hdu1848Fibonacci again and again

链接：http://acm.hdu.edu.cn/showproblem.php?pid=1848

题意：中文题。

分析：基础的SG函数的应用。

代码：

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<bitset>
#include<math.h>
#include<cstdio>
#include<vector>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N=1010;
const int MAX=1000000100;
const int mod=100000000;
const int MOD1=1000000007;
const int MOD2=1000000009;
const double EPS=0.00000001;
typedef long long ll;
const ll MOD=1000000007;
const int INF=1000000010;
typedef double db;
typedef long double ldb;
typedef unsigned long long ull;
int a[14]={1,2,3,5,8,13,21,34,55,89,144,233,377,610};
int q
,p
;
void deal(int n) {
int i,j;
memset(q,0,sizeof(q));
for (i=1;i<=n;i++) {
for (j=0;j<14;j++)
if (i<a[j]) break ;
else q[p[i-a[j]]]=1;
for (j=0;j<=1000;j++)
if (!q[j]) { p[i]=j;break ; }
for (j=0;j<14;j++)
if (i<a[j]) break ;
else q[p[i-a[j]]]=0;
}
}
int main()
{
int n,m,o;
deal(1000);
while (scanf("%d%d%d", &m, &n, &o)&&(n+m+o)) {
if (p
^p[m]^p[o]) printf("Fibo\n");
else printf("Nacci\n");
}
return 0;
}