Bear and Candies 123 Codechef Solution: 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
Sample Input 1
10 3 2 4 2 1 1 1 2 1 3 9 3 9 11 9 12 9 1000 8 11
Sample Output 1
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() {
// your code goes here
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 Chase2learn.This tutorial is only for Educational and Learning purpose.
