HDU 1535 S-Nim（SG函数）

S-Nim

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8729    Accepted Submission(s): 3660

Problem 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.

 

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

 

题意

首先给出k，表示有几种每次取石子个数的集合，即给出123，每次可以取1,2,3个。然后给出询问次数m。每次询问给出n堆石子，然后询问当前状态是P还是N。

分析

SG函数。两种求法，都写了一遍。耗时间差不多

code

预处理

#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

const int N = 10000;
int sg[10100],f[1010];
int k,n,m;
bool Hash[1010];

void get_SG() {
    memset(sg,0,sizeof(sg));
    for (int i=1; i<=N; ++i) {
        memset(Hash,false,sizeof(Hash));
        for (int j=1; j<=k&&f[j]<=i; ++j) 
            Hash[sg[i-f[j]]] = true;
        for (int j=0; j<=N; ++j) 
            if (!Hash[j]) {sg[i] = j;break;}        
    }
}

int main () {
    while (~scanf("%d",&k) && k) {
        for (int i=1; i<=k; ++i) 
            scanf("%d",&f[i]);
        sort(f+1,f+k+1);
        get_SG();
        scanf("%d",&m);
        for (int i=1; i<=m; ++i) {
            scanf("%d",&n);
            int ans = 0;
            for (int t,j=1; j<=n; ++j) {
                scanf("%d",&t);
                ans ^= sg[t];
            }
            if (ans == 0) printf("L");
            else printf("W");
        }
        puts("");
    }
    return 0;
}

dfs记忆化搜索

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int sg[10100],f[1010];
int k,n,m;

int get_SG(int x) {
    if (sg[x] != -1) return sg[x];
    bool Hash[110]; //不能再外面开 
    memset(Hash,false,sizeof(Hash));
    for (int i=1; i<=k; ++i) {
        if (f[i] > x) break;
        get_SG(x-f[i]);
        Hash[sg[x-f[i]]] = true;
    }
    for (int i=0; ; ++i) 
        if (!Hash[i]) {sg[x] = i;break;}
    return sg[x];
}
int main () {
    while (~scanf("%d",&k) && k) {
        for (int i=1; i<=k; ++i) 
            scanf("%d",&f[i]);
        sort(f+1,f+k+1);
        memset(sg,-1,sizeof(sg));
        sg[0] = 0;
        scanf("%d",&m);
        for (int i=1; i<=m; ++i) {
            scanf("%d",&n);
            int ans = 0;
            for (int t,j=1; j<=n; ++j) {
                scanf("%d",&t);
                ans ^= get_SG(t);
            }
            if (ans == 0) printf("L");
            else printf("W");
        }
        puts("");
    }
    return 0;
}