Hdu 4111 Alice and Bob (game)

Hdu 4111 Alice and Bob (game)

Link: hdu 4111 Alice and Bob

Alice and Bob play games with N piles of stones. Each time they take a rock from a pile or merge the two stones, Alice first asks

The final winner.

Solution: NP theorem, write some NP theorem to find the law, and then use the inductive method to prove it.

Count the number c of 1 and the number s of steps in the case of not 1, including merging.

• C is an odd number, and s is not equal to 2: The first hand wins.
• S is a multiple of 2 or 0: c is 3.
• Otherwise, if s is an odd number, the first hand wins.

  If there is no 1 in the first place, it is proved that when the total number of steps is an odd number, the first hand wins, N points. On the contrary, if it is an even number, it is a pvertex. (In this case

  To win the next game, you only need to ensure that the number of any pile of stones is no less than 2 until there is only one pile at the end, because 1 is a special one)
  

  In addition, it is proved that an odd number of 1 and s is not 2, it is a winning state. Starting from a 1, suppose there is a 1 and s. If s is an odd number, it can be merged first.

  Two piles of stones; if s is an even number, the first hand can take 1; so in this case, the first hand must win, because move 1 can choose to drop one or two steps.

  Now there are two cases of 1 and s. If one of them consumes 1 and merges with s, the remaining State is (1, s + 1) N points; remove 1,

  The remaining (1, s) is also N points. merge two 1, (2, s) => (s + 3) States. If s is an odd number, that is, P-state. The first hand is the winner, and vice versa.

  This is a defeat. This situation is summarized into the third article in the conclusion. In the case of three values, one value can be changed to another value.
  

  However, when s = 2 or s = 0, it is a special case, because after 2 is removed, it becomes 1, and when s is 0, it is all 1. Through the above rule, we already know

  1 is special. Now there is an NP Graph
  
  It is not hard to find that if c is a multiple of 3, it must be a first-hand defeat.

  #include 
      
       #include 
       
        #include using namespace std;bool judge (int c, int s) {    if (c&1 && s > 2)        return true;    if (s == 0 || s == 2)        return c % 3;    return s&1;}int main () {    int cas;    scanf("%d", &cas);    for (int kcas = 1; kcas <= cas; kcas++) {        int n, s = 0, c = 0, x;        scanf("%d", &n);        for (int i = 0; i < n; i++) {            scanf("%d", &x);            if (x == 1)                c++;            else {                if (s == 0)                    s = x;                else                    s += x + 1;            }        }        printf("Case #%d: %s\n", kcas, judge(c, s) ? "Alice" : "Bob");    }    return 0;}