Discussion Detail - XAZXOJ

题解
为啥错了 T1058
LZH LV 8 @ 2023-11-4 16:03:53

# include <bits/stdc++.h>
using namespace std;

int main ()
{
	double a,b,c;
	cin>>a>>b>>c;
	if(sqrt(b*b-4*a*c)>0)
	{if((-b+sqrt(b*b-4*a*c))/(2*a)<(-b-sqrt(b*b-4*a*c))/(2*a))
	
	printf("x1=%.5lf;x2=%.5lf",(-b+sqrt(b*b-4*a*c))/(2*a),(-b-sqrt(b*b-4*a*c))/(2*a));}
	if(sqrt(b*b-4*a*c)<0)
	cout<<"No answer!";
	if(sqrt(b*b-4*a*c)==0)
	printf("x1=x2=%.5lf",(-b+sqrt(b*b-4*a*c))/(2*a));
	
	
	


	
	
	
	return 0; 
}

changmengyu LV 6 @ 2024-1-28 15:42:02

@UID:2 sfasfasdfads
changmengyu LV 6 @ 2024-1-28 15:39:14

@2 dfasdf
changmengyu LV 6 @ 2024-1-28 15:38:48

@34 fdsv
changmengyu LV 6 @ 2024-1-28 15:36:50

@47 的撒
Csvoner LV 3 SU @ 2024-1-28 15:36:09

@34 的撒旦
Csvoner LV 3 SU @ 2024-1-28 15:31:01

d

Y_ZL LV 8 @ 2023-11-5 11:52:21

1.判别式( $\Delta$ )是 $b^2-4ac$ 不能带根号。根号下不能是负数。
2. $\Delta = 0$ 时输出 $\frac{-b}{2a}$ 就行了。

# include <bits/stdc++.h>
using namespace std;

int main ()
{
	double a,b,c;
	cin>>a>>b>>c;
	if(b*b-4*a*c>0)
        if((-b+sqrt(b*b-4*a*c))/(2*a) > (-b-sqrt(b*b-4*a*c))/(2*a))
	        printf("x1=%.5lf;x2=%.5lf",(-b-sqrt(b*b-4*a*c))/(2*a),(-b+sqrt(b*b-4*a*c))/(2*a));
        else printf("x1=%.5lf;x2=%.5lf",(-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)
	    cout<<"No answer!";
	if(b*b-4*a*c == 0)
	    printf("x1=x2=%.5lf",(-b)/(2*a));
	return 0; 
}