$HDU1848\ Fibonacci\ again\ and\ again$ 博弈论

正解:博弈论

解题报告:

传送门!

首先按照套路显然是考虑先预处理出所有数的$SG$函数值然后全局的$SG$就是$SG(n)$^$SG(m)$^$SG(p)$,这儿应该麻油问题$QwQ$?

然后就考虑怎么求$SG$函数?于是就直接$SG(x)=mex\{SG(x-y),y\in Fib,x\ge y\}$就好鸭$QwQ$

$over！$

#include<bits/stdc++.h>
using namespace std;
#define il inline
#define gc getchar()
#define ri register int
#define rb register bool
#define rc register char
#define rp(i,x,y) for(ri i=x;i<=y;++i)
 
const int N=+;
int f[N],sg[N];
bool gdgs=,vis[N];
 
il int read()
{
    rc ch=gc;ri x=;rb y=;
    while(ch!='-' && (ch>'' || ch<''))ch=gc;
    if(ch=='-')ch=gc,y=;
    while(ch>='' && ch<='')x=(x<<)+(x<<)+(ch^''),ch=gc;
    return y?x:-x;
}
il void pre()
{
    f[]=;f[]=;rp(i,,)f[i]=f[i-]+f[i-];
    rp(i,,){for(ri j=;f[j]<=i;++j)vis[sg[i-f[j]]]=;ri tmp=;while(vis[tmp])++tmp;sg[i]=tmp;for(ri j=;f[j]<=i;++j)vis[sg[i-f[j]]]=;}
}
 
int main()
{
//    freopen("1848.in","r",stdin);freopen("1848.out","w",stdout);
    pre();
    while(gdgs){ri m=read(),n=read(),p=read();if(!m && !n && !p)return ;printf(sg[m]^sg[n]^sg[p]?"Fibo\n":"Nacci\n");}
    return ;
}