POJ1143:Number Game(狀態壓縮)
來源：程序員人生 發布時間：2015-05-13 08:15:34 閱讀次數：3704次
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:
- A number which has already been selected by Christine or Matt, or a multiple of such a number,cannot be chosen.
- A sum of such multiples cannot be chosen, either.
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.
- A winning move is a move after which the game position is a losing position.
- A winning position is a position in which a winning move exists. A losing position is a position in which no winning move exists.
- In particular, the position in which all numbers are forbidden is a losing position. (This makes sense since the player who would have to move in that case loses the game.)
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
題意:兩人玩游戲，有1些列數，1旦選了某些數字，那末這些數字的倍數和這些數字通過相加能組成的數字都不能選了，當輪到某個人選的時候，沒有數字可以選了，那末這個人就必輸
思路：以狀態緊縮記錄枚舉所有情況來進行搜索便可
#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,x,y) for(i=x;i>=y;i--)
#define mem(a,x) memset(a,x,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,2,20)
{
if(a[i] && !dfs(a,i))
ans[len++] = i;
}
printf("Test Case #%d
",cas++);
if(len)
{
printf("The winning moves are:");
up(i,0,len⑴)
printf(" %d",ans[i]);
}
else
{
printf("There's no winning move.");
dp[s] = ⑴;
}
printf("
");
}
return 0;
}
生活不易，碼農辛苦
如果您覺得本網站對您的學習有所幫助,可以手機掃描二維碼進行捐贈
