HDU Steps 2.2.1

參考：http://blog.csdn.net/niushuai666/article/details/7013352

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對數的性質,loga(b^c)=c*loga(b),loga(b*c)=loga(b)+loga(c);
假設給出一個數10234432,那麼log10(10234432)=log10(1.0234432*10^7)=log10(1.0234432)+7;

log10(1.0234432)就是log10(10234432)的小數部分.

log10(1.0234432)=0.010063744
10^0.010063744=1.023443198

本題取四位

這題要利用到數列的公式:an=(1/√5) * [((1+√5)/2)^n-((1-√5)/2)^n](n=1,2,3.....)

取完對數

由於log10(1-((1-√5)/(1+√5))^n)->0

所以log10(an)=-0.5*log10(5.0)+((double)n)*log(f)/log(10.0);

#include<cstdio>
#include<cmath>
int a[21]={0,1,1};
int main()
{
	int n;
	for(int i=3;i<21;i++)
		a[i]=a[i-1]+a[i-2];
	while(scanf("%d",&n)!=EOF)
	{
		if(n<21)
			printf("%d\n",a[n]);
		else
		{
			double b=-0.5*log(5.0)/log(10.0)+((double)n)*log(0.5+sqrt(5.0)/2.0)/log(10.0);
			b-=floor(b);
			b=pow(10.0,b);//一開始遺漏了= =
			while(b<1000)b*=10;
			printf("%d\n",(int)b);
		}
	}
	return 0;
}

複習

log(n) 即 log2(n)

log(m)(n) = log(n) / log(m) 

pow(10.0, b) = 10.0^b

floor(b)捨去小數部分

ceil(b) = floor(b) + 1