;
int t2=0;
while(t1<=a){
counter1++;
i=i+2;
t1=t1+i;
}
while(t2<=b){
counter2++;
j=j+2;
t2=t2+j;
}
k=counter2-1;
if(counter1>k){
cout<<"Limak"<<endl;
}
else {
cout<<"Bob"<<endl;
}
}
return 0;
}
```

### Bear and Candies 123 CodeChef Solution in Python

```t = int(input())
for i in range(t):
a,b = map(int,input().split())
a_1 = 0
b_1 = 0
temp = 1
while True:
if temp%2 == 1:
if a_1+temp > a:
print('Bob')
break
a_1 += temp
temp += 1
else:
if b_1 + temp > b:
print('Limak')
break
b_1 += temp
temp +=1
```

Disclaimer: The above Problem (Bear and Candies 123 ) is generated by CodeChef but the solution is provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.

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