傻傻得看两个函数谁大,理解错了一段文字,走了很多弯路,还好最后终于理解了。
#include "stdafx.h"
#include <iostream>
#include <math.h>
using namespace std;
double quicklog(__in double base, __in double value);
int _tmain(int argc, _TCHAR* argv[])
{
double n = 100000000;
cout<<pow(n,quicklog(3.0f/2,2))<<"\t"<<n*quicklog(2,n);
return 0;
}
double quicklog(__in double base, __in double value)
{
return log(value)/log(base);
}