zoj 1241 Geametry Made Simple

Geometry Made Simple

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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.

分析：

此题需分情况讨论，要对读入的三个数做判断。

当c存在时，要判断此三角形是否存在。

代码：

#include <cstdio>#include <cmath>int a, b, c;int flag;void solve() {    if (a == -1) {        if (b >= c)            puts("Impossible.");        else            printf("a = %.3lf\n", sqrt(c * 1.0 * c - b * 1.0 * b));    }    else if (b == -1) {        if (a >= c)            puts("Impossible.");        else            printf("b = %.3lf\n", sqrt(c * 1.0 * c - a * 1.0 * a));    }    else if (c == -1)        printf("c = %.3lf\n", sqrt(a * 1.0 * a + b * 1.0 * b));}int main() {    flag = 1;    while (~scanf("%d%d%d", &a, &b, &c)) {        if (a == 0 || b == 0 || c == 0) break;        printf("Triangle #%d\n", flag++);        solve();        puts("");    }return 0;}