poj 3299 Humidex(模拟)

题意：三个量，三个公式，知二求一；

思路：求每个量时一口气求出；

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int t,n,m;
double tem,dew,hum;
char str1[3],str2[3];
int main()
{
    int i,j,k,flag1,flag2,flag3;
    while(scanf("%s",str1)!=EOF)
    {
        if(str1[0]=='E') break;
        flag1=0,flag2=0,flag3=0;
        tem=dew=hum=200;
        if(str1[0]=='T') flag1=1,scanf("%lf",&tem);
        else if(str1[0]=='D') flag2=1,scanf("%lf",&dew);
        else flag3=1,scanf("%lf",&hum);
        scanf("%s",str2);
        if(str2[0]=='T') flag1=1,scanf("%lf",&tem);
        else if(str2[0]=='D') flag2=1,scanf("%lf",&dew);
        else flag3=1,scanf("%lf",&hum);
        if(flag2&&flag1)
             hum=tem+0.5555*(6.11*exp(5417.7530*(1/273.16-1/(dew+273.16)))-10);
        else if(flag2&&flag3)
             tem=hum-0.5555*(6.11*exp(5417.7530*(1/273.16-1/(dew+273.16)))-10);
        else 
            dew=1/((1/273.16)-((log((((hum-tem)/0.5555)+10.0)/6.11))/5417.7530))-273.16;
        printf("T %.1f D %.1f H %.1f\n",tem,dew,hum);
    }
    return 0;
}