SG函数找规律练习-HDU3032- Nim or not Nim?

问题描述：

两人博弈，N堆石子，可以选择移走某一堆的任意数量的石子，也可以选择将这一堆分成任意两堆东西

解法：
1、SG函数求解

void init()
{
memset(sg,0,sizeof(sg));
//  int m;
sg[0] = 0;
sg[1] = 1;
for(int i = 1 ; i < maxn;i++)
{
memset(vis,0,sizeof(vis));
for(int j = 1; j <= i;j++)
{
vis[sg[i-j]] = 1;//移走任意个数石子
}
for(int j = 1; j < i;j++)
{
vis[sg[j]^sg[i-j]] = 1;//将石子分为两堆
}
for(int j = 0 ;; j++)
{
if(!vis[j])
{
sg[i] = j;
break;
}
}
//if(sg[i]==0)
cout<<"i:"<<i<<" sg[i]"<<sg[i]<<endl;
}
}

2、找SG函数的规律

int fsg(int n)
{
if(n%4==3)
return n+1;
else if(n%4==0)
return n-1;
else
return n;
}

AC代码

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
const int maxn = 1004;
int vis[maxn];
int sg[maxn];
int N;
int a[maxn];
void init()
{
memset(sg,0,sizeof(sg));
//  int m;
sg[0] = 0;
sg[1] = 1;
for(int i = 1 ; i < maxn;i++)
{
memset(vis,0,sizeof(vis));
for(int j = 1; j <= i;j++)
{
vis[sg[i-j]] = 1;
}
for(int j = 1; j < i;j++)
{
vis[sg[j]^sg[i-j]] = 1;
}
for(int j = 0 ;; j++)
{
if(!vis[j])
{
sg[i] = j;
break;
}
}
//if(sg[i]==0)
cout<<"i:"<<i<<" sg[i]"<<sg[i]<<endl;
}
}
int fsg(int n)
{
if(n%4==3)
return n+1;
else if(n%4==0)
return n-1;
else
return n;
}
int main()
{
//init();
int n;
cin>>N;
while(N--)
{
cin>>n;
for(int i = 0 ; i < n ; i++)
cin>>a[i];
int res;
res = fsg(a[0]);
for(int i = 1 ; i < n ; i++)
{
res ^= fsg(a[i]);
}
//cout<<res<<endl;
if(res==0)
cout<<"Bob"<<endl;
else
cout<<"Alice"<<endl;
}
}

Nim or not Nim?

Problem 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

break; } } //if(sg[i]==0) cout<<"i:"<<i<<" sg[i]"<<sg[i]<<endl; }}