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KALTSUM - k Alternating Sum |
Sameen has:
- An array having N integers,
- Q friends.
His friends are curious about the array. So, each of his friends asks Sameen a question about the array. Every question is described by 3 integers: i, j and k. In reply to a question, Sameen has to say the “k alternating sum” of the subarray starting at position i and ending at position j [1 based indexing].
“k alternating sum” of a subarray starting at position i and ending at position j can be calculated in the following way:
Add the first k numbers [starting from position i]
Subtract the second k numbers [starting from position i+k]
Add the third k numbers [starting from position i+2*k]
Subtract the fourth k numbers [starting from position i+3*k]
And so on till adding/subtracting the j-th number…
(j-i+1) will be divisible by k.
[See sample Input/output and explanation section for more details]
Can you help Sameen in answering the questions?
Input
The first line of input contains two integers N and Q. The next line contains N integers, the numbers in the array. Then each of the following Q lines contains 3 integers i, j & k.
Output
For each query output an integer in a separate line, the answer for that query. Queries should be answered in the order given in the input.
Constraints:
1 ≤ k ≤ 100000
1 ≤ N ≤ 100000
1 ≤ Q ≤ 100000
-1000000000 ≤ Value of a number in the array ≤ 1000000000
(j-i+1) will be divisible by k.
Example
Input: 6 6 4 1 -2 -3 4 5 2 5 2 1 6 1 1 6 3 1 6 6 3 3 1 3 4 1 Output: -2 3 -3 9 -2 1
Explanation
In the first query, the subarray is [ 1, -2, -3, 4].
So “2 alternating sum” is equal to: [1-2]-[-3+4] = -2
For the second query, we get [4]-[1]+[-2]-[-3]+[4]-[5] = 3
N.B: Dataset is huge. Use faster I/O method.
Added by: | Bertho Coder |
Date: | 2016-04-27 |
Time limit: | 1s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ASM64 GOSU JS-MONKEY |
Resource: | National High School Programming Contest 2016 |
hide comments
2019-04-03 10:47:53 Tanmoy Tapos Datta
Perfect example for heavy light trick. |
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2016-10-26 21:52:25
This is one badass problem!..just think of the kind of time complexity ur program needs in order to avoid TLEs..and after that think of how you can implement the code of that time complexity.. U will definitely find some way.. |
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2016-10-04 17:39:01
O((r-l+1)*q) is TLE ? I am definitely missing something obvious.. :/ Reply: You need to do better. Last edit: 2016-12-30 01:16:09 |
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2016-08-14 16:03:43 Rishav Goyal
easy ..... just one trick. < 1 minute solution Last edit: 2016-08-14 16:04:00 |
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2016-05-23 13:15:26
getting TLE :( how to overcome it ?? |