C语言编程问题,高手求助!浮点数能不能显示小数点6位以后的数字?

2025-02-09 02:03:57
推荐回答(5个)
回答1:

用double类型 别用float

而且这个交点应该一堆吧...

这个只是x > 0的部分,x < 0的部分关于原点对称。

#include
#include

#define __pi 3.1415926535897932384626433832795
#define __x_max 1.4451326206513048896928159563086
#define __max_depth 50

int compare(double x)
{
if (sin(50.0f / 3 * x) > 50.0f / (23 * __pi) * x)
{
return 1;
}

return 0;
}

double get_cross(double x0, double x1)
{
int f = compare(x0);

for (int i = 0; i < __max_depth; i++)
{
double t = (x0 + x1) * 0.5;

if (compare(t) != f)
{
x1 = t;
}
else
{
x0 = t;
}
}

return (x0 + x1) * 0.5;
}

void main()
{
double step = __pi * 0.5 / (50.0f / 3);

for (double i = 0; i < __x_max;)
{
double x = get_cross(i, i + step);

printf("x: %.12f (%.12f %.12f)\n", x, sin(50.0f / 3 * x), 50.0f / (23 * __pi) * x);

i += step;

x = get_cross(i, i + step);

printf("x: %.12f (%.12f %.12f)\n", x, sin(50.0f / 3 * x), 50.0f / (23 * __pi) * x);

i += 3 * step;
}
}

回答2:

浮点型数也有单精度和双精度之分,单精度float只能正确显示小数部分7位(不包括小数点),而双精度double则可以正确显示小数部分15位(不包括小数点)。意思是说,不管单精度还是双精度,小数点后几十位也是可以显示的,只是在其精度范围之外输出的是不确定的数字罢了

回答3:

/*=======================================================
*Author :wacs5
*Date :20090210(YYYYMMDD)
*Function :牛顿迭代法求方程的根
*=======================================================*/
#include
#include
#include
#define MAX_DIEDAI_TIME 200

main()
{
int n=0;
double x=0.1; /*初值*/
double jingdu=1e-8; /*精度*/
double function(double x);
double d2function(double x);
double newton_diedai(double x0,int *n,double jingdu);

system("cls");
x=newton_diedai(x,&n,jingdu);
printf("x=%.8lf\ty=%lf\n",x,function(x));
getch();
return 0;
}

/*====================
想要求解的方程的表达式
======================*/
double function(double x)
{
return sin(50*x/3)-50*x/(23*M_PI);
}

/*===========================
想要求解的方程的表达式的导数
=============================*/
double d2function(double x)
{
return 50/3*cos(50*x/3)-50/(23*M_PI);
}

/*=============================================
牛顿迭代法解方程组的解
x0为迭代的初值,n为迭代次数,jingdu为精度
function为求根代数式,d2functoin为其导数
返回最终符合一定精度的根
*/
double newton_diedai(double x0,int *n,double jingdu)
{
double x,temp;
temp=d2function(x0);
if (fabs(temp)>1e-10) /*防止除数为0*/
{
x=x0-function(x0)/temp;
printf("n=%d\tx=%.5lf\n",*n,x);
}
else
{
printf("error:div 0:\nPress any key to exit:");
getch();
exit(1);
}
if (++(*n)>MAX_DIEDAI_TIME)
{
printf("diedai time:%d > MAX_DIEDAI_TIME:\nPress any key to exit:",*n);
getch();
exit(1);
}
temp=function(x);
if (fabs(temp)return x;
else
return newton_diedai(x,n,jingdu);
}

/*******以上程序能得到一个解0.54250152,你改变一下初值
即main函数中double x=0.1;这一行
可能可以得到其它的解
做图估计,解的个数应当是15个左右*/

回答4:

要用double或者,float只能精确到小数点后六位左右,这是在保存小数用的位数决定的,然后牛顿迭代法求解方程,你可以去搜一搜牛顿迭代法,都是类似的方程求解问题

回答5:

double dLeft = sin(50*x/3);
double dRight = 50*x*(23*PI);
double dDiff = dLeft - dRight;
if((dDiff > -0.000001) && (dDiff < 0.000001))
{
printf("a root found, which is: %f\n", x);
}

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