HOJ 2695 Humidex

最新推荐文章于 2016-11-23 22:41:24 发布

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本文提供了一个程序实现，通过给定的公式计算三个数字中的任意两个值来得出第三个值。该程序包括解决温度、密度和热力等参数之间的关系，并通过输入字符指令来选择所需的操作。

http://acm.hit.edu.cn/hoj/problem/view?id=2695

给三个数字中的两个根据公式计算出第三个

#include <stdio.h>
#include <math.h>

double solve_t(double d, double h);
double solve_d(double t, double h);
double solve_h(double t, double d);

int main()
{
    double num1, num2;
    char ch1, ch2;
    double t, d, h;

    while (scanf("%c", &ch1)!=EOF)
    {
        getchar();
        if (ch1 == 'E')
            break;
        scanf("%lf %c %lf", &num1, &ch2, &num2);
        getchar();
        if (ch1 == 'T' && ch2 == 'D')
        {
            t = num1, d = num2;
            h = solve_h(t, d);
        }
        else if(ch1 == 'D' && ch2 == 'T')
        {
            t = num2, d = num1;
            h = solve_h(t, d);
        }

        else if(ch1 == 'T' && ch2 == 'H')
        {
            t = num1, h = num2;
            d = solve_d(t, h);
        }
        else if(ch1 == 'H' && ch2 == 'T')
        {
            t = num2, h = num1;
            d = solve_d(t, h);
        }

        else if(ch1 == 'D' && ch2 == 'H')
        {
            d = num1, h = num2;
            t = solve_t(d, h);
        }
        else if(ch1 == 'H' && ch2 == 'D')
        {
            d = num2, h = num1;
            t = solve_t(d, h);
        }
        printf("T %.1lf D %.1lf H %.1lf\n", t, d, h);
    }

    return 0;
}


double solve_t(double d, double h)
{
    double e, t;

    e = 6.11 * exp (5417.7530 * ((1/273.16) - (1/(d+273.16))));
    t = h - (0.5555)*(e - 10.0);
    return t;
}

double solve_d(double t, double h)
{
    double e, d;

    e = (h - t)/0.5555 +10;
    d = 1 / (1 / 273.16 - ( log(e) - log(6.11) ) / 5417.7530) - 273.16;
    return d;
}

double solve_h(double t, double d)
{
    double e, h;

    e = 6.11 * exp (5417.7530 * ((1/273.16) - (1/(d+273.16))));
    h = t + (0.5555)*(e - 10.0);
    return h;
}