poj 2960 S-Nim Nim博弈变形 有限取法
A - S-Nim
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Description
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.
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
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.
Output
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.
Sample Input
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
Sample Output
LWW
WWL
Nim博弈的一种变形，实际上题目中把Nim博弈整个描述了一遍，连解法都说了，只是在Nim博弈的基础上增加了每次取石子的个数受到限制，但是，我们依然可以将其转换为SG函数的问题，详情可以看这篇文章，总的来说，就是将n堆石子看成n个Nim游戏，针对每一个Nim游戏，SG值就反应了其结果，而n个游戏的和我们就可以用这n个SG值相异或得到，实际上这是一道模板题。
/***********************
功能：n堆石子，每一堆每次只能取固定的个数，给出一种局面，最最后谁会赢
参数：堆数以及每一堆的数量
返回值：输赢
***********************/
#include <iostream>
#include <stdio.h>
#include <string.h>
#define maxn 11000
using namespace std;
int sg[maxn];
bool Hash[maxn];
void sg_solve(int *s,int t,int range) //range求解范围 S[]数组是可以每次取的值，t是s的长度。
{
int i,j;
memset(sg,0,sizeof(sg));
for(i=1;i<=maxn;i++)
{
memset(Hash,0,sizeof(Hash));
for(j=0;j<t;j++)
if(i - s[j] >= 0)
Hash[sg[i-s[j]]] = 1;
for(j=0;j<=maxn;j++)
if(!Hash[j])
break;
sg[i] = j;
}
}
int main()
{
int i,j;
int k,m,l;
int s[maxn],hi;
int ans;
char res[maxn];
while(~scanf("%d",&k)&&k)
{
for(i=0;i<k;i++)
scanf("%d",&s[i]);
sg_solve(s,k,10005);
scanf("%d",&m);
for(j=0;j<m;j++)
{
ans=0;
scanf("%d",&l);
for(i=0;i<l;i++)
{
scanf("%d",&hi);
ans^=sg[hi];
}
res[j]=ans==0?'L':'W';
}
for(i=0;i<m;i++)
printf("%c",res[i]);
printf("\n");
}
return 0;
}
/*****************************
输入：
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
输出：
LWW
WWL
*****************************/