Solution: the sum of any two sides is greater than the third side to form a triangle. With three sides of a triangle area formula to seek called: Heron's formula.
Helen pushed to the following formula:
#include <stdio.h>
#include <math.h>
int main(void)
{
float x1, y1, x2, y2, x3, y3;
float a, b, c, L, A, p;
scanf("%f%f%f%f%f%f", &x1, &y1, &x2, &y2, &x3, &y3);
a = sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
b = sqrt((x1 - x3) * (x1 - x3) + (y1 - y3) * (y1 - y3));
c = sqrt((x2 - x3) * (x2 - x3) + (y2 - y3) * (y2 - y3));
if (a + b > c && a + c > b && b + c > a)
{
L = a + b + c;
p = L / 2;
A = sqrt(p * (p - a) * (p - b) * (p - c));
printf("L = %.2f, A = %.2f\n", L, A);
}
else
printf("Impossible\n");
return 0;
}
Note: The middle of the side length abc into pow likely to cause errors.
Code Inspection:
#include <stdio.h>
#include <math.h>
int main(void)
{
float x1, y1, x2, y2, x3, y3;
float a, b, c, L, A, p;
scanf("%f%f%f%f%f%f", &x1, &y1, &x2, &y2, &x3, &y3);
a = sqrt(pow((x1 - x2), 2) + pow((y1 - y2), 2));
b = sqrt(pow((x1 - x3), 2) + pow((y1 - y3), 2));
c = sqrt(pow((x2 - x3), 2) + pow((y2 - y3), 2));
if (a + b > c && a + c > b && b + c > a)
{
L = a + b + c;
p = L / 2;
A = sqrt(p * (p - a) * (p - b) * (p - c));
printf("L = %.2f, A = %.2f\n", L, A);
}
else
printf("Impossible\n");
return 0;
}
Results are as follows:
Analyze the reasons:
() function is used to find the x y power (power), x, and y are double type pow function value, which is a prototype: double pow (double x, double y);
likely caused the error:
- If the base x is negative and y is not an integer index, it will lead to error domain error.
- If the index base x and y are 0, it may result in an error domain error, or may not; this is related with the realization of the library.
- If the base x is 0, y index is negative, could result in domain error pole error or error, it may not; this is related with the realization of the library.
- If the return value ret is too big or too small, it will result in an error range error.