/*
用do...while 计算一元二次方程。并证明do...while 主要用于人机交互
*/

#include<stdio.h>
#include<math.h>
int main(void)
{
    double a, b, c;
    double x1, x2;
    double delta;
    char ch;

    do
    {   
        printf("请输入一元二次方程的三个系数:\n");
        printf("a = ");
        scanf("%lf", &a);
        
        printf("b = ");
        scanf("%lf", &b);
        
        printf("c = ");
        scanf("%lf", &c);
        
        delta = b * b - 4 * a * c;
        
        if(delta > 0)
        {
            x1 = (-b + sqrt(delta))/(2 * a);
            x2 = (-b - sqrt(delta))/(2 * a);
            printf("有两个解，x1 = %lf, x2 = %lf\n", x1, x2);
        }
        else if(0 == delta)
        {
            x1 = x2 = (-b) /(2 * a);
            printf("有唯一解，x1 = x2 = %lf",x1, x2);
        }
        else
        {
            printf("该函数没有解\n");
        }
        printf("是否需要继续验证？（Y/N）");
        scanf(" %c\n", &ch);
    }
    while('y'==ch || 'Y'==ch);

    return 0;
}

输出结果如下：

请输入一元二次方程的三个系数:
a = 1
b = 5
c = 6
有两个解，x1 = -2.000000, x2 = -3.000000
是否需要继续验证？（Y/N）y
1
请输入一元二次方程的三个系数:
a = b = 5
c = 6
有两个解，x1 = -2.000000, x2 = -3.000000
是否需要继续验证？（Y/N）n
1

--------------------------------
Process exited after 13.94 seconds with return value 0
请按任意键继续. . .

怎么不符了
a=1 b=5 c=6 没错啊
(x+2)(x+3)=0
x=-2,x=-3

printf("有唯一解，x1 = x2 = %lf",x1, x2);