Nim or not Nim?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 623 Accepted Submission(s): 288
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.
Question meaning: Alice and Bob take turns to take N piles of stones, each pile is S[i], Alice first, each time can take any stone from any pile, or can divide a pile of stones into two small piles. The first to finish wins.
Data range: (1 ≤ N ≤ 10^6, 1 ≤ S[i] ≤ 2^31 - 1).
Idea: This title is in the game- take-split game ( this game allows to take something, and then divide the original game into several identical games)
Due to the range of data, the sg value cannot be directly calculated, but the table can only be used to find the law ;
There is the SJ theorem:
For any Anti-SG game, if we stipulate that the game ends when the SG value of all single games in the situation is 0, the first player will win if and only if either of the following two conditions are satisfied:
(1) The SG function of the game is not 0, and the SG function of a single game in the game is greater than 1.
(2) The SG function of the game is 0, and there is no single game in the game whose SG function is greater than 1.
Lasker's Nim游戏：每一轮允许两会中操作之一：①、从一堆石子中取走任意多个，②、将一堆数量不少于2的石子分成都不为空的两堆。
分析：很明显：sg(0) = 0，sg(1) = 1。
状态2的后继有：0，1和（1，1），他们的SG值分别为0，1，0，所以sg(2) =2。
状态3的后继有：0、1、2、（1，2），他们的SG值分别为0、1、2、3，所以sg(3) = 4。
状态4的后继有：0、1、2、3、（1，3）和（2，2），他们的SG值分别为0，1，2，4，5，0，所以sg(4) = 3.
再推一些，推测得到：对于所有的k >= 0，有 sg( 4k+1 ) = 4k+1； sg( 4k+2 ) = 4k+2； sg( 4k+3 ) = 4k+4； sg( 4k+4 ) = 4k+3。
假设游戏初始时有3堆，分别有2、5和7颗石子。三堆的SG函数值分别为2、5、8，他们的Nim和等于15.所以要走到P状态，就要使得第三堆的SG值变成7，可以将第三对按1和6分成两堆。
SG打表代码：
#include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> using namespace std; const int N=1000010; int sg[N]; int g(int x){ int mex[1010]; memset(mex,0,sizeof(mex)); if(sg[x]!=-1) return sg[x]; for(int i=x-1;i>=0;i--) mex[g(i)]=1; for(int i=1;i<=x/2;i++){ int ans=0; ans^=g(i); ans^=g(x-i); mex[ans]=1; } for(int i=0;;i++) if(!mex[i]) return sg[x]=i; } int main(){ freopen("input.txt","r",stdin); int t,n; scanf("%d",&t); memset(sg,-1,sizeof(sg)); while(t--){ scanf("%d",&n); int x; for(int i=0;i<n;i++){ scanf("%d",&x); g(x); printf("sg[%d]=%d\n",x,sg[x]); } for(int i=0;i<=100;i++){ printf("%d ",sg[i]); //if(i%10==0) //system("pause"); } printf("\n"); } return 0; }
可得规律：sg(4k)=4k-1;sg(4k+1)=4k+1;sg(4k+2)=4k+2;sg(4k+3)=4k+4;
本题AC代码：
#include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> using namespace std; int main(){ //freopen("input.txt","r",stdin); int t,n; scanf("%d",&t); while(t--){ scanf("%d",&n); int ans=0,x; for(int i=0;i<n;i++){ scanf("%d",&x); if(x%4==0) ans^=(x-1); else if(x%4==1 || x%4==2) ans^=x; else ans^=(x+1); } if(ans!=0) puts("Alice"); else puts("Bob"); } return 0; }