POJ 3299

//题目类型：水题（数论）
//解题思路：看清题意，直接推导公式即可
#include <iostream>
#include <stdio.h>
//#include <conio.h>
#include <math.h>
using namespace std;
double temper,humidex,dew;
void solvedew()
{
    dew = 1/(1/273.16-log((humidex-temper+5.555)/(0.5555*6.11))/5417.7530)-273.16;
}
void solvetemper()
{
    temper = humidex-(0.5555)*(6.11*exp(5417.7530*((1/273.16) - (1/(dew+273.16)))) - 10.0);
}
void solvehumidex()
{
    humidex = temper + (0.5555)*(6.11*exp(5417.7530*((1/273.16) - (1/(dew+273.16)))) - 10.0);     
}
int main()
{
    //freopen("1.txt","r",stdin);
    bool botemper,bohumidex,bodew;
    char ch[5];
    while(1)
    {
        botemper = false;
        bohumidex = false;
        bodew = false;
        scanf("%s",ch);   //尽量采用字符串读字符
        if(ch[0]=='E') break;
        else if(ch[0]=='T')
        {
             scanf("%lf",&temper);        //G++中双精度浮点数使用scanf读取时依然采用%lf
             botemper = true;
        }
        else if(ch[0]=='D')
        {
             scanf("%lf",&dew);        
             bodew = true;
        }
        else if(ch[0]=='H')
        {
             scanf("%lf",&humidex);
             bohumidex = true;
        }
        scanf("%s",ch);
        if(ch[0]=='T')
        {
             scanf("%lf",&temper);
             botemper = true;
        }
        else if(ch[0]=='D')
        {
             scanf("%lf",&dew);
             bodew = true;
        }
        else if(ch[0]=='H')
        {
             scanf("%lf",&humidex);
             bohumidex = true;
        }
        if(bodew && botemper) solvehumidex();
        else if(bodew && bohumidex) solvetemper();
        else if(botemper && bohumidex) solvedew();
        //printf("T %.1lf D %.1lf H %.1lf\n",temper,dew,humidex);
        printf("T %.1f D %.1f H %.1f\n",temper,dew,humidex);     //特别注意G++中不支持%lf
    }
    //getch();
    return 0;
}

转载于:https://www.cnblogs.com/north_dragon/archive/2010/09/29/1838304.html