关于C语言求一元二次方程根的算法,运行怎么这样?#include<stdio.h>#include <math.h>int main(){double a,b,c,x1,x2,disc,p,q;scanf("%1f%1f%1f",&a,&b,&c);disc=b*b-4*a*c;p=-b/(2*a);q=sqrt(disc)/(2*a);x1=p
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![关于C语言求一元二次方程根的算法,运行怎么这样?#include<stdio.h>#include <math.h>int main(){double a,b,c,x1,x2,disc,p,q;scanf("%1f%1f%1f",&a,&b,&c);disc=b*b-4*a*c;p=-b/(2*a);q=sqrt(disc)/(2*a);x1=p](/uploads/image/z/5077934-62-4.jpg?t=%E5%85%B3%E4%BA%8EC%E8%AF%AD%E8%A8%80%E6%B1%82%E4%B8%80%E5%85%83%E4%BA%8C%E6%AC%A1%E6%96%B9%E7%A8%8B%E6%A0%B9%E7%9A%84%E7%AE%97%E6%B3%95%2C%E8%BF%90%E8%A1%8C%E6%80%8E%E4%B9%88%E8%BF%99%E6%A0%B7%3F%23include%26lt%3Bstdio.h%26gt%3B%23include+%26lt%3Bmath.h%26gt%3Bint+main%28%29%7Bdouble+a%2Cb%2Cc%2Cx1%2Cx2%2Cdisc%2Cp%2Cq%3Bscanf%28%26quot%3B%251f%251f%251f%26quot%3B%2C%26amp%3Ba%2C%26amp%3Bb%2C%26amp%3Bc%29%3Bdisc%3Db%2Ab-4%2Aa%2Ac%3Bp%3D-b%2F%282%2Aa%29%3Bq%3Dsqrt%28disc%29%2F%282%2Aa%29%3Bx1%3Dp)
关于C语言求一元二次方程根的算法,运行怎么这样?#include<stdio.h>#include <math.h>int main(){double a,b,c,x1,x2,disc,p,q;scanf("%1f%1f%1f",&a,&b,&c);disc=b*b-4*a*c;p=-b/(2*a);q=sqrt(disc)/(2*a);x1=p
关于C语言求一元二次方程根的算法,运行怎么这样?
#include<stdio.h>
#include <math.h>
int main()
{double a,b,c,x1,x2,disc,p,q;
scanf("%1f%1f%1f",&a,&b,&c);
disc=b*b-4*a*c;
p=-b/(2*a);
q=sqrt(disc)/(2*a);
x1=p+q;x2=p-q;
printf("x1=%7.2f\nx2=%7.2f\n",x1,x2);
return 0;
}
关于C语言求一元二次方程根的算法,运行怎么这样?#include<stdio.h>#include <math.h>int main(){double a,b,c,x1,x2,disc,p,q;scanf("%1f%1f%1f",&a,&b,&c);disc=b*b-4*a*c;p=-b/(2*a);q=sqrt(disc)/(2*a);x1=p
#include
#include
int main()
{
double a,b,c,x1,x2,disc,p,q;
scanf("%1lf%1lf%1lf",&a,&b,&c);//scanf中double应对应lf,而不是f
disc=b*b-4*a*c;
p=-b/(2*a);
q=sqrt(disc)/(2*a);
x1=p+q;x2=p-q;
printf("x1=%7.2f\nx2=%7.2f\n",x1,x2);
return 0;
}