【POJ3299】Humidex（简单的数学推导）

公式题中已经给出，直接求解即可。

 

 1 #include <iostream>

 2 #include <cstdlib>

 3 #include <cstdio>

 4 #include <cstdlib>

 5 #include <cstring>

 6 #include <cmath>

 7 #include <cctype>

 8 #include <algorithm>

 9 #include <numeric>

10 #include <sstream>

11 #include <map>

12 #include <iomanip>

13 using namespace std;

14 

15 map<char, double> data;

16 

17 int main () {

18     ios :: sync_with_stdio (false);

19     char C1, C2;

20     double X1, X2;

21     while (cin >> C1) {

22         if (C1 == 'E') break;

23         cin >> X1 >> C2 >> X2;

24         data[C1] = X1; data[C2] = X2;

25         if ( (C1 == 'T' && C2 == 'H') || (C1 == 'H' && C2 == 'T') ) {

26             double h = data['H'] - data['T'];

27             double e = h / 0.5555 + 10.0;

28             double ln_ = log (e / 6.11);

29             data['D'] = 1.0 / (1.0 / 273.16 - ln_ / 5417.7536) - 273.16;

30         }

31         else {

32 

33             double e = 6.11 * exp (5417.7530 * (1.0 / 273.16 - (1.0 / (data['D'] + 273.16))));

34             //cout << "e = " << e << endl;

35             double h = (0.5555) * (e - 10.0);

36             //cout << "h = " << h << endl;

37             if ((C1 == 'T' && C2 == 'D' ) || (C1 == 'D' && C2 == 'T')) {

38                 //cout << "T D -> H" << endl;

39                 data['H'] = data['T'] + h;

40             }

41             if ((C1 == 'H' && C2 == 'D' ) || (C1 == 'D' && C2 == 'H')) {

42                 //cout << "H D -> T" << endl;

43                 data['T'] = data['H'] - h;

44             }

45         }

46         printf ("T %.1f D %.1f H %.1f\n", data['T'], data['D'], data['H']);

47     }

48     return 0;

49 }