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.
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.
样例输入
3 4 -1
-1 2 7
5 -1 3
0 0 0
样例输出
Triangle #1
c = 5.000
Triangle #2
a = 6.708
Triangle #3
Impossible.
参考代码
#include <stdio.h>
#include <math.h>
int main()
{
int a,b,c,k=1;
double m=0;
while (scanf("%d%d%d",&a,&b,&c))
{
if (a==0&&b==0&&c==0)
break;
printf("Triangle #%dn",k);
if ((c<a||c<b)&&c!=-1)
printf("Impossible.n");
else if (a==-1)
{
m=sqrt(c*c-b*b);
printf("a = %.3lfn",m);
}
else if (b==-1)
{
m=sqrt(c*c-a*a);
printf("b = %.3lfn",m);
}
else if (c==-1)
{
m=sqrt(a*a+b*b);
printf("c = %.3lfn",m);
}
printf("n");
k++;
}
return 0;
}
解析
暂无
