HDU Steps 2.2.1

来源：互联网发布：算法的基本特征编辑：程序博客网时间：2024/06/10 03:52

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

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对数的性质,loga(b^c)=c*loga(b),loga(b*c)=loga(b)+loga(c);
假设给出一个数10234432,那么log10(10234432)=log10(1.0234432*10^7)=log10(1.0234432)+7;

log10(1.0234432)就是log10(10234432)的小数部分.

log10(1.0234432)=0.010063744
10^0.010063744=1.023443198

本题取四位

这题要利用到数列的公式:an=(1/√5) * [((1+√5)/2)^n-((1-√5)/2)^n](n=1,2,3.....)

取完对数

由于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);//一开始遗漏了= =while(b<1000)b*=10;printf("%d\n",(int)b);}}return 0;}

复习

log(n) 即 log2(n)

log(m)(n) = log(n) / log(m)

pow(10.0, b) = 10.0^b

floor(b)舍去小数部分

ceil(b) = floor(b) + 1