# Chef & Easy Queries(Solution)-Codechef October Long Challenge Solution

**Question**

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

# Input:

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

# Output:

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

# Constraints

- 1≤T≤10⁵
- 1≤ sum of n over all testcases ≤10⁵
- 1≤k≤10⁸
- 0≤Qi≤10⁸

# Subtasks

- Subtask 1–20% points — Sum of Qi over all testcases and days ≤3.106≤3.106
- 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.

**Solution**

import java.util.*;

import java.lang.*;

import java.io.*;class Codechef

{

public static void main (String[] args) throws java.lang.Exception

{

Scanner sc = new Scanner(System.in);

long t = sc.nextLong();

while(t > 0){

int n = sc.nextInt();

long k = sc.nextLong();

long arr[] = new long[n];

long sum =0,day=0,flag=0;

for(int i=0;i<n;i++){

arr[i] = sc.nextLong();

sum += arr[i];

}

for(int i=0;i<n-1;i++){

if(arr[i] < k){

flag =1;

day =i;

break;

}

arr[i+1] += (arr[i]-k);

}

if(flag == 1){

System.out.println(day+1);

}

else{

day = (sum / k)+1;

System.out.println(day);

}

t--;

}

}

}

This solution is Just for the understanding of logic after the contest is over. In no way I urge you to cheat in the contest. Remember these challenges are only for learning.

Question Link

https://www.codechef.com/OCT20B/problems/CHEFEZQ