##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
最后一项小于0并且很小可以不用计算