Question

C语言 输入参数a,b,c，求一元二次方程a*x*x＋b*x＋c＝0的根，结果保留2位小数。

程序填空，不要改变与输入输出有关的语句。
输入一个正整数repeat (0<repeat<10)，做repeat次下列运算：
输入参数a,b,c，求一元二次方程a*x*x＋b*x＋c＝0的根，结果保留2位小数。
输出使用以下语句：
printf("参数都为零，方程无意义!\n");
printf("a和b为0，c不为0，方程不成立\n");
printf("x=%0.2f\n", -c/b);
printf("x1=%0.2f\n", (-b+sqrt(d))/(2*a));
printf("x2=%0.2f\n", (-b-sqrt(d))/(2*a));
printf("x1=%0.2f+%0.2fi\n", -b/(2*a), sqrt(-d)/(2*a));
printf("x2=%0.2f-%0.2fi\n", -b/(2*a), sqrt(-d)/(2*a));
输入输出示例：括号内为说明
输入：
5 (repeat=5)
0 0 0 (a=0,b=0,c=0)
0 0 1 (a=0,b=0,c=1)
0 2 4 (a=0,b=2,c=4)
2.1 8.9 3.5 (a=2.1,b=8.9,c=3.5)
1 2 3 (a=1,b=2,c=3)
输出：
参数都为零，方程无意义!
a和b为0，c不为0，方程不成立
x = -2.00
x1 = -0.44
x2 = -3.80
x1 = -1.00+1.41i
x2 = -1.00-1.41i

#include <stdio.h>
#include <math.h>
int main(void)
{
int repeat, ri;
double a, b, c, d;

scanf("%d", &repeat);
for(ri = 1; ri <= repeat; ri++){
scanf("%lf%lf%lf", &a, &b, &c);
/*---------*/
}
}

Answer 1

#include <stdio.h>
#include <math.h>
int main(void)
{
int repeat, ri;
double a, b, c, d;

scanf("%d", &repeat);
for(ri = 1; ri <= repeat; ri++){
scanf("%lf%lf%lf", &a, &b, &c);
d=b*b-4*a*c;
if (a==0&&b==0&&c==0) printf("参数都为零，方程无意义!\n");
else if (a==0&&b==0&&(c!=0)) printf("a和b为0，c不为0，方程不成立\n");
else if (a==0) printf("x = %0.2f\n", -c/b);
else if (d>=0)
{
printf("x1 = %0.2f\n", (-b+sqrt(d))/(2*a));
printf("x2 = %0.2f\n", (-b-sqrt(d))/(2*a));
}
else if (d<0)
{
printf("x1 = %0.2f+%0.2fi\n", -b/(2*a), sqrt(-d)/(2*a));
printf("x2 = %0.2f-%0.2fi\n", -b/(2*a), sqrt(-d)/(2*a));
}

}
}

Answer 2

walk

Answer 3

aaaaasadsccvvbbnn

Answer 4

喜欢不劳而获的人。。。。。。。。

Answer 5

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