## Algorithm to improve programming to find the roots of a quadratic equation

``````问题描述
编写一个求解一元二次方程的实数根的程序，方程的系数由用户在运行xh

输入一行三个整数分别为一元二次方程的三个系数，数据之间以空格隔开

输出一行为方程的实数根（若两根不同 较大的在前 两根以空格隔开 若两根相同 输出一个 若无根 输出 NO ）

1 -5 4

4 1

1 -2 1

1

1 0 1

NO
``````

Problem solution ideas:

Formula method:

code show as below:

``````#include<stdio.h>
#include<math.h>
int main() {

double a,b,c;
scanf("%lf%lf%lf",&a,&b,&c);
double root=0,x1=0,x2=0;
root=b*b-4*a*c;
if(root>0){

x1=(-b+sqrt(root))/(2*a);
x2=(-b-sqrt(root))/(2*a);
if(x1>x2){

printf("%g %g",x1,x2);
}
}else if(root==0){

x1=(-b)/(2*a);
x2=x1;
printf("%g",x1);
}else{

printf("NO");
}

return 0;
}
``````

This question should pay attention to a precision output, otherwise it will be sad for some data. For example, 2 5 1.

``````printf("%g %g",x1,x2);
``````

What does %g mean in this line of code.
%g is an output format type of the C language printf() function. It indicates that single and double-precision real numbers are output with the shorter output width in %f%e. The %e format is used when the exponent is less than -4 or greater than or equal to the precision.

for example

``````printf("%g\n", 0.00001234);

printf("%g\n", 0.0001234);

printf("%.2g\n", 123.45);

printf("%.2g\n", 23.45);
``````

result:

``````1.234e-05
0.0001234
1.2e+02
23
``````

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Origin blog.csdn.net/mjh1667002013/article/details/115037588
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