292. Nim Game

 

Problem:

You are playing the following Nim Game with your friend: There is a heap of stones on the table, each time one of you take turns to remove 1 to 3 stones. The one who removes the last stone will be the winner. You will take the first turn to remove the stones.

Both of you are very clever and have optimal strategies for the game. Write a function to determine whether you can win the game given the number of stones in the heap.

For example, if there are 4 stones in the heap, then you will never win the game: no matter 1, 2, or 3 stones you remove, the last stone will always be removed by your friend.

 

Analysis:

if there is 1 stone, I win

if there are 2 stones, I win

if there are 3 stones, I win

if there are 4 stones, I can only take 1 or 2 or 3 stones, which means there will be 3 or 2 or 1 stones left, and I will loose.

if there are 5 stones, i can take 1,2,3, and there will be 4,3,2 stones left. So, I will win

if there are 6 stones, I can take 2, and there will be 4 stones left. I also will win.

if there are 7 stones , I can take 3, and there will be 4 stones left.  I win again.

if there are 8 stones, I will lose.

In conclusion, if the number of the stone can be divide by 4, I will lose the game, except for the case that the number is less than 4.

 

Solusion:

bool canWinNim(int n) {
    if(n<4 || n%4 !=0)
        return true;
    else
        return false;
}

　　

A more efficient way: save energy and increase IPC of microarchitecture(combine the branch instructions and void miss predict of branch).

bool canWinNim(int n) {
    //if(n<4 || n%4 !=0)
    //    return true;
    //else
    //    return false;
    return ((n<4||n%4!=0)&&true);
}

　　

A more efficient solution:

bool canWinNim(int n) {
    return (n%4);
}