c++求二元一次方程的根#include#includevoid main(){printf(" ★☆欢迎使用二元一次方程求根工具☆★\n\n\n");float a,b,c,q,p,x1,x2;printf(" ★请输入a,b,c的值★\n\n\n\n\n\n");printf("a=");scanf("%f",&a);printf("b=");scanf("
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![c++求二元一次方程的根#include#includevoid main(){printf(](/uploads/image/z/1599139-19-9.jpg?t=c%2B%2B%E6%B1%82%E4%BA%8C%E5%85%83%E4%B8%80%E6%AC%A1%E6%96%B9%E7%A8%8B%E7%9A%84%E6%A0%B9%23include%23includevoid+main%28%29%7Bprintf%28%22+%E2%98%85%E2%98%86%E6%AC%A2%E8%BF%8E%E4%BD%BF%E7%94%A8%E4%BA%8C%E5%85%83%E4%B8%80%E6%AC%A1%E6%96%B9%E7%A8%8B%E6%B1%82%E6%A0%B9%E5%B7%A5%E5%85%B7%E2%98%86%E2%98%85%5Cn%5Cn%5Cn%22%29%3Bfloat+a%2Cb%2Cc%2Cq%2Cp%2Cx1%2Cx2%3Bprintf%28%22+%E2%98%85%E8%AF%B7%E8%BE%93%E5%85%A5a%2Cb%2Cc%E7%9A%84%E5%80%BC%E2%98%85%5Cn%5Cn%5Cn%5Cn%5Cn%5Cn%22%29%3Bprintf%28%22a%3D%22%29%3Bscanf%28%22%25f%22%2C%26a%29%3Bprintf%28%22b%3D%22%29%3Bscanf%28%22)
c++求二元一次方程的根#include#includevoid main(){printf(" ★☆欢迎使用二元一次方程求根工具☆★\n\n\n");float a,b,c,q,p,x1,x2;printf(" ★请输入a,b,c的值★\n\n\n\n\n\n");printf("a=");scanf("%f",&a);printf("b=");scanf("
c++求二元一次方程的根
#include
#include
void main()
{
printf(" ★☆欢迎使用二元一次方程求根工具☆★\n\n\n");
float a,b,c,q,p,x1,x2;
printf(" ★请输入a,b,c的值★\n\n\n\n\n\n");
printf("a=");
scanf("%f",&a);
printf("b=");
scanf("%f",&b);
printf("c=");
scanf("%f",&c);
q=-b/2*a;p=sqrt(b*b-4*a*c);
x1=q-p;x2=q+p;
printf("x1=%f,x2=%f,&x1&x2\n");
}
系统给了一个warnings :warning C4244:'=' :conversion from 'double' to 'float',possible loss of data
而且我用这个运行后,计算结果,总是得x1=0.000000,x2=一个负的可大的数!
我的公式逻辑上有错误吗?还有,我想加一个判定条件
如果b^2-4ac
c++求二元一次方程的根#include#includevoid main(){printf(" ★☆欢迎使用二元一次方程求根工具☆★\n\n\n");float a,b,c,q,p,x1,x2;printf(" ★请输入a,b,c的值★\n\n\n\n\n\n");printf("a=");scanf("%f",&a);printf("b=");scanf("
警告很正常 输出有问题
改了 功能完善了
#include
#include
#include
void main()
{
printf(" ★☆欢迎使用二元一次方程求根工具☆★\n\n\n");
float a,b,c,x1,x2;
while(1)
{
printf(" ★请输入a,b,c的值★\n\n\n\n\n\n");
printf("a=");
scanf("%f",&a);
printf("b=");
scanf("%f",&b);
printf("c=");
scanf("%f",&c);
x1=(-b-sqrt(b*b-4*a*c))/2*a;
x2=(-b+sqrt(b*b-4*a*c))/2*a;
if(b*b-4*a*c >= 0)
printf("x1=%f,x2=%f\n",x1,x2);
else
printf("没有实数解!\n");
printf("继续计算?");
char s[4];
scanf("%s",s);
if(s == "yes")
continue;
else if(s == "no")
break;
}
}