POJ 3299

一道确实很简单的一题，却浪费了我大量的时间，一开始题目看了很久，看完后，正如网上的资料所说是一道相当水的题，特别适宜刷题，可是自己被while(scanf("%c",&T)!=EOF)和scanf("%f %c %f",&a,&D,&b);ch = getchar();卡住了，对于字符的输入要特别小心，因为“回车”也会被看做字符，而被输进去，故要设置一些特殊字符来消除“回车”：ch = getchar();

 

 

#include<stdio.h> 

#include<math.h>

int main(){

    float e,y,h,temp,dew,humidex,a,b;

    char T,D, ch;

while(scanf("%c",&T)!=EOF){

 if(T=='E') break;

      scanf("%f %c %f",&a,&D,&b);ch = getchar();

 //printf("AA/n");

 switch(T+D){

 case 'T'+'D':

     {

     temp = (T=='T') ? a:b;

 dew = (D=='D') ? b:a;

  y = pow(2.718281828,5417.7530*((1/273.16) - (1/(dew+273.16))));

           e = 6.11*y;

      h = (0.5555)*(e - 10.0);

      humidex = temp + h;

      break;

      }

 case 'T'+'H':

{

temp = (T=='T') ? a:b;

humidex = (D=='H') ? b:a;

h = humidex - temp;

e = (float) (h / 0.5555 + 10.0) ;

dew = (float) (1/(-log(e/6.11)/5417.7530 + (1/273.16))-273.16);

break;

}

 

case 'H'+'D':

{

humidex = (T=='H') ? a:b;

dew = (D=='D') ? b:a;

e = (float) (6.11 * exp (5417.7530 * ((1/273.16) - (1/(dew+273.16)))));

h = (float) ((0.5555) * (e - 10.0));

temp = humidex - h;

break;

 

        }

 }

 printf("T %.1f D %.1f H %.1f/n",temp,dew,humidex);

 

}

return 0;

}