先看对数的性质,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)=0.01006374410^0.010063744=1.023443198那么要取几位就很明显了吧~先取对数(对10取),然后得到结果的小数部分bit,pow(10.0,bit)以后如果答案还是<1000那么就一直乘10。注意偶先处理了0~20项是为了方便处理~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 log10(an)=-0.5*log10(5.0) +(( double )n)*log(f)/log(10.0)+log10(1-((1-√5 )/(1+√5 ))^n)其中f= (sqrt(5.0)+1.0)/2.0; log10(1-((1-√5 )/(1+√5 ))^n)->0所以可以写成log10(an)=-0.5*log10(5.0) +(( double )n)*log(f)/log(10.0);最后取其小数部分。#include<iostream>#include<cmath>using namespace std;int fac[21]={0,1,1};const double f=(sqrt(5.0)+1.0)/2.0;int main(){ double bit; int n,i; for(i=3;i<=20;i++)fac[i]=fac[i-1]+fac[i-2];//求前20项 while(cin>>n) { if(n<=20) { cout<<fac[n]<<endl; continue; } bit=-0.5*log(5.0)/log(10.0)+((double)n)*log(f)/log(10.0);//忽略最后一项无穷小 bit=bit-floor(bit); bit=pow(10.0,bit); while(bit<1000)bit=bit*10.0; printf("%d/n",(int)bit); } return 0;}