# PROBLEM LINK:

Div1, Div2Practice

**Author:**Praveen Dhinwa

**Tester:**Triveni Mahatha

**Editorialist:**Adarsh Kumar

Practice

I have a doubt.

For the following test case:

`5 7 9 3 8 10 5`

The answer should be 6. So taking k=6, the above inequality will be calculated as following:

`9/6 + 3/6 + 8/6 + 10/6 + 5/6 (Ceil of all the terms are added) 2 + 1 + 2 + 2 + 1 =8`

Hour available is 7, so am I doing something. If there’s any problem in logic, please point out the correct solution to above test case and logic.

Thanks

@philomath_mht the ans. will be 8 as you can only divide maximum (7-5)=2 bunch of bananas and the minimum required bananas that should be eaten is 8.

9/8 + 3/8 + 8/8 + 10/8 + 5/8 = 2 + 1 + 1 + 2 + 1 = 7

Simple binary search : Link to solution https://www.codechef.com/viewsolution/17582805

Change int to long long && float to double. Here’s your modified solution :

you can look at my code. its almost similar to your code(99%same)

link -**https://www.codechef.com/viewsolution/17841451**

Do upvote if you find useful.

https://www.codechef.com/viewsolution/17841507

Can someone tell me what’s wrong with this???

https://www.codechef.com/viewsolution/17889934

can someone tell me what’s wrong with this??

Can someone tell me what is the mistake in my code:

https://www.codechef.com/viewsolution/17768075

I was getting a PA with my solution and just after I changed

```
**
ans += std::ceil((float)a*/k); to
ans += (a* + hr - 1)/hr ;
**
```

it go AC…**will someone please explain how the latter is different then the first one?**

what if we sum all the piles and divide them with the no. of hours given and

take the upper bound of the quotient.