1. We can solve the problem for each connected component independently.

2. If there is an odd number of vertices in a connected component, there is no answer (otherwise the sum of all degrees in the resulting graph is odd for this component, which is impossible).

3. Let's find an arbitrary (rooted) spanning tree. Now we can build the answer starting from leaves. For each node, we will do the following: if the number of edges from kids to it is even, we add an edge from this vertex to its parent. Otherwise, we don't do anything. Why does it work? For all vertices, except, possibly, for the root, the answer is correct by design. The sum of all degrees must be even. The sum of all degrees of all vertices except for the root is odd (as the number of such vertices is odd and their degrees are odd). Thus, the degree of the root is also odd.

Problem M: Idea: Count of reachable numbers less than 10 ^ 9 is small, around 100,000.

So you can either do a dynamic programming using map<int,bool> dp where dp[i] denotes whether it is a losing state or winning state. A state is winning iff you can reach some losing state.
DP Using Map
DP Using Array — since 2, 3, 5, 7 are the possible prime factors of reachable numbers.

I think I have a better solution. Consider the smallest numbers of the form (9^i * 2^i) and (9^j*2^(j-1) which are >= given number n. Take their min and if (9^i * 2^i) is lesser Ollie wins, else Stan wins.

This works since whenever Ollie wins, steps will be 2*9*2*9*...*2*9 and when Stan wins, steps will be 9*2*9*2*...*2*9.

So, you can go from right to left in the sequence i.e. start from num=1. Do num*9 and num*2 alternatively till (num < n). If the last step was num*2, Ollie wins, else Stan wins.

int curr = 1;
bool flag = 1;
while (curr < n)
{
	if(flag)
	{
		curr *= 9;
		flag = 0;
	}
	else
	{
		curr *= 2;
		flag = 1;
	}
}
if (!flag)
	cout << "Stan wins.\n";
else
	cout << "Ollie wins.\n";

You can find the probability "No 2 man chose the same color" that probability equals to product of (2 * C — 2 * i) / (2 * C — i) with 1 <= i < M

Answer doesn't depend on count of ladies.

Imagine that gentelmens choose colors. Calculate dp_i — probability of first i gentelmens to have different colors. Suppose that gentelmens choose from two variants of each color.

dp_i = dp_i - 1·p_i

where p_i is probability i-th gentelmen to choose new color

p_i = (2·с - 2·(i - 1)) / (2·c - (i - 1))

Answer is 1 - dp_m. We don't need multiply all p_i by large m because answer is very small. In my code I calculated i from m - 300000 to m.