bzoj千题计划260：bzoj2940: [Poi2000]条纹

Posted 2020-10-26 日拱一卒 功不唐捐

http://www.lydsy.com/JudgeOnline/problem.php?id=2940

 

SG

博弈入门推荐张一飞的《由感性认识到理性认识 ——透析一类搏弈游戏的解答过程》

 

#include<cstdio>
#include<cstring>

using namespace std;

int a,b,c;
int sg[1001];

int Sg(int n)
{
    if(sg[n]!=-1) return sg[n];
    bool vis[1001]={0};
    for(int i=1;i+a-1<=n;++i) vis[Sg(i-1)^Sg(n-i-a+1)]=true;
    for(int i=1;i+b-1<=n;++i) vis[Sg(i-1)^Sg(n-i-b+1)]=true;
    for(int i=1;i+c-1<=n;++i) vis[Sg(i-1)^Sg(n-i-c+1)]=true;
    for(int i=0;i<=n;++i)
        if(!vis[i]) return sg[n]=i;
}

int main()
{
    scanf("%d%d%d",&a,&b,&c);
    int T,n;
    scanf("%d",&T);
    memset(sg,-1,sizeof(sg));
    sg[0]=0;
    while(T--)
    {
        scanf("%d",&n);
        if(Sg(n)) printf("1\n");
        else printf("2\n");
    }
    return 0;
}