Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

In the figure below you can see triangle ABC and its in-苞ircle (Circle that touches all the sides of a triangle internally). The radius of this in circle is r. Three other circles are drawn. Each of them touches two sides of this triangle and the in circle of ABC. The radiuses of these circles are r₁, r₂ and r₃.

Given the values of r, r₁, r₂ and r₃ you will have to find the area of triangle ABC.

Input 

The input file can contain up to 1000 lines of inputs. Each line contains four positive floating-計oint numbers which denotes the values of r, r₁, r₂ and r₃ respectively.

Input is terminated by a line containing four negative integers.

Output 

For each line of input produce one line of output. This line contains serial of output followed by a floating-計oint number which denotes the area of triangle ABC. This floating-計oint number may have two digits after the decimal point. You can assume that for the given values of r, r₁, r₂ and r₃ it will always be possible to construct a triangle ABC. If required you can assume that = 3.141592653589793 and also use double precision floating-計oint numbers for floating-計oint calculations. You can assume that there will be no such input for which small precision errors will cause difference in printed output. Look at the output for sample input for details.

Sample Input 

49.1958415692 5.3025839959 20.7869367050 31.8019699761 
186.6830516757 71.9474500429 84.8796672233 
37.6219288070 
-1 -1 -1 -1

Sample Output 

Case 1: 18237.14 
Case 2: 195777.32

---

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

int main() {
    double r, r1, r2, r3;
    int cases = 0;
    while(scanf("%lf %lf %lf %lf", &r, &r1, &r2, &r3) == 4) {
        if(r < 0)   break;
        double p1, p2, p3, a, b, c;
        p1 = r/(r-r1)*sqrt(pow(r+r1, 2)-pow(r-r1, 2));
        p2 = r/(r-r2)*sqrt(pow(r+r2, 2)-pow(r-r2, 2));
        p3 = r/(r-r3)*sqrt(pow(r+r3, 2)-pow(r-r3, 2));
        a = p1+p2, b = p2+p3, c = p3+p1;
        printf("Case %d: %.2lf", ++cases, (a+b+c)*r/2);
        puts("");
    }
    return 0;
}