2次方程式の解を求める

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#include "stdafx.h"
#include <math.h>
int main(void)
{
int a,b,c;
double D,alpha,beta,x;
scanf_s("%d %d %d",&a,&b,&c);
printf("a=%d b=%d c=%d\n",a,b,c);
if (a!=0)
{
D = b*b-4.0*a*c;
if (D==0)
{
x = -b/(2*a);
printf("実数の重根 %g\n",x);
}
else if (D>0)
{
alpha = -b/(2*a);
beta = sqrt(D)/(2*a);
printf("実数の解 %g±%g, %g %g\n",
alpha,beta,alpha+beta,alpha-beta);
}
else /* D<0 */
{
alpha = -b/(2*a);
beta = sqrt(-D)/(2*a);
printf("複素数の解 %g±i%g\n",alpha,beta);
}
}
else
{
if (b!=0)
{
x = -(c/b);
printf("１次⽅方程式の解 %g\n",x);
}
else if (c==0) /* b==0 && c==0 */
{
printf("解は不定（任意の値）\n");
}
else /* b==0 && c!=0 */
{
printf("不能\n");
}
}
quadequ.c
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51 }
2
return 0;
quadequ.c

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