acdream.18.KIDx's Triangle(数学推导）

KIDx's Triangle

Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)

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Problem Description

One day, KIDx solved a math problem for middle students in seconds! And than he created this problem.

Now, give you the degree of a, b, c, d, please calculate the degree of ∠AED.

Input

There are multiple test cases.

Each case contains one line 4 integers formatted as: "a b c d"

0 ≤ a, b, c, d < 90°.

0 < a+b < 90°, 0 < c+d < 90°.

Output

For each test case, output the answer in one line, rounded to 2 decimal places.

Sample Input

10 70 60 20
10 70 70 0

Sample Output

20.00
140.00

#include<stdio.h>
#include<math.h>
#define pi 3.141592653589793
double a , b , c , d ;
double BE , BD , DE , bata , AD , AE ;

inline double f (double k)
{
    return 1.0 * k * pi / 180 ;
}

int main ()
{
    freopen ("a.txt" , "r" , stdin ) ;
    while (~ scanf ("%lf%lf%lf%lf" , &a , &b , &c , &d)) {
        if (a == 0) {
            printf ("0.00
") ;
            continue ;
        }
        AD = sin (f(c)) / sin (f(a + b + c)) ;
        BD = sin (f(c + d)) / sin (f(a + b + c + d)) - sin (f(c)) / sin (f(a + b + c)) ;
        BE = sin (f(a + b)) / sin (f(a + b + c + d)) - sin (f(b)) / sin (f(b + c + d)) ;

        DE = sqrt (1.0 * (BE * BE + BD * BD + 2 * BE * BD * cos (f(a + b + c + d)))) ;
        AE = sin (f(c + d)) / sin (f(b + c + d)) ;
        bata =  acos (1.0 * (DE * DE + AE * AE - AD * AD) / (2.0 * DE * AE)) * 180 / pi;
            //    printf ("DE = %.2f , EF = %.2f , DF = %.2f
" , DE , EF , DF) ;
     //   printf ("bata = %.2f
" , bata * 180 / pi) ;
        if (bata < 0)
            printf ("%.2f
" ,180.0 - bata) ;
        else
            printf ("%.2f
" , bata) ;
    }
    return 0 ;
}

标程中的大写字母表示角度，小写字母表示边长，注意0的特殊情况即可。

设CE = a = 1，AE = b，AC = c，BE = d，AB = e，AD = f，BD = g，DE = h

使用正弦定理可依次求出AE，AB，AD

然后利用余弦定理依次求出DE，∠AED

其他解法：建立坐标系搞，或者二分搞