找DEV C++ 这个软件，把下面的代码复制粘贴编译保存可以用来求2元1次函数式的X1 X2值（输入a.b.c敲回车就可以了）#include
#include
main(){ float a,b,c,disc,x1,x2,p,q; printf ("input a,b,c:"); scanf ("%f,%f,%f",&a,&b,&c); if (a==0) if (b==0) if (c==0) printf("it is trivial,\n"); else printf("it is impossible,\n"); else { printf("it has none solution:\n"); printf("x=%6.2f\n",-c/b); } else { disc=b*b-4*a*c; if(disc>=0) if(disc>0) { printf("it has two real solutions :\n"); x1=(-b+sqrt(disc))/(2*a); x2=(-b-sqrt(disc))/(2*a); printf("x1=%6.2f,x2=%6.2f\n",x1,x2); } else { printf("it has two same real solutions:\n"); printf("x1=x2=%6.2f\n",-b/(2*a)); }else { printf("it has two complex solutions:\n"); p=-b/(2*a); q=sqrt(-disc)/(2*a); printf("x1=%6.2f+%6.2fi,x2=%6.2f-%6.2fi\n",p,q,p,q); } } getchar(); return 0 ; }