Geometry Made Simple

Geometry Made Simple

Geometry Made Simple

Description

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.

Sample Input

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

Sample Output

Triangle #1
c = 5.000

Triangle #2
a = 6.708

Triangle #3
Impossible.
題意分析：這是一道三角形的勾股定理的題目，這道題目的題意大體是按順序輸入a、b、c三條邊， c為斜邊，如果輸入-1表示該邊為未知邊，求此邊長度，如果沒有，輸出不可能。
下面給一下代碼：

#include
#include
int main()
{
    int i=1;
    int  a, b, c;
    double x;
    while(scanf("%d %d %d",&a, &b, &c)==3)
    {
        x=0;
        if(a==0&&b==0&&c==0)
        {
            break;
        }
        printf("Triangle #%d\n",i);
        if(a==-1)
        {
            x=c*c-b*b;
            if(x<0)
            {
                printf("Impossible.\n");
            }
            else
            {
                printf("a = %.3f\n",(double)sqrt(x));
            }

        }
        else if(b==-1)
        {
            x=c*c-a*a;
            if(x<0)
            {
                printf("Impossible.\n");
            }
            else
            {
                printf("b = %.3f\n",(double)sqrt(x));
            }

        }
        else
        {
            x=a*a+b*b;
            if(x<0)
            {
                printf("Impossible.\n");
            }
            else
            {
                printf("c = %.3f\n",(double)sqrt(x));
            }

        }
         i++;
         printf("\n");
    }
    return 0;
}