题意解释
这道题是给出了3个值的任意两个值，求剩下的那个。题中给出了公式，所以直接自己反推下公式就好了。
收获
大概就是读题耐心的重要性，最开始想成了求H的过程，WA了好久。
AC代码
#include<iostream>
#include<math.h>
#include<iomanip>
#include<algorithm>
using namespace std;
int main(void)
{
double T, D, H, E, h, t1, t2;
char c1, c2;
while (cin >> c1&&c1 != 'E')
{
cin >> t1 >> c2 >> t2;
if (c1 == 'T'&&c2 == 'D' || c1 == 'D'&&c2 == 'T')
{
T = t1, D = t2;
if (c1 == 'D') swap(T, D);
H = T + 0.5555*(6.11*exp(5417.7530*((1 / 273.16) - (1 / (D + 273.16)))) - 10.0);
}
else if (c1 == 'T'&&c2 == 'H' || c1 == 'H'&&c2 == 'T')
{
T = t1, H = t2;
if (c1 == 'H') swap(T, H);
D = 1 / (1 / 273.16 - log(((H - T) / 0.5555 + 10.0 )/ 6.11) / 5417.7530) - 273.16;
}
else
{
D = t1, H = t2;
if (c1 == 'H') swap(D, H);
T = H - 0.5555*(6.11*exp(5417.7530*((1 / 273.16) - (1 / (D + 273.16)))) - 10.0);
}
cout << setiosflags(ios::fixed) << setprecision(1) << "T " << T << " D " << D << " H " << H << endl;
}
return 0;
}
