正解:博弈论
解题报告:
传送门!
威佐夫博弈板子昂$QwQ$
关于这一类问题也有个结论,是说,先手必败的状态一定形如$(\left \lfloor i+\phi \right \rfloor,\left \lfloor i+\phi^{2} \right \rfloor)$,然后$\phi=\frac{\sqrt{5}+1}{2}$
证是不会证的了,但找到了一篇证明看这个趴$QAQ$
#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 double phi=(sqrt(5)+1)/2; il int read() { rc ch=gc;ri x=0;rb y=1; while(ch!='-' && (ch>'9' || ch<'0'))ch=gc; if(ch=='-')ch=gc,y=0; while(ch>='0' && ch<='9')x=(x<<1)+(x<<3)+(ch^'0'),ch=gc; return y?x:-x; } int main() { ri a=read(),b=read();if(a>b)swap(a,b);ri tmp=(b-a)*phi; if(tmp==a)printf("0\n");else printf("1\n"); return 0; }