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# C++ code to find who cannot give sufficient candies

Suppose we have two numbers a and b. There are a and b number of candies in Amal's and Bimal's hand. Amal offered 1 candy to Bimal and Bimal gave two candies to Amal, in the next turn Amal gave 3 candies and Bimal gave 4 and so on. This continued until the moment when one of them couldn’t give the right amount of candy. They don’t consider the candies they have got from the opponent, as their own. We have to find, who is the first can’t give the right amount of candy.

So, if the input is like a = 7; b = 6, then the output will be Amal, because initially Amal gave 1, Bimal gave 2, then Amal gave 3 and Bimal gave 4, now in this turn Amal has to give 5 candies but he has only 4.

## Steps

To solve this, we will follow these steps −

x := square root of a if x * (x + 1) > b, then: return "Bimal" Otherwise return "Amal"

## Example

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h> using namespace std; string solve(int a, int b){ int x = sqrt(a); if (x * (x + 1) > b) return "Bimal"; else return "Amal"; } int main(){ int a = 7; int b = 6; cout << solve(a, b) << endl; }

## Input

7, 6

## Output

Amal

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