Description
Arthur and his sister Caroll have been playing a game called Nim for some time now. Nim is played as follows:
recently learned an easy way to always be able to find the best move:
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.
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.
- 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.
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.
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.
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
题意:
n堆石子,每个玩家现在只允许在,预定义的集合S中移除指定数量的石子。
取完石头者胜。
思路:
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用记忆化搜索,或者枚举的方法,得到sg函数值,求单个游戏sg值的nim和,即为整个游戏的sg值。
注意标记数组开成105,因为最多只有100种取法,那么sg值不会超过100。如果开成10000,速度大打折扣
代码:
枚举+预处理 90+ms
#include<stdio.h>
#include<string.h>
#define For(a,b,c) for(int a = b; a <= c; a++)
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
int f[105], sg[10005], book[105];
void get_sg(int limit)
{
for(int i = 0; i < 10001; i++)
{
mem(book,0);
For(j,1,limit)
{
int k = i - f[j];
if(k < 0) continue;
book[sg[k]] = 1;
}
for(int j = 0; ;j++)
{
if(!book[j])
{
sg[i] = j;
break;
}
}
}
}
int main()
{
int limit;
while(scanf("%d",&limit), limit)
{
For(i,1,limit) scanf("%d",&f[i]);
get_sg(limit);
int t;
scanf("%d",&t);
while(t--)
{
int n;
scanf("%d",&n);
int ans = 0, a;
while(n--)
{
scanf("%d",&a);
ans ^= sg[a];
}
printf("%c", ans ? 'W' : 'L');
}
printf("\n");
}
return 0;
}
dfs 90+ms
#include<stdio.h>
#include<string.h>
#define For(a,b,c) for(int a = b; a <= c; a++)
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
int f[105], sg[10005];
int limit;
void get_sg(int a)
{
if(sg[a] != -1) return;
bool book[105];
mem(book,false);
For(i,1,limit)
{
int k = a - f[i];
if(k < 0) continue;
if(sg[k] == -1) get_sg(k);
book[sg[k]] = 1;
}
for(int i = 0; ; i++) if(!book[i]) {sg[a] = i; return;}
}
int main()
{
while(scanf("%d",&limit), limit)
{
For(i,1,limit) scanf("%d",&f[i]);
mem(sg,-1);
sg[0] = 0;
int t;
scanf("%d",&t);
while(t--)
{
int n;
scanf("%d",&n);
int ans = 0, a;
while(n--)
{
scanf("%d",&a);
if(sg[a] == -1) get_sg(a);
ans ^= sg[a];
}
printf("%c", ans ? 'W' : 'L');
}
printf("\n");
}
return 0;
}