题意：
有n堆石子，alice先取，每次可以选择拿走一堆石子中的1~x(该堆石子总数) ，也可以选择将这堆石子分成任意的两堆。alice与bob轮流取，取走最后一个石子的人胜利。
思路：
因为数的范围比较大，所以最好通过SG打表的结果找出规律在解。
sg(4k+1)=4k+1;sg(4k+2)=4k+2;sg(4k+3)=4k+4; sg(4k)=4k-1;
1 2 4 3 5 6 8 7
Sample Input
2
3
2 2 3
2
3 3
Sample Output
Alice
Bob
SG打表找规律
1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <algorithm> 5 # include <string> 6 # include <cmath> 7 # include <queue> 8 # include <list> 9 # define LL long long 10 using namespace std ; 11 12 13 int sg[1010]; 14 bool vis[1010]; 15 16 int mex(int n) //求N的SG值 17 { 18 if(sg[n] != -1)return sg[n]; 19 memset(vis,false,sizeof(vis)); 20 for(int i=n-1;i>=0;i--) 21 { 22 vis[mex(i)]=1; 23 } 24 for(int i=1;i<=n/2;i++) 25 { 26 vis[mex(i)^mex(n-i)]=1; 27 } 28 for(int i = 0;;i++) 29 if(vis[i] == false) 30 { 31 sg[n] = i; 32 break; 33 } 34 return sg[n]; 35 } 36 37 int main() 38 { 39 40 memset(sg,-1,sizeof(sg)); 41 for(int i = 0;i <= 200;i++) 42 sg[i] = mex(i); 43 int t , n ,x ; 44 45 for(int i=0;i<=100;i++) 46 { 47 cout<<sg[i]<<" "; 48 if(i%10==0) 49 cout<<endl; 50 } 51 52 return 0; 53 }
题解代码
1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 using namespace std ; 5 6 7 int main () 8 { 9 int T ; 10 scanf("%d" , &T) ; 11 while(T--) 12 { 13 int n ; 14 scanf("%d" , &n) ; 15 int x ; 16 int sum = 0 ; 17 while(n--) 18 { 19 scanf("%d" , &x) ; 20 if (x % 4 == 1) 21 sum ^= x ; 22 else if (x % 4 == 2) 23 sum ^= x ; 24 else if (x % 4 == 3) 25 sum ^= (x+1) ; 26 else 27 sum ^= (x-1) ; 28 } 29 if (sum == 0) 30 printf("Bob ") ; 31 else 32 printf("Alice ") ; 33 } 34 35 36 return 0 ; 37 }