zoj1241 Geometry Made Simple

Geometry Made Simple

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.

Input

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.

Output

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.

題目大意就是三條邊，第三邊是直角三角形的斜邊，計算沒給的那條邊。

輸入時沒給的那條邊用-1

輸出時每組資料後有一空行，若不存在，則輸出Impossible.

輸出格式為：

Triangle #1

c = 5.000

水題，不多說

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