brute force with twist

http://codeforces.com/contest/104/problem/E

The problem asks all sum a+b*k k=0...(a+b*k<=n) for p different pairs of (a,b).
The size of array, n<=3*10^5. Also p<=3*10^5.

Naive impl has to TLE, so some cleverness needs to be there.

Since we can compute sum a+b*k for same b, different a in one scan of the array, we can take care of all b<=1000 queries. For others we have b>=1000, then the number of summands in one (a,b) pair is at most 3*10^5/10^3=300, and for p=3*10^5 we have total being 300*3*10^5=10^8 which is fast enough.

It is a bit strange that prob E in CF is usually easier or less messy than prob D for div2, if you see what the problem is really asking for.

A few notes about implementation:
1. CF machine favors scanf/printf, the same code runs 1970ms with cin/cout and 890ms with scanf/printf
2. CF machine requires %I64d for long long, do NOT use %lld as this will cause WA sometimes.
3. use vector instead of plain array is not much slower, runs in 1080ms. Even repeated push_back for 10^5 times is not slow at all.
4. accessing elements in large array is cheap, repeatedly add to a[i] is as fast as make a local tmp += d; and then a[i]=tmp.