spoj#KALTSUM. k Alternating Sum
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.