【HDU1848】Fibonacci again and again (博弈论)
2017-02-10 18:47
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题目大意:有3堆石头,两个人轮流取,每次取的数量只能是Fibonacci中的数,取完所有的人获胜,求先手胜负情况。
题解:SG定理裸题SG定理
代码:
题解:SG定理裸题SG定理
代码:
#include<cstdio> #include<cstring> #include<algorithm> using std::max; using std::swap; #define MAXN 1005 int g[MAXN]; int fib[20],fib_cnt; bool vis[MAXN]; int main() { int m,n,p; int a=1,b=1; while(a<MAXN) { fib[fib_cnt++]=a; b=a+b; swap(a,b); } g[0]=0; for(int i=1,j;i<MAXN;i++) { memset(vis,0,sizeof vis); for(j=0;j<fib_cnt&&fib[j]<=i;j++) vis[g[i-fib[j]]]=1; for(j=0;vis[j];j++); g[i]=j; } while(1) { scanf("%d%d%d",&m,&n,&p); if(!m&&!n&&!p)break; if(((g[m]^g )^g[p])==0) printf("Nacci\n"); else printf("Fibo\n"); } return 0; }
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