我指的并不是这段代码
时间: 2023-08-06 18:15:48 浏览: 35
非常抱歉,我理解错了。以下是将您之前提供的问题中的中文字符替换为英文字符后的代码:
```c
#include <stdio.h>
#include <math.h>
void roots1(double a, double b, double c) {
double x1, x2, d;
d = b * b - 4 * a * c;
x1 = (-b + sqrt(d)) / (2 * a);
x2 = (-b - sqrt(d)) / (2 * a);
printf("When b2-4ac is greater than 0, the two roots of the equation are: x1 = %f, x2 = %f\n", x1, x2);
}
void roots2(double a, double b, double c) {
double x;
x = -b / (2 * a);
printf("When b2-4ac is equal to 0, the only root of the equation is: %f\n", x);
}
void roots3(double a, double b, double c) {
double p, q;
p = -b / (2 * a);
q = sqrt(4 * a * c - b * b) / (2 * a);
printf("When b2-4ac is less than 0, the two complex roots of the equation are: x1 = %f + %fi, x2 = %f - %fi\n", p, q, p, q);
}
int main() {
double a, b, c, d;
printf("Please enter the coefficients of the equation a, b, c:\n");
scanf("%lf%lf%lf", &a, &b, &c);
d = b * b - 4 * a * c;
if (d > 0) {
roots1(a, b, c);
} else if (d == 0) {
roots2(a, b, c);
} else {
roots3(a, b, c);
}
return 0;
}
```
在这个更改后的程序中,我们将中文字符替换为英文字符,使程序更加规范和易懂。同时,程序的功能没有发生变化,仍然可以求出方程 `ax2+bx+c=0` 当 `b2-4ac` 大于 0、等于 0 和小于 0 时的根。