Nim or not Nim?
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Description
Nim is a two-player mathematic game of strategy in which players take turns removing objects from distinct heaps. On each turn, a player must remove at least one object, and may remove any number of objects provided they all come from the same heap. Nim is usually played as a misere game, in which the player to take the last object loses. Nim can also be played as a normal play game, which means that the person who makes the last move (i.e., who takes the last object) wins. This is called normal play because most games follow this convention, even though Nim usually does not. Alice and Bob is tired of playing Nim under the standard rule, so they make a difference by also allowing the player to separate one of the heaps into two smaller ones. That is, each turn the player may either remove any number of objects from a heap or separate a heap into two smaller ones, and the one who takes the last object wins.
Input
Input contains multiple test cases. The first line is an integer 1 ≤ T ≤ 100, the number of test cases. Each case begins with an integer N, indicating the number of the heaps, the next line contains N integers s[0], s[1], ...., s[N-1], representing heaps with s[0], s[1], ..., s[N-1] objects respectively.(1 ≤ N ≤ 10^6, 1 ≤ S[i] ≤ 2^31 - 1)
Output
For each test case, output a line which contains either "Alice" or "Bob", which is the winner of this game. Alice will play first. You may asume they never make mistakes.
Sample Input
2 3 2 2 3 2 3 3
Sample Output
Alice Bob
题意：给定n堆石子，两人轮流操作，每次选一堆石子，取任意石子或则将石子分成两个更小的堆(非0)，取得最后一个石子的为胜。
题解：比较裸的SG定理，用sg定理打表，得到表1,2,4,3,5,6,8,7,9,10,12,11...可以发现当x%4==0时sg[x]=x-1；当x%4==3时sg[x]=x+1；其余sg[x]=x。然后异或下就出来结果了。
SG定理打表+找规律：
#include#include #include #include using namespace std; const int maxn=1e4+10; int sg[maxn],vis[maxn]; void init() { int i,j,k; sg[0]=0,sg[1]=1; for(i=2;i<=1000;i++) { memset(vis,0,sizeof(vis)); for(j=1;j
AC代码：
#include#include #include #include using namespace std; const int maxn=1e6+10; int find(int x) { if(x%4==0)return x-1; else if(x%4==3)return x+1; return x; } int main() { int T; scanf("%d",&T); while(T--) { int a,n,i,j,ans=0; scanf("%d",&n); for(i=0;i