codeforces 255E Furlo and Rublo and Game(SG函数)

题目链接：http://codeforces.com/problemset/problem/255/E

题意：n堆石子，每次从一堆（设这堆的石子数为x）中必须拿到剩下y个，y满足:0<=y<x,且y*y*y*y>=x且y*y<=x。

思路：求每个状态的SG函数。

const int MAX=900005;

int SG[MAX],sum[4][MAX],visit[MAX];



i64 sqrt4(i64 x)

{

    i64 low=0,high=1000,mid;

    while(low<=high)

    {

        mid=(low+high)>>1;

        if(sqr(sqr(mid))>=x) high=mid-1;

        else low=mid+1;

    }

    if(sqr(sqr(high))>=x) return high;

    return low;

}



i64 sqrt2(i64 x)

{

    i64 low=0,high=10000000,mid;

    while(low<=high)

    {

        mid=(low+high)>>1;

        if(sqr(mid)>x) high=mid-1;

        else low=mid+1;

    }

    if(sqr(low)>x) return high;

    return low;

}





void init()

{

    SG[0]=0;

    clr(visit,-1);

    int i,j,k=0;

    for(i=1;i<MAX;i++)

    {

        for(j=sqrt4(i);j*j<=i&&j<i;j++) visit[SG[j]]=k;

        SG[i]=0;

        while(visit[SG[i]]==k) SG[i]++;

        k++;

    }

    for(i=MAX-2;i>=0;i--)

    {

        for(j=0;j<4;j++) sum[j][i]=sum[j][i+1];

        sum[SG[i]][i]++;

    }

}



int getSG(i64 x)

{

    if(x<MAX) return SG[x];

    i64 L=sqrt4(x),R=sqrt2(x),t=0;

    while(t<4&&sum[t][L]-sum[t][R+1]) t++;

    return t;

}



int n;



int main()

{

    init();

    scanf("%d",&n);

    i64 x,ans=0;

    while(n--)

    {

        cin>>x;

        ans^=getSG(x);

    }

    if(ans) puts("Furlo");

    else puts("Rublo");

    return 0;

}