Yet Another Game of Stones ZOJ - 3964

情况

1. 如果B[i]==2 ，但A[i] %2 ==1 ,对于Alice来说肯定没法取完，所以必输
2. 考虑B[i] == 1,B[i] == 2,这两种情况，如果B[i]==1，A[i] == 1，这就相当于nim博弈了，如果B[i]==1&A[i] >1的个数加上B[i] ==2 &&B[i] > 1的个数大于等于2，那么这种情况无论nim_sum ==0 ,或者不等于零，都是必输的
3. 如果B[i] == 1&&A[i] > 1的个数为1，那么肯定要先取这个，考虑取奇数个之后的nim值
4. 如果B[i] == 2&&A[i] > 1的个数为1，那么肯定要取完，考虑剩下来的nim值
5. 其他情况退化成nim博弈

const int maxn = 1e6+100;

int A[maxn];
int B[maxn];
  int n,T;

bool solve(){
    scanf("%d",&n);
    int sum = 0;
    int c1 = 0,c2 =0;
    rep(i,0,n) scanf("%d",&A[i]),sum ^= A[i];
    rep(i,0,n) {
      scanf("%d",&B[i]);
     
    }
    rep(i,0,n){
      if(B[i] ==2 && A[i] %2 ==1) return 0;
      if(B[i] == 1&&A[i] > 1) c1++;
      if(B[i] == 2&&A[i] > 1) c2++;
    }
    if(c1 +c2 > 1) return 0;
    if(c1 == 1){
       // for(auto c:)
       rep(i,0,n)
        {
          if(A[i] > 1&&B[i] == 1){
              sum ^= A[i];
            sum ^= (A[i]%2==0);
            return !sum;
          }
        }
     }
     
     if(c2 == 1){
          rep(i,0,n){
              if(A[i] > 1&&B[i] == 2){
                      sum ^= A[i];
                      return !sum;
                }
        }
     }
     
     return sum;
}
int main(void)
{

  cin>>T;
  while(T--){
    printf("%s\n", solve() ? "Alice" : "Bob");
  }
   return 0;
}