hdu 1568 Fibonacci

##代码

#include<cstdio>
#include<cstdlib>
#include<cmath>
const double constant_a = (1 + sqrt(5)) / 2;
const double constant_b = (1 - sqrt(5)) / 2;
const double constant_c = sqrt(5) / 5;
int main(){
	int n;
	while(~scanf("%d",&n)){
		if(n==0){
			printf("0\n");
			continue;	
		}
		int ans;
		if(n<=20)
			ans=(int)(constant_c * (pow(constant_a, n) - pow(constant_b, n)));
		else
			ans=(int)pow(10.0, 3.0 + fmod(-0.5*log10(5.0)+n*log10(constant_a), 1));;
		printf("%d\n",ans);		
	}
	return 0;	
}

##总结

1. 
2. 
   最后一项小于0并且很小可以不用计算