UVA10236 斐波那契素数

题意：任取斐波那契数列中一项f[i],若对于所有j
解法：这题的理论分析在黑书上有，结论是从第五项开始下标为素数的斐波那契数都是斐波那契素数

#include <stdio.h>
#include <string.h>
const int MAXN = 250010;;
int prime[25010];
bool isprime[MAXN];
long double fib[MAXN];
int main(){
	int i, j;
	for(i=0; i<MAXN; i++)
		isprime[i] = true;
	for(i=2; i<=500; i++){
		for(j=i*i; j<=MAXN; j+=i)
			isprime[j] = false;
	}
	int idx = 1;
	for(i=2; i<=MAXN; i++){
		if(isprime[i])
			prime[idx++] = i;
	}
	prime[1] = 3;
	prime[2] = 4;
	fib[0] = fib[1] = 1;
	bool flag = false;
	for(i=2; i<=MAXN; i++){
		if(flag){
			fib[i] = fib[i-1] + fib[i-2]/10;
			flag = 0;
		}
		else {
			fib[i] = fib[i-1] + fib[i-2];
		}
		if(fib[i]>1e9){
			fib[i]/=10;
			flag = true;
		}
	}
	int n;
	while(scanf("%d", &n)!=EOF)
		printf("%d\n", (int)fib[prime[n]-1]);
}