POJ 3299 Humidex 杂题

题意：略。按公式便好。

#include <cmath>
#include <iomanip>
#include <iostream>
using namespace std;

int main()
{
	double one, two;
	char first, second;
	cout << setprecision ( 1 )  << fixed;

	while ( cin >> first && first != 'E' )
	{
		cin >> one >> second >> two;
		if ( first == 'T' || second == 'T' )
		{
			if ( first != 'T' )
			{
				swap(first,second);
				swap(one,two);
			}
			if ( second == 'D' )
			{
				double h = ( 0.5555 ) * (  6.11 * exp( 5417.7530 * ( 1 / 273.16 -  1 / ( two + 273.16 ) ) ) - 10.0 );
				cout << "T " << one << " D " << two << " H " << one + h << endl;
			}
			else
			{
				double d = -273.16 - 1 / (log ( ((two - one )/ 0.5555 + 10.0) / 6.11 ) / 5417.7530 - 1 / 273.16);
				cout << "T " << one << " D " << d << " H " << two << endl;
			}
		}
		else
		{
			if ( first != 'H' )
			{
				swap(first,second);
				swap(one,two);
			} 
			double t = one - ( 0.5555 ) * (  6.11 * exp( 5417.7530 * ( 1 / 273.16 -  1 / ( two + 273.16 ) ) ) - 10.0 );
			cout << "T " << t << " D " << two << " H " << one << endl;
		}
	}
	return 0;
}