Add Tutorial for LOJ-1369: Answering Queries (en) (#443)

* Added tutorial for LOJ-1369 Answering Queries

* Format 1369

---------

Co-authored-by: Rafid Bin Mostofa <[email protected]>

Author: Arup43

1369/en.md

@@ -0,0 +1,111 @@
+# LOJ 1369 - Answering Queries
+
+First let's understand what the given code is doing.
+There will be two types of queries, one is calculating the sum and the other 
+one is to change the value of an element of the given array `A[]`.
+If you calculate the sum in the way the given code does, the complexity of 
+the code will be O(t * q * n * n), which will not fit in the Time limit.
+So, we need to do the same thing what the given code does but in an efficient
+way. But what that way could be? Let's see!
+
+We can use _contribution technique_ to solve this problem. Firstly, we will 
+determine how much contribution an element of the array does to the `sum`.
+
+For example, let's take an array of four elements {1, 4, -2, 8}.
+The `sum` will be the summation of (1-4) + (1-(-2)) + (1-8), (4-(-2)) + (4-8) and (-2-8).
+Here first element is summed for 3 times and subtracted for 0 times, 
+so the contribution of 1 to the `sum` is 3 - 0 = 3.
+Again second element is summed for 2 times and subtracted for 1 times,
+so the contribution of 4 to the `sum` is 2 - 1 = 1.
+In the same way, the contributions of the third element -2
+and fourth element 8 to the `sum` are -1 and -2 respectively.
+
+So, in this way we will calculate the contribution of each element of the
+array to the `sum` and store it to an array named `contribution[n]`.
+_How can we do that?_
+
+Observe the pattern: each element is summed for the exact same number of
+elements it has after it and that number will be `(n - index - 1)`.
+In the same way, each element is subtracted for the exact same number of
+elements it has before it and that number is the index of that element `index`.
+So we can calculate the contribution of each element in this way:
+```
+contribution[index] = (n - index - 1) - index;
+```
+
+Now we can calculate the `sum` by going through every element of the array
+and add `A[index] * contribution[index]` to the sum.
+Even if the value of an element changes, the corresponding value in the
+`contribution` array of that element will still remain the same, because
+`contribution` array has nothing to do with elements' value rather it
+tells us how `sum` is calculated.
+
+So, when the query will be to change the value of an element, we will just
+recalculate the `sum` in the following manner in constant time,
+no need to loop over the whole array.
+
+```
+sum -= arr[x] * contribution[x];
+arr[x] = v;
+sum += arr[x] * contribution[x];
+```
+
+The time complexity of this approach will be O(t * (n + q)). _How?_
+
+### C++ Code
+
+```cpp
+
+/*
+    Author: Arup Debnath
+    Date: 24 March, 2023
+
+    Happy Coding!!
+*/
+
+#include <bits/stdc++.h>
+#define N ((int)1e5 + 5)
+
+using namespace std;
+
+int A[N];
+int contribution[N];
+
+int main()
+{
+    ios_base::sync_with_stdio(false);
+    cin.tie(NULL);
+
+    int t;
+    cin>>t;
+    for(int i = 1; i <= t; i++) {
+        int n, q;
+        cin>>n>>q;
+
+        long long sum = 0;
+
+        for(int j = 0; j < n; j++) {
+            cin>>A[j];
+            contribution[j] = (n-j-1) - j;
+            sum += 1LL * A[j] * contribution[j]; 
+            // multiplying by 1LL is important because A[j] * contribution[j]
+            // can be greater than INT_MAX (2^31 - 1)
+        }
+
+        cout << "Case " << i << ":\n";
+        while(q--) {
+            int type;
+            cin>>type;
+            if(type == 0) {
+                int idx, val;
+                cin>>idx>>val;
+                sum -= 1LL * A[idx] * contribution[idx];
+                A[idx] = val;
+                sum += 1LL * A[idx] * contribution[idx];
+            } else {
+                cout << sum << "\n";
+            }
+        }
+    }
+}
+```