Geometry Made Simple

古城微笑少年丶 2022-08-10 04:44 145阅读 0赞

## Geometry Made Simple ##

##### Time Limit: 1000ms Memory limit: 65536K 有疑问？点这里^\_^ #####

## 题目描述 ##

Mathematics can be so easy when you have a computer. Consider the following example. You probably know that in a right-angled triangle, the length of the three sides a, b, c (where c is the longest side, called the hypotenuse) satisfy the relation a\*a+b\*b=c\*c. This is called Pythagora's Law.  
  
Here we consider the problem of computing the length of the third side, if two are given.  
  
![1054.gif][]

## 输入 ##

The input contains the descriptions of several triangles. Each description consists of a line containing three integers a, b and c, giving the lengths of the respective sides of a right-angled triangle. Exactly one of the three numbers is equal to -1 (the 'unknown' side), the others are positive (the 'given' sides).  
  
A description having a=b=c=0 terminates the input.

## 输出 ##

For each triangle description in the input, first output the number of the triangle, as shown in the sample output. Then print "Impossible." if there is no right-angled triangle, that has the 'given' side lengths. Otherwise output the length of the 'unknown' side in the format "s = l", where s is the name of the unknown side (a, b or c), and l is its length. l must be printed exact to three digits to the right of the decimal point.  
  
Print a blank line after each test case.

## 示例输入 ##

3 4 -1
    -1 2 7
    5 -1 3
    0 0 0

## 示例输出 ##

Triangle #1
    c = 5.000
    
    Triangle #2
    a = 6.708
    
    Triangle #3
    Impossible.

## 提示 ##

## 来源 ##

面向对数据结构和算法不太懂的同学

## 示例程序 ##

#include <stdio.h>   
    #include<math.h>   
    int main()   
    {   
        double a,b,c,d;   
        int k=0;   
        while(scanf("%lf%lf%lf",&a,&b,&c),(a||b||c))   
        {   
            k++;   
            if(a==-1)   
            {   
                d=sqrt(c*c-b*b);   
                if(d>c-b&&d<c+b)   
                    printf("Triangle #%d\na = %.3lf\n\n",k,d);   
                else  
                    printf("Triangle #%d\nImpossible.\n\n",k);   
            }   
            if(b==-1)   
            {   
                d=sqrt(c*c-a*a);   
                if(d>c-a&&d<c+a)   
                    printf("Triangle #%d\nb = %.3lf\n\n",k,d);   
                else  
                    printf("Triangle #%d\nImpossible.\n\n",k);   
            }   
            if(c==-1)   
            {   
                d=sqrt(a*a+b*b);   
                if(d>fabs(b-a)&&d<a+b)   
                    printf("Triangle #%d\nc = %.3lf\n\n",k,d);   
                else  
                    printf("Triangle #%d\nImpossible.\n\n",k);   
            }   
        }   
        return 0;   
    }

[1054.gif]: /images/20220731/217f34c4f2e3477e891af3215c3e2bd2.png