Chef and Easy Queries codechef October long challenge problem solution editorial

Chef and Easy Queries CodeChef October long challenge problem solution -

This problem (Chef and Easy Queries) is taken from CodeChef October long challenge.

Problem Statement-

Chef published a blog post, and is now receiving many queries about it. On day i, he receives Qi$Q_i$ queries. But Chef can answer at most k$k$ queries in a single day.

Chef always answers the maximum number of questions that he can on any given day (note however that this cannot be more than k). The remaining questions (if any) will be carried over to the next day.

Fortunately, after n$n$ days, the queries have stopped. Chef would like to know the first day during which he has some free time, i.e. the first day when he answered less than k$k$ questions.

Input:

• First line will contain T$T$, the number of testcases. Then the testcases follow.
• The first line of each testcase contains two space separated integers n$n$ and k$k$.
• The second line of each testcase contains n$n$ space separated integers, namely Q1,Q2,...Qn$Q_1,Q_2,...Q_n$.

Output:

For each testcase, output in a single line the first day during which chef answers less than k$k$ questions.

Constraints

• 1≤T≤105$1\le T\le {10}^5$
• 1≤$1\le$ sum of n$n$ over all testcases ≤105$\le {10}^5$
• 1≤k≤108$1\le k\le {10}^8$
• 0≤Qi≤108$0\le Q_i\le {10}^8$

Subtasks

• Subtask 1 - 20% points - Sum of Qi$Q_i$ over all testcases and days ≤3.106$\le 3.{10}^6$
• Subtask 2 - 80% points - Original constraints

Sample Input:

2 
6 5 
10 5 5 3 2 1 
1 1
100

Sample Output:

6
101

Explanation:

Test Case 1

On the first day, chef answers 5 questions and leaves the remaining 5 (out of the 10) for the future days.

On the second day, chef has 10 questions waiting to be answered (5 received on the second day and 5 unanswered questions from day 1). Chef answers 5 of these questions and leaves the remaining 5 for the future.

On the third day, chef has 10 questions waiting to be answered (5 received on the third day and 5 unanswered questions from earlier). Chef answers 5 of these questions and leaves the remaining 5 for later.

On the fourth day, chef has 8 questions waiting to be answered (3 received on the fourth day and 5 unanswered questions from earlier). Chef answers 5 of these questions and leaves the remaining 3 for later.

On the fifth day, chef has 5 questions waiting to be answered (2 received on the fifth day and 3 unanswered questions from earlier). Chef answers all 5 of these questions.

On the sixth day, chef has 1 question, which he answers. This is the first day he answers less than 5 questions, and so the answer is 6.

Test Case 2

Chef answers 1 question a day for the first 100 days. On day 101, he is free

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Chef and easy queries is also a simple problem from october long challenge so i will explain this problem into very simple language.This problem can be done by using brute froce approach.

Logic -

Logic of this problem divided into 5 step they are mentioned below.

step-1 take t,n and k;

step -2 maintain sum variable, prev variable and day variable. sum stores sum of all quaries and prev store sum of remaining queries and 

day store very first day when chef has free time.They sould be long long int otherwise your program show wrong answer do care about long long int.

step -3 maintain a check flag which tells does chef have free time or not in very first-day chef has free time and so mark flag as true 

initially.

step-4 while iterating trough i=0 to n take input 

long long  int d;

        cin>>d;

and add d into sum variable and add d to remaining queries which is stored in prev variable.

sum+=d;

 prev+=d;

and check is previous is less then k if yes mark check as false  so that we can not enter 

in the if statement again because here we 

got very first day when chef has free time.

if(prev<k&&check){

            day=i+1;

            check=false;

            

        }

step -5 if it is not so subtract k queries from previous queries because he will solve k queries on ith day.

reapeat all step till we reach the answer.

after iterating frolm i=0 to n we can decide wether chef can have free time or not 

if our check is marked as false it mean chef got free time before all nth day  

so i will print day 

other wise devide total sum by k because chef will solve average k questies per day and print (sum/k +1)th day.

Code for Problem mentioned below-