学步园

大意略。

思路：

素数定理：GCD(F_n, F_m) = F_GCD(n, m)，则Fibonacci_Prime[k] = Fibonacci[Prime[k]],（ k >= 3）

http://en.wikipedia.org/wiki/Fibonacci_prime

一开始用大数算出了14位左右的FIbonacci素数在9位，15就是10位，然后用Fibonacci通项公式去算>=15的素数时，精度丢失比较严重，精确的范围在20000左右，我怀疑它是估计把数据出到22000，泪奔。

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
using namespace std;

typedef unsigned long long ULL;

const int MAXN = 250100;
const int maxn = 1100;

ULL prime[MAXN];
bool vis[MAXN] = {0};

int tot;
int n;

struct bign  
{  
    int len, s[maxn];  
    bign ()  
    {  
        memset(s, 0, sizeof(s));  
        len = 1;  
    }  
    bign (int num) { *this = num; }  
    bign (const char *num) { *this = num; }  
    bign operator = (const int num)  
    {
        char s[maxn];  
        sprintf(s, "%d", num);  
        *this = s;  
        return *this;  
    }
    bign operator = (const char *num)  
    {
        len = strlen(num);  
        for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';  
        return *this;  
    }
    bign operator + (const bign &b) const //+  
    {  
        bign c;  
        c.len = 0;  
        for(int i = 0, g = 0; g || i < max(len, b.len); i++)  
        {  
            int x = g;  
            if(i < len) x += s[i];  
            if(i < b.len) x += b.s[i];  
            c.s[c.len++] = x % 10;  
            g = x / 10;  
        }  
        return c;  
    }
	void print()
	{
		for(int i = len-1; i >= 0; i--) printf("%d", s[i]);
		printf("\n");
	}
};

bign f[2200];

void init()
{
	tot = 1;
	for(int i = 2; i < MAXN; i++) if(!vis[i])
	{
		prime[tot++] = i;
		for(ULL j = i*i; j < MAXN; j += i) vis[j] = 1;
	}
	prime[1] = 3, prime[2] = 4;
	f[1] = 1, f[2] = 1;
    for(int i = 3; i <= prime[31]; i++) f[i] = f[i-1] + f[i-2];
}

ULL cal(int n)
{
    double f = (1.0+sqrt(5.0))/2.0;
    double a = -0.5*log10(5.0)+((double)n)*log10(f);  
    double p =  a - (ULL)a;
    ULL t =  ULL(pow(10.0, p) * pow(10.0, 8.0));
    return t;
}

void solve()
{
	if(n <= 14)
	{
		f[prime[n]].print();
	}
	else
	{
		ULL ans = cal(prime[n]);
		printf("%llu\n", ans);
	}
}

int main()
{
	init();
	while(~scanf("%d", &n))
	{
		solve();
	}
	return 0;
}