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#include<stdio.h>
#include<math.h>
int main()
{
 double a,b,c;
 while(scanf("%lf%lf%lf",&a,&b,&c)==3){
  if(a>0){
   if((b*b-4*a*c)>0){
   printf("Two different roots x1=%f , x2=%f\n",(-b+sqrt(b*b-4*a*c))/(2*a),(-b-sqrt(b*b-4*a*c))/(2*a));}
   if((b*b-4*a*c)==0)
   printf("Two same roots x=%f\n",(-b)/(2*a));
   if((b*b-4*a*c)<0)
    printf("No real root\n");
   }
  if(a<0){
   if((b*b-4*a*c)>0){
   printf("Two different roots x1=%f , x2=%f\n",(-b-sqrt(b*b-4*a*c))/(2*a),(-b+sqrt(b*b-4*a*c))/(2*a));}
   if((b*b-4*a*c)==0)
   printf("Two same roots x=%f\n",(-b)/(2*a));
   if((b*b-4*a*c)<0)
    printf("No real root\n");
   }
  }
  return 0;   
}
  

---

 

---

#include<stdio.h>
#include<math.h>
int main()
{
 int a,b,c;
 while(scanf("%d%d%d",&a,&b,&c)==3){
  if(a>0){
   if((b*b-4*a*c)>0)
   printf("Two different roots x1=%d , x2=%d\n",(-b+sqrt(b*b-4*a*c))/(2*a),(-b-sqrt(b*b-4*a*c))/(2*a));
   if((b*b-4*a*c)==0)
   printf("Two same roots x=%d\n",(-b)/(2*a));
   if((b*b-4*a*c)<0)
    printf("No real root\n");
   }
  if(a<0){
   if((b*b-4*a*c)>0)
   printf("Two different roots x1=%d , x2=%d\n",(-b-sqrt(b*b-4*a*c))/(2*a),(-b+sqrt(b*b-4*a*c))/(2*a));
   if((b*b-4*a*c)==0)
   printf("Two same roots x=%d\n",(-b)/(2*a));
   if((b*b-4*a*c)<0)
    printf("No real root\n");
   }
  }
  return 0;   
}
抱歉應該是這個，就是如果輸入1 4 3會出現x1=0,x2=-1074.....

 不知道是為什麼= =

 

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只能請更強的高手來解惑~~

目前我只知道是printf或是sqrt的問題....它會變的怪怪的

解決方法就是強制轉型....(其實他的名子我不知道是啥,看看就好)

printf("Two different roots x1=%d , x2=%d\n",(int)(-b+sqrt(b*b-4*a*c))/(2*a),(int)(-b-sqrt(b*b-4*a*c))/(2*a)); 

 

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噢~謝謝囉

雖然還不是很清楚

但應該可以慢慢理解=ˇ=