The Archeologist's Dilemma

算法：设x为要求的指数的幂，i为位数,n为给出的数，则n*10^i<2^x<(n+1)*10^i. 对小于号两边取对数，得

log2(n)+i*log2(10)<x<log2(n+1)+i*log2(10), 所以只要求出这个范围内的最小正整数就是答案了。

 

// The Archeologist's Dilemma.cpp : Defines the entry point for the console application.//#include "stdafx.h"using namespace std;int _tmain(int argc, _TCHAR* argv[]){double logNA(double a,double n);double data=0;while(cin>>data){if(data==EOF)break;else{bool flag=false;for(long i=long(logNA(10,data)+2);flag==false;++i){double before=logNA(2,data)+i*logNA(2,10);double after=logNA(2,data+1)+i*logNA(2,10);for(long j=(long)before;(j<=(long)after)&&(flag==false);++j){if((j>(long)before)&&(j<=(long)after)) {cout<<j<<"/n";flag=true;break;}}}}}system("pause");return 0;}double logNA(double a,double n){return log10(n)/log10(a);}