运用printf的输出机制

解题思路:

注意事项:

参考代码:

#define _CRT_SECURE_NO_DEPRECATE

#pragma warning(disable:4996)

#include<stdio.h>

#include<math.h>

void root(float a, float b, float c)

{

float temp = b * b - 4 * a * c;

if (temp >= 0)

{

printf("x1=%.3f x2=%.3f\n", (-b + sqrt(temp)) / (2 * a), (-b - sqrt(temp)) / (2 * a));

}

else

{

printf("x1=%.3f+%.3fi x2=%.3f-%.3fi", -b / (2 * a), sqrt(-temp) /( 2 * a), -b / (2 * a), sqrt(-temp) / (2 * a));

}

}

int main()

{

float  a, b, c;

scanf("%f%f%f", &a, &b, &c);

root(a, b, c);

return 0;

}