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If I want 1/i for 1<=i<=5000000, i.e. inv[i]
inv[1] = 1;
for (int i=2; i<p; ++i)
inv[i] = (p - (p/i) * inv[p%i] % p) % p;
I used this but it times out. So I wanted a way for this.
Besides that, if I use fact[i-1]*invfact[i] : would that solve? It is not giving me correct answer so plz tell me what is wrong over here?
If p is prime then
. You can use BigMod to pre-calculate inverse of factorials in 
complexity.
Also you can calculate inverse of number from 1 to n is O(n) time and then multiply to pre-calculate inverse of factorials, this will give O(n) complexity.