06 习题B
第六题。
#include <stdio.h>
#include <math.h>
int main()
{
double a, b, x;
printf("一元一次方程：ax+b=0\n");
printf("a, b = ?");
scanf("%lg%lg", &a, &b);
if (a != 0)
{
x = -b/a;
printf("唯一解：x = %g",x);
}
else
{
if (b == 0)
{
printf("无穷多解");
}
else
{
printf("无解");
}
}
return 0;
}

#include <stdio.h>
#include <math.h>
int main()
{
double a, b, c, delta, d, e;
printf("一元二次方程：ax2 + bx + c = 0\n");
printf("a, b, c = ?");
scanf("%lg%lg%lg", &a, &b, &c);
delta = b * b - 4 * a * c;
if (fabs(delta) <= 1e-8)
{
printf("x1 = x2 = %g\n", -b/(2*a) );
}
else if ( delta > 0)
{
printf("x1 = %g\n", (-b + sqrt(delta)) / (2 * a));
printf("x2 = %g\n", (-b - sqrt(delta)) / (2 * a));
}
else
{
d = -b / (2 * a);
e = sqrt(-delta) /(fabs(2 * a));
printf("x1 = %g + %gi \n", d, e);
printf("x2 = %g - %gi \n", d, e);
}
return 0;
}
讨厌负的=A=

#include <stdio.h>
#include <math.h>
#define dblpi = 6.2831853071795865
#define epsilon 1e-8
int main ()
{
double x, p, y;
int d = 0;
printf("x = ?");
scanf("%lg", &x);
x = fmod(x, dblpi);
p = 1;
y = p;
do
{
d += 2;
p *= -x * x /((d-1) * d);
y += p;
}
while (fabs(p) >= epsilon);
printf("y = %g\n", y);
return 0;
}