HDU 1848 -Fibonacci again and again 博弈论
2014-08-11 17:33
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题目链接:
http://acm.hdu.edu.cn/showproblem.php?pid=1848
参考思路:
http://blog.csdn.net/u014135961/article/details/38495837
http://acm.hdu.edu.cn/showproblem.php?pid=1848
参考思路:
http://blog.csdn.net/u014135961/article/details/38495837
#include<cstdio> #include<cstring> const int N=1001; int judge ; int fib[30];//存1 2 4 8 16 .. bool vis ; void Initial() { fib[0]=1,fib[1]=1; for(int i=2;i<30;++i){ fib[i]=fib[i-1]+fib[i-2]; if(fib[i]>1000) break; } //置一表示为必胜点 for(int i=0;i<N;++i){ judge[i]=i; memset(vis,false,sizeof(vis)); for(int j=0;fib[j]<=i;++j) vis[judge[i-fib[j]]]=true; for(int j=0;j<N;++j) if(!vis[j]){ judge[i]=j; break; } } } int main() { int n,m,p; Initial(); while(scanf("%d %d %d",&m,&n,&p)){ if(m==0&&n==0&&p==0) break; if(judge ^judge[m]^judge[p]) puts("Fibo"); else puts("Nacci"); } }
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