<=<<x/y break case"%: cout<<endk<x<<%<<y<<=<<x%y; break; default: cout<<endk<wrong 8编程求方程ax2+bx+c=0的解,其中的a,b,c由键盘输入。 #include <stdio. h> #include <math. h> c,disc,xl,x2,p, scanf(%f, %f, %f", &a, &b, &c); if(a==0) {if(c==0) printf("该方程无穷解n"); else printi("该方程无解"); else print"单根x=%rn",-ba): i disc=b=b-4"a*c; if(disc==O) printf(xl=x2=%7.2fn",-b/(2*a) else if (disc>0 ixl=(-b+sqrt(disc)(2*a); (b-sqrt(disc))/(2*a); printf("x1=%7.2f,x2=%7.2fn",x1,x2); el sqrt(fabs(disc)/(2*a); printf(x1=%7. 2f+%7.2fi\n",p, q); printf("x1=%7.2f -%7.2f in",P, q);case'/': cout<<endl<<x<<"÷"<<y<<"="<<x/y; break; case'%': cout<<endl<<x<<"%"<<y<<"="<<x%y; break; default: cout<<endl<<"Wrong"; } } ⒏编程求方程 ax2+bx+c=0 的解，其中的 a,b,c 由键盘输入。 #include <stdio.h> #include <math.h> void main() {float a,b,c,disc,x1,x2,p,q; scanf("%f,%f,%f",&a,&b,&c); if (a==0) {if (b==0) { if(c==0) printf("该方程无穷解\n"); else printf("该方程无解\n"); } else printf("单根 x＝%f\n",-b/a); } else { disc=b*b-4*a*c; if (disc==0) printf("x1=x2=%7.2f\n", -b/(2*a)); else {if (disc>0) { x1=(-b+sqrt(disc))/(2*a); x2=(-b-sqrt(disc))/(2*a); printf("x1=%7.2f,x2=%7.2f\n", x1, x2); } else {p=-b/(2*a); q=sqrt(fabs(disc))/(2*a); printf("x1=%7.2f + %7.2f i\n", p,q); printf("x1=%7.2f — %7.2f i\n", p,q); } } } }