POJ1143:Number Game(状态压缩)
发布时间：2020-12-13 20:09:36 所属栏目：PHP教程 来源：网络整理
导读：Description Christine and Matt are playing an exciting game they just invented: the Number Game. The rules of this game are as follows. The players take turns choosing integers greater than 1. First,Christine chooses a number,then Matt cho
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Description
Christine and Matt are playing an exciting game they just invented: the Number Game. The rules of this game are as follows.
The players take turns choosing integers greater than 1. First,Christine chooses a number,then Matt chooses a number,then Christine again,and so on. The following rules restrict how new numbers may be chosen by the two players:
If a player cannot choose any new number according to these rules,then that player loses the game. Here is an example: Christine starts by choosing 4. This prevents Matt from choosing 4,8,12,etc.Let's assume that his move is 3. Now the numbers 3,6,9,etc. are excluded,too; furthermore,numbers like: 7 = 3+4;10 = 2*3+4;11 = 3+2*4;13 = 3*3+4;... are also not available. So,in fact,the only numbers left are 2 and 5. Christine now selects 2. Since 5=2+3 is now forbidden,she wins because there is no number left for Matt to choose. Your task is to write a program which will help play (and win!) the Number Game. Of course,there might be an infinite number of choices for a player,so it may not be easy to find the best move among these possibilities. But after playing for some time,the number of remaining choices becomes finite,and that is the point where your program can help. Given a game position (a list of numbers which are not yet forbidden),your program should output all winning moves. A winning move is a move by which the player who is about to move can force a win,no matter what the other player will do afterwards. More formally,a winning move can be defined as follows.
Input
The input consists of several test cases. Each test case is given by exactly one line describing one position.
Each line will start with a number n (1 <= n <= 20),the number of integers which are still available. The remainder of this line contains the list of these numbers a1;...;an(2 <= ai <= 20). The positions described in this way will always be positions which can really occur in the actual Number Game. For example,if 3 is not in the list of allowed numbers,6 is not in the list,either. At the end of the input,there will be a line containing only a zero (instead of n); this line should not be processed. Output
For each test case,your program should output "Test case #m",where m is the number of the test case (starting with 1). Follow this by either "There's no winning move." if this is true for the position described in the input file,or "The winning moves are:
w1 w2 ... wk" where the wi are all winning moves in this position,satisfying wi < wi+1 for 1 <= i < k. After this line,output a blank line.
Sample Input 2 2 5
2 2 3
5 2 3 4 5 6
0
Sample Output Test Case #1
The winning moves are: 2
Test Case #2
There's no winning move.
Test Case #3
The winning moves are: 4 5 6
Source
Mid-Central European Regional Contest 2000
题意:两人玩游戏，有1些列数，1旦选了某些数字，那末这些数字的倍数和这些数字通过相加能组成的数字都不能选了，当轮到某个人选的时候，没有数字可以选了，那末这个人就必输
思路：以状态紧缩记录枚举所有情况来进行搜索便可
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <algorithm>
using namespace std;
#define ls 2*i
#define rs 2*i+1
#define up(i,x,y) for(i=x;i<=y;i++)
#define down(i,y) for(i=x;i>=y;i--)
#define mem(a,x) memset(a,sizeof(a))
#define w(a) while(a)
#define LL long long
const double pi = acos(⑴.0);
#define Len 200005
#define mod 19999997
const int INF = 0x3f3f3f3f;
#define exp 1e⑹
int cas = 1;
int n,a[25],dp[600000],ans[25],len;
int getnum(int s[])
{
int ret = 0,i;
for(int i = 2;i<=20;i++)
{
if(s[i])
ret|=1;
ret<<=1;
}
return ret;
}
int dfs(int s[],int st)
{
int cpy[25],i,j;
memcpy(cpy,s,25*sizeof(int));
cpy[st] = 0;
for(i = 2; i+st<=20; i++)
{
if(!cpy[i])
cpy[i+st] = 0;
}
int ss = getnum(cpy);
if(dp[ss]!=0)
{
if(dp[ss]>0) return 1;
else return 0;
}
up(i,2,20)
{
if(cpy[i] && !dfs(cpy,i))
{
dp[ss] = 1;
return 1;
}
}
dp[ss] = ⑴;
return 0;
}
int main()
{
int i,j,k;
mem(dp,0);
w((scanf("%d",&n),n))
{
mem(a,0);
up(i,1,n)
{
scanf("%d",&k);
a[k] = 1;
}
int s = getnum(a);
len = 0;
up(i,20)
{
if(a[i] && !dfs(a,i))
ans[len++] = i;
}
printf("Test Case #%d
",cas++);
if(len)
{
printf("The winning moves are:");
up(i,len⑴)
printf(" %d",ans[i]);
}
else
{
printf("There's no winning move.");
dp[s] = ⑴;
}
printf("
");
}
return 0;
}
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