这很容易的啊
#include
#include
double funcx( double x , double a)
{
double y;
if ( x==0 )
return 1;
y=0.5*(x+ a/x ) ;//改此函数可以求各种迭代
return y;
}
int main()
{
double x1,x2,e=1e-5 ,a;
printf("input a : ");
scanf("%lf", &a );
x2=1;
do {
x1=x2;
x2=funcx( x1, a );
} while( fabs(x2-x1) > e );
printf("%g\n", x2 );
return 0;
}
# include
# include
main()
{
double m,n=1.0,a;
printf("请输入a:");
scanf("%lf",&a);
do
{
m=n;
n= (m + a/m) / 2;
}
while(fabs(n-m)>1e-5);
printf("结果为:%7.5f\n",n);
}
# include
# include
main()
{
double m,n=1.0,a;
printf("请输入a:");
scanf("%lf",&a);
do
{
m=n;
n= (m + a/m) / 2;
}
while(fabs(n-m)>1e-5);
printf("结果为:%7.5f\n",n);
}。
最佳答案由提问者推荐
#include
#include
void main()
{
float a,x0,x1;
scanf("%f",&a);
x0=a/2;
x1=(x0+a/x0)/2;
do{
x0=x1;
x1=(x0+a/x0)/2;
}while(fabs(x0-x1)>1.0e-5);
printf("%.5f",x1);
}。