S-Nim Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3077 Accepted Submission(s): 1361 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 Source Norgesmesterskapet 2004 Recommend LL 題意： 首先輸入K 表示一個集合的大小 之後輸入集合 表示對於這對石子只能去這個集合中的元素的個數 之後輸入 一個m 表示接下來對於這個集合要進行m次詢問 之後m行 每行輸入一個n 表示有n個堆 每堆有n1個石子 問這一行所表示的狀態是贏還是輸 如果贏輸入W否則L 思路： 對於n堆石子 可以分成n個游戲 之後把n個游戲合起來就好了
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
//注意 S數組要按從小到大排序 SG函數要初始化為-1 對於每個集合只需初始化1邊
//不需要每求一個數的SG就初始化一邊
int SG[10100],n,m,s[102],k;//k是集合s的大小 S[i]是定義的特殊取法規則的數組
int dfs(int x)//求SG[x]模板
{
if(SG[x]!=-1) return SG[x];
bool vis[110];
memset(vis,0,sizeof(vis));
for(int i=0;i<k;i++)
{
if(x>=s[i])
{
dfs(x-s[i]);
vis[SG[x-s[i]]]=1;
}
}
int e;
for(int i=0;;i++)
if(!vis[i])
{
e=i;
break;
}
return SG[x]=e;
}
int main()
{
int cas,i;
while(scanf("%d",&k)!=EOF)
{
if(!k) break;
memset(SG,-1,sizeof(SG));
for(i=0;i<k;i++) scanf("%d",&s[i]);
sort(s,s+k);
scanf("%d",&cas);
while(cas--)
{
int t,sum=0;
scanf("%d",&t);
while(t--)
{
int num;
scanf("%d",&num);
sum^=dfs(num);
// printf("SG[%d]=%d\n",num,SG[num]);
}
if(sum==0) printf("L");
else printf("W");
}
printf("\n");
}
return 0;
}
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int K=101;
const int H=10001;//H是我們要打表打到的最大值
int k,m,l,h,s[K],sg[H],mex[K];///k是集合元素的個數 s[]是集合 mex大小大約和集合大小差不多
///注意s的排序
void sprague_grundy()
{
int i,j;
sg[0]=0;
for (i=1;i<H;i++){
memset(mex,0,sizeof(mex));
j=1;
while (j<=k && i>=s[j]){
mex[sg[i-s[j]]]=1;
j++;
}
j=0;
while (mex[j]) j++;
sg[i]=j;
}
}
int main(){
int tmp,i,j;
scanf("%d",&k);
while (k!=0){
for (i=1;i<=k;i++)
scanf("%d",&s[i]);
sort(s+1,s+k+1); //這個不能少
sprague_grundy();
scanf("%d",&m);
for (i=0;i<m;i++){
scanf("%d",&l);
tmp=0;
for (j=0;j<l;j++){
scanf("%d",&h);
tmp=tmp^sg[h];
}
if (tmp)
putchar('W');
else
putchar('L');
}
putchar('\n');
scanf("%d",&k);
}
return 0;}
