#include<stdio.h>
#include<math.h>
int main()
{
int a,b,c,x;
double disc,x1,x2;
scanf("%d %d %d",&a,&b,&c);
disc=b*b-4*a*c;
if (fabs(disc)<1e-6)
printf("\"NO\"");
else
{
if (fabs(disc)>1e-6)
{
x1=(-b+sqrt(disc))/(2*a);
x2=(-b-sqrt(disc))/(2*a);
if (x1==(int)x1&&x2==(int)x2)
printf("%d %d",x1,x2);
else printf("\"NO\"");
}
else
{
x=(-b)/2*a;
if (x=(int)x)
printf("%d",x);
else printf("NO");
}
}
return 0;
}
1 回答
已采納
qq_莫非
TA貢獻(xiàn)257條經(jīng)驗(yàn) 獲得超93個(gè)贊
#include?<stdio.h>??
#include?<math.h>?
void?main?()?
{?
?double?sqrt(double?x);?
?int?a,b,c;?
?double?x1,x2,x,e,d,g,f;?
scanf("%d?%d?%d",&a,&b,&c);?
??d=b*b-4*a*c;??
if?(a==0)?
{??????if?(b!=0)?
{x=-(double)c/(double)b;?
if?(x==0)?
{x=-x;?
printf("x=%.6lf\n",x);}?
else?
printf("x=%.6lf\n",x);?
}?
????else?
?????printf("Input?error!\n");}?
else?if?(d<0)?
{d=-d;?
e=sqrt(d);?
g=-(double)b/(2*(double)a);?
f=e/(2*a);?
????if?(g!=0)?
?????printf("x1=%.6lf+%.6lfi\nx2=%.6lf-%.6lfi\n",g,f,g,f);?
????else?
????printf("x1=%.6lfi\nx2=-%.6lfi\n",f,f);}?
else?if?(d==0)?
{?????x1=x2=-b/(2*a);?
??????printf("x1=x2=%.6lf\n",x1);}?
??????else?
??{????e=sqrt(d);?
???????x1=(-b+e)/(2*a);?
??????????x2=(-b-e)/(2*a);?
??????????printf("x1=%.6lf\nx2=%.6lf\n",x1,x2);}?
}
- 1 回答
- 0 關(guān)注
- 1248 瀏覽
添加回答
舉報(bào)
0/150
提交
取消