[ACM] hdu 1536 S-Nim（Nim组合博弈 SG函数打表）

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S-Nim

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4004 Accepted Submission(s): 1732

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 2 532 5 123 2 4 74 2 3 7 125 1 2 3 4 532 5 123 2 4 74 2 3 7 120

Sample Output

LWWWWL

解题思路：

这个题折腾了两三天，参考了两个模板，在这之间折腾过来折腾过去，终于把用法和需要注意的地方弄清楚了，汗。注意的是： bool类型的数组比int类型的数组快，不超时与超时的区别，在sg组合博弈时，只能在（a1,a2,a3,a4）中取，要特别注意这里面的数字是否是有序的，这个特别重要，下面贴出的两个模板对应了两种情况。最后，大数据输入还得用scanf。。

模板1（该模板注意的是一定要把 f 数组中的数字从小到大排序）：

const int N=10008;//N为所有堆最多石子的数量int f[108],sg[N];//f[]用来保存只能拿多少个，sg[]来保存SG值bool hash[N];//mex{}void sg_solve(int t,int N)//t指f[]中的个数{    int i,j;    memset(sg,0,sizeof(sg));    for(i=1;i<=N;i++)    {        memset(hash,0,sizeof(hash));        for(j=1;j<=t&&f[j]<=i;j++)        {            hash[sg[i-f[j]]]=1;        }        for(j=0;j<=N;j++)            if(!hash[j])                break;        sg[i] = j;    }}

模板2（该模板不需要进行排序）

const int N=10008;//N为所有堆最多石子的数量int f[108],sg[N];//f[]用来保存只能拿多少个，sg[]来保存SG值bool hash[N];//mex{}void sg_solve(int t,int N){    int i,j;    memset(sg,0,sizeof(sg));    for(i=1;i<=N;i++)    {        memset(hash,0,sizeof(hash));        for(j=1;j<=t;j++)            if(i - f[j] >= 0)               hash[sg[i-f[j]]] = 1;        for(j=0;j<=N;j++)            if(!hash[j])                break;        sg[i] = j;    }}

代码（使用第二个模板）

#include <iostream>#include <algorithm>#include <stdio.h>#include <string.h>using namespace std;const int N=10008;//N为所有堆最多石子的数量int f[108],sg[N];//f[]用来保存只能拿多少个，sg[]来保存SG值bool hash[N];//mex{}void sg_solve(int t,int N){    int i,j;    memset(sg,0,sizeof(sg));    for(i=1;i<=N;i++)    {        memset(hash,0,sizeof(hash));        for(j=1;j<=t;j++)            if(i - f[j] >= 0)               hash[sg[i-f[j]]] = 1;        for(j=0;j<=N;j++)            if(!hash[j])                break;        sg[i] = j;    }}int main(){    int k,m,l,num,i,j;    while(scanf("%d",&k),k)    {        for(i=1;i<=k;i++)            scanf("%d",&f[i]);        sg_solve(k,N);        scanf("%d",&m);        string ans="";        for( i=1;i<=m;i++)        {            int sum=0;            scanf("%d",&l);            for( j=1;j<=l;j++)            {                scanf("%d",&num);                sum^=sg[num];            }            if(sum==0)                ans+="L";            else                ans+="W";        }        cout<<ans<<endl;    }    return 0;}

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