poj3299 Humidex

The humidex formula is as follows:

humidex = temperature + h
h = (0.5555)× (e - 10.0)
e = 6.11 × exp [5417.7530 × ((1/273.16) - (1/(dewpoint+273.16)))]

where  exp(x) is 2.718281828 raised to the exponent  x.

While humidex is just a number, radio announcers often announce it as if it were the temperature, e.g. "It's 47 degrees out there ... [pause] .. with the humidex,". Sometimes weather reports give the temperature and dewpoint, or the temperature and humidex, but rarely do they report all three measurements. Write a program that, given any two of the measurements, will calculate the third.

You may assume that for all inputs, the temperature, dewpoint, and humidex are all between -100°C and 100°C.

Input

Input will consist of a number of lines. Each line except the last will consist of four items separated by spaces: a letter, a number, a second letter, and a second number. Each letter specifies the meaning of the number that follows it, and will be either T, indicating temperature, D, indicating dewpoint, or H, indicating humidex. The last line of input will consist of the single letter E.

Output

For each line of input except the last, produce one line of output. Each line of output should have the form:

 T number D number H number

where the three numbers are replaced with the temperature, dewpoint, and humidex. Each value should be expressed rounded to the nearest tenth of a degree, with exactly one digit after the decimal point. All temperatures are in degrees celsius.

Sample Input

T 30 D 15

T 30.0 D 25.0

E

Sample Output

T 30.0 D 15.0 H 34.0

T 30.0 D 25.0 H 42.3

#include <iostream>

#include <cmath>

using namespace std;

 

const double e = 2.718281828;

 

double fun1(double h, double d)//求T

{

 return h - (0.5555 * ((6.11 * pow(e, 5417.7530*(1/273.16-1/(d+273.16)))) - 10.0));

}

 

double fun2(double t, double d)//求H

{

    return t + (0.5555 * ((6.11*pow(e,5417.7530*(1/273.16-1/(d+273.16)))) - 10.0));

}

 

double fun3(double t, double h)//求D

{

    return 1/(1/273.16 - log(((h-t)/0.5555+10.0)/6.11)/5417.7530) - 273.16;

}

 

int main()

{

    char a1,a2;

    double t1,t2;

    while(scanf("%c", &a1) && a1 != 'E')

    {

        scanf("%lf %c %lf", &t1, &a2, &t2);

        //printf("%c %lf %c %lf\n", a1, t1, a2, t2);

        //输出

        //数据前后顺序，如T 30 D 15和D 15 T 30答案应该一样

        if(a1=='D'&&a2=='H')

            printf("T %.1f D %.1f H %.1f\n",fun1(t2,t1),t1,t2);

        else if(a1=='H'&&a2=='D')

            printf("T %.1f D %.1f H %.1f\n",fun1(t1,t2),t2,t1);

        else if(a1=='T'&&a2=='D')

            printf("T %.1f D %.1f H %.1f\n",t1,t2,fun2(t1,t2));

        else if(a1=='D'&&a2=='T')

            printf("T %.1f D %.1f H %.1f\n",t2,t1,fun2(t2,t1));

        else if(a1=='T'&&a2=='H')

            printf("T %.1f D %.1f H %.1f\n",t1,fun3(t1,t2),t2);

        else if(a1=='H'&&a2=='T')

            printf("T %.1f D %.1f H %.1f\n",t2,fun3(t2,t1),t1);   

    }

    return 0;

}