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这程序那错了吖解决方案

2012-02-16 
这程序那错了吖#includestdio.h #includemath.h main(){floata,b,c,dprintf( \na )scanf( %f ,

这程序那错了吖
#include   "stdio.h "
#include   "math.h "
main()
{
        float   a,b,c,d;
        printf   ( "\n   a= ");
        scanf   ( "%f ",&a);
        printf   ( "\n   b= ");
        scanf   ( "&f ",&b);
        printf   ( "\n   c= ");
        scanf   ( "&f ",&c);
        printf   ( "\n   d= ");
        scanf   ( "&f ",&d);
        d=   b*b-4*a*c;
        if   (d> 0)
        {
        printf( "\n   x1=%f ",(-b+sart(d))/(2*a));
        printf( "\n   x2+%f ",(-b-sart(d))/(2*a));
        }
        if   (d==0)   printf( "\n   x1=x2%f ",(-b)/(2*a));
        if   (d <0)   printf( "\n   The   equation   has   no   real   root! ");
}

[解决办法]
#include "stdio.h "
#include "math.h "
void main()
{
float a,b,c,d;
printf ( "a= ");
scanf ( "%f ",&a);
printf ( "\nb= ");
scanf ( "%f ",&b);
printf ( "\nc= ");
scanf ( "%f ",&c);

d = b*b-4*a*c;
printf ( "\nd=%f ",d);
if (d> 0)
{
printf( "\nx1=%f\n ",(-b+sqrt(d))/(2*a));
printf( "\nx2=%f\n ",(-b-sqrt(d))/(2*a));
}
if (d==0) printf( "\nx1=x2=%f\n ",(-b)/(2*a));
if (d <0) printf( "\nThe equation has no real root!\n ");
}

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