Every 4-tuple (A_i-1, A_i, B_i-1, B_i) limits feasible K by upper or lower bound using following formula:

A_i-1 + K*B_i-1 >= A_i + K*B_i

K*(B_i-1 - B_i) >= A_i - A_i-1

K>= (A_i - A_i-1)/(B_i-1 - B_i) if (B_i-1 - B_i) is positive

K <= (A_i - A_i-1)/(B_i-1 - B_i) if (B_i-1 - B_i) is negative

At 4th problem: If the total amount of meat the butcher distribute is T and we denote X = a convenable "division" of T, such that each person receive X * B[i] kilos of meat (this is the explanation of that ratio, mentioned in the text), we can transform the problem into a system of inequalities: A[1] + X * B[1] >= A[2] + X * B[2] >= ... >= A[N] + X * B[N]. From these inequalities, we can obtain a lowerbound and an upperbound for X, and the answer is LowerBound_for_X * (B[1] + B[2] + ... + B[N]). Time complexity: O(N ^ 2) :)

LE: I haven't seen Dixtosa's post.

LLE: There is another blogpost about COCI: http://codeforces.me/blog/entry/9867

To me, the problems seemed to be reasonably hard this time (I found the 4th around as hard as the 3rd, though). At least compared to the 1st contest of this year, where 1-5 were really easy and the 6th was really hard and without balanced test cases (a bruteforce was given the same amount of points as a much more advanced KMP-using solution). I didn't solve the 6th problem this time either because I didn't have enough time, but I had a complete idea and it wasn't that hard to implement.

I solved it with binary search.
Here is my code :

int isit(double x)
{
	for(int i=0; i<n; i++)
		N[i] = A[i] + B[i]*(x/all);
	
	for(int i=1; i<n; i++)
			if(N[i-1] < N[i])
				return 0;
	
	return 1;
	
}

int tap(double x)
{
	for(int i=0; i<n; i++)
		N[i] = A[i] + B[i]*(x/all);
	
	for(int i=0; i<n; i++)
		for(int j=i+1; j<n; j++)
			if(N[i] < N[j]  &&  A[i] < A[j])
				return 1;
	
	return 0;
}

double galany(double b,double l,double r,int cnt = 0)
{
	if(cnt > 40)	return r;
	if(isit(b+l))		return l;
	if(isit(b+r))		return r;
	
	double mid = l + (r-l)/2.0;
	
	if(tap(double(b + mid)))	return galany(b, mid, r, cnt+1);
	return galany(b, l, mid, cnt+1);
}

double b_s(int l,int r)
{
	if(r-l < 2)
	{
		double a = (double)l + galany((double)l, 0.0, 1.0);
		double b = (double)r + galany((double)r, 0.0, 1.0);
		if(isit(a))	return a;
		if(isit(b))	return b;
		return -1.0;
	}
	
	int mid = (l+r)/2;
	
	if(tap(double(mid)))	return b_s(mid, r);
	return b_s(l, mid);
}