POJ 3299 Humidex(简单公式递推)

//纯粹公式递推，注意题目要求，任意知2求1//别递推错就OK了#include<iostream>#include<cmath>#include<iomanip>using namespace std;const double Exp = 2.718281828;char a,b;double aa,bb;double var[100];double T,H,D;double e,h;void print(){cout << "T ";cout << fixed << setprecision(1) << T;cout << " D ";cout << fixed << setprecision(1) << D;cout << " H ";cout << fixed << setprecision(1) << H << endl;}int main(){while(cin >> a){if(a == 'E')break;cin >> aa >> b >> bb;memset(var,0,sizeof(var));var[a] = aa;var[b] = bb;T = H = D = 0;//注意初始化if(var['T'] != 0)T = var['T'];if(var['D'] != 0)D = var['D'];if(var['H'] != 0)H = var['H'];//通过这种形式赋值可以简化判断结构if(T != 0 && D != 0){e = 6.11 * pow(Exp,5417.7530 * ((1/273.16) - (1/(D+273.16))));h = (0.5555) * (e - 10.0);H = T + h;print();}else if(T != 0 && H != 0){h = H - T;e = h / 0.5555 + 10.0;D = 1/(log(e/6.11)/((-1)*5417.7530) + 1/273.16) - 273.16;print();}else if(H != 0 && D != 0){e = 6.11 * pow(Exp,5417.7530 * ((1/273.16) - (1/(D+273.16))));h = (0.5555) * (e - 10.0);T = H - h;print();}}return 0;}