递推dp（uva10404）

Bachet's Game

Time Limit:6666MS Memory Limit:Unknown 64bit IO Format:%lld & %llu

SubmitStatus

Description

Problem B: Bachet's Game

Bachet's game is probably known to all but probably not by this name. Initially there aren stones 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 thank stones 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 ofm 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 isn <= 1000000 the number of stones on the table; the second number ism <= 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：两个人做游戏，有n块石字，m堆，每个人每次只能取m堆数量中的个数，问谁会赢

思路：dp[i]表示还剩i个是S能不能赢，那么当dp[i-a[j]]为偶数，S就可以赢

#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;const int maxn=1000100;int dp[maxn];int a[maxn];int N,M;int main(){    while(scanf("%d",&N)!=EOF)    {        scanf("%d",&M);        for(int i=1;i<=M;i++)scanf("%d",&a[i]);        memset(dp,0,sizeof(dp));        for(int i=1;i<=N;i++)            for(int j=1;j<=M;j++)                if(i>=a[j]&&dp[i-a[j]]==0)dp[i]=1;        if(dp[N])printf("Stan wins\n");        else printf("Ollie wins\n");    }    return 0;}