UVa 10404 Bachet's Game(DP)

Bachet's Game

Bachet's game is probably known to all but probably not by this name. Initially there are nstones on the table. There are two players Stan and Ollie, who move alternately. Stan always starts. The legal moves consist in removing at least one but not more than kstones from the table. The winner is the one to take the last stone.

Here we consider a variation of this game. The number of stones that can be removed in a single move must be a member of a certain set of m numbers. Among the m numbers there is always 1 and thus the game never stalls.

Input

The input consists of a number of lines. Each line describes one game by a sequence of positive numbers. The first number is n <= 1000000 the number of stones on the table; the second number is m <= 10 giving the number of numbers that follow; the last m numbers on the line specify how many stones can be removed from the table in a single move.

Input

For each line of input, output one line saying either Stan wins or Ollie wins assuming that both of them play perfectly.

Sample input

20 3 1 3 821 3 1 3 822 3 1 3 823 3 1 3 81000000 10 1 23 38 11 7 5 4 8 3 13999996 10 1 23 38 11 7 5 4 8 3 13

Output for sample input

Stan winsStan winsOllie winsStan winsStan winsOllie wins

题意   给你n个小石头  和一个数组a[m]   然后两个人轮流取石头  stan先取  olive后取  他们都完美发挥    谁取完最后一个石头谁就是赢家

感觉不是很容易看出来是dp题  令d[i]表示只有i个石子时谁赢   1表示stan赢  0表示olive赢   

i-a[j]表示从i个石子一次取走a[j]个还剩下的    所以有  当(i-a[j]>0&&d[i-a[j]]=0)是   d[i]=1; 

#include<cstdio>#include<cstring>using namespace std;int d[1000005],a[12],ans,n,m;int main(){    while(scanf("%d",&n)!=EOF)    {        scanf("%d",&m);        for(int i=1;i<=m;++i)                scanf("%d",&a[i]);        memset(d,0,sizeof(d));        for(int i=1;i<=n;++i)            for(int j=1;j<=m;++j)            {                if(i-a[j]>=0&&d[i-a[j]]==0)                {                    d[i]=1;                    break;                }            }        printf(d[n]?"Stan wins\n":"Ollie wins\n");    }}