2013
12-12

# S-Nim

Arthur and his sister Caroll have been playing a game called Nim for some time now. Nim is played as follows:

The starting position has a number of heaps, all containing some, not necessarily equal, number of beads.

The players take turns chosing a heap and removing a positive number of beads from it.

The first player not able to make a move, loses.

Arthur and Caroll really enjoyed playing this simple game until they recently learned an easy way to always be able to find the best move:

Xor the number of beads in the heaps in the current position (i.e. if we have 2, 4 and 7 the xor-sum will be 1 as 2 xor 4 xor 7 = 1).

If the xor-sum is 0, too bad, you will lose.

Otherwise, move such that the xor-sum becomes 0. This is always possible.

It is quite easy to convince oneself that this works. Consider these facts:

The player that takes the last bead wins.

After the winning player’s last move the xor-sum will be 0.

The xor-sum will change after every move.

Which means that if you make sure that the xor-sum always is 0 when you have made your move, your opponent will never be able to win, and, thus, you will win.

Understandibly it is no fun to play a game when both players know how to play perfectly (ignorance is bliss). Fourtunately, Arthur and Caroll soon came up with a similar game, S-Nim, that seemed to solve this problem. Each player is now only allowed to remove a number of beads in some predefined set S, e.g. if we have S =(2, 5) each player is only allowed to remove 2 or 5 beads. Now it is not always possible to make the xor-sum 0 and, thus, the strategy above is useless. Or is it?

your job is to write a program that determines if a position of S-Nim is a losing or a winning position. A position is a winning position if there is at least one move to a losing position. A position is a losing position if there are no moves to a losing position. This means, as expected, that a position with no legal moves is a losing position.

Input consists of a number of test cases. For each test case: The first line contains a number k (0 < k ≤ 100 describing the size of S, followed by k numbers si (0 < si ≤ 10000) describing S. The second line contains a number m (0 < m ≤ 100) describing the number of positions to evaluate. The next m lines each contain a number l (0 < l ≤ 100) describing the number of heaps and l numbers hi (0 ≤ hi ≤ 10000) describing the number of beads in the heaps. The last test case is followed by a 0 on a line of its own.

For each position: If the described position is a winning position print a ‘W’.If the described position is a losing position print an ‘L’. Print a newline after each test case.

2 2 5
3
2 5 12
3 2 4 7
4 2 3 7 12
5 1 2 3 4 5
3
2 5 12
3 2 4 7
4 2 3 7 12
0

LWW
WWL

sg函数：sg函数是博弈中的确定一个position性质的一个函数，全称是sprague-grundy。
/*

*/
#include<iostream>
#include<cstdlib>
#include<vector>
#include<map>
#include<cstring>
#include<set>
#include<string>
#include<algorithm>
#include<sstream>
#include<ctype.h>
#include<fstream>
#include<string.h>
#include<stdio.h>
#include<math.h>
#include<stack>
#include<queue>
#include<ctime>
//#include<conio.h>
using namespace std;

const int INF_MAX=0x7FFFFFFF;
const int INF_MIN=-(1<<31);

const double eps=1e-10;
const double pi=acos(-1.0);

#define pb push_back   //a.pb( )
#define chmin(a,b) ((a)<(b)?(a):(b))
#define chmax(a,b) ((a)>(b)?(a):(b))

template<class T> inline T gcd(T a,T b)//NOTES:gcd(
{if(a<0)return gcd(-a,b);if(b<0)return gcd(a,-b);return (b==0)?a:gcd(b,a%b);}
template<class T> inline T lcm(T a,T b)//NOTES:lcm(
{if(a<0)return lcm(-a,b);if(b<0)return lcm(a,-b);return a*(b/gcd(a,b));}

typedef pair<int, int> PII;
typedef vector<PII> VPII;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef long long LL;
int dir_4[4][2]={{0,1},{-1,0},{0,-1},{1,0}};
int dir_8[8][2]={{0,1},{-1,1},{-1,0},{-1,-1},{0,-1},{1,-1},{1,0},{1,1}};
//下，左下，左，左上，上，右上，右，右下。

//******* WATER ****************************************************************

const int MAXN = 10500;
bool judge[150];
int sg[MAXN];
int M[150];
const int Init = 1e7;
int Num;

void input_m()
{
for(int i = 0; i < Num; i++)
{
cin>>M[i];
}
return ;
}

void debug()
{
cout<<"sg function"<<endl;
for(int i = 0; i < 100; i++)
{
cout<<i<<" "<<sg[i]<<endl;
}
return ;
}

void getsg()
{
for(int i = 0; i < MAXN; i++)
{
memset(judge, false, sizeof(judge));
//int tsg = Init;
for(int j = 0; j < Num; j++)
{
int ps = i - M[j];
if(ps >= 0) judge[sg[ps]] = true;
}
//if(tsg == Init) tsg = 0;
for(int j = 0; j < Num + 1; j++)
{
if(judge[j] == false)
{
sg[i] = j;
break;
}
}
}
//debug();
return ;
}
int main()
{
//freopen("input.txt","r",stdin);
//freopen("output.txt","w",stdout);
while(cin>>Num, Num)
{
input_m();
getsg();
int num;
cin>>num;
while(num--)
{
int nn, tp;
cin>>nn;
int ret = 0;
for(int i = 0; i < nn; i++)
{
cin>>tp;
ret ^= sg[tp];
}
if(ret == 0) cout<<"L";
else cout<<"W";
}
cout<<endl;
}
return 0;
//printf("%.6f\n",(double)clock()/CLOCKS_PER_SEC);
}

1. 为什么for循环找到的i一定是素数叻，而且约数定理说的是n=p1^a1*p2^a2*p3^a3*…*pk^ak，而你每次取余都用的是原来的m，也就是n

2. 约瑟夫也用说这么长……很成熟的一个问题了，分治的方法解起来o(n)就可以了，有兴趣可以看看具体数学的第一章，关于约瑟夫问题推导出了一系列的结论，很漂亮

3. 嗯 分析得很到位，确实用模板编程能让面试官对你的印象更好。在设置辅助栈的时候可以这样：push时，比较要push的elem和辅助栈的栈顶，elem<=min.top()，则min.push(elem).否则只要push（elem）就好。在pop的时候，比较stack.top()与min.top(),if(stack.top()<=min.top()),则{stack.pop();min.pop();}，否则{stack.pop();}.

4. 代码是给出了，但是解析的也太不清晰了吧！如 13 abejkcfghid jkebfghicda
第一步拆分为 三部分 (bejk, cfghi, d) * C(13,3)，为什么要这样拆分，原则是什么？