if N%4 != 1 Alice will win otherwise she will lose

This isn't dp. If n % 4 != 0 then Alice will win. If n%4 !=0, first move you chose number n % 4. After that, if Bob remove x, Alice remove 4-x. The similar thnig if n % 4=0, only Bob will win.

So, there is common dp solution for this kind of games.

let's have array bool dp[MAX_N]

• Function: dp[i] determines if player whose turn wins when i stones left. (true=guarantee win, false=guarantee lose)
• Base values: by the rules, if 0 stones left, player whose turn wins: dp[0]=true
• Transitions: consider dp[i]. We have i stones left. After our turn our opponent will be in situation dp[i-1], dp[i-2] or dp[i-3]. So if one of that situations leads to lose, we choose it, and dp[i] leads to win. If they all lead to win, then we lose anyway. Thus dp[i] = !(dp[i-1]&&dp[i-2]&&dp[i-3])
• Answer: obviously dp[N]

Example:

   i   0 1 2 3 4 5 6 7 8 
dp[i]  1 0 1 1 1 0 1 1 1

dp[0] = 1;
dp[1] = 0; //We have only one move, and it lead to 1, so it is losing position
dp[2] = 1; //We can take 1 stone to put opponent in losing position dp[1]
dp[3] = 1; //We can take 2 stones to put opponent in losing position dp[1]
dp[4] = 1; //We can take 3 stones to put opponent in losing position dp[1]
dp[5] = 0; //Doesn't matter how many stones we take we and up in winning position for opponent
dp[6] = 1; //We can take 1 stone to put opponent in losing position dp[5]
dp[7] = 1; //We can take 2 stones to put opponent in losing position dp[5]
dp[8] = 1; //We can take 3 stones to put opponent in losing position dp[5]

As you can see in this particular problem there is obvious pattern, which was mentioned by Outsider above, but as I said, this kind of dp approach(finding losing position for opponent) can be used in large amount of more complicated game problems.