-use双精度次使用的sqrt()和指数函数EXP() - 使用*计算方 - 不要使用POW()
我得到他们只是没有什么,什么我的预期值。我试图把他们都签订了,但它并没有改变任何东西,我试着打印出与12位小数并没有什么似乎working.I有联系的数学库,并将其定义为好。
double normal(double x, double sigma, double mu)
{
double func = 1.0/(sigma * sqrt(2.0*M_PI));
double raise = 1.0/2.0*((x-mu)/sigma);
double func1 = func * exp(raise);
double comp_func = (func1 * func1);
return comp_func;
}
int main(void)
{
// create two constant variables for μ and σ
const double sigma, mu;
//create a variable for x - only dynamic variable in equation
unsigned int x;
//create a variable for N values of x to use for loop
int no_x;
//scaniing value into mu
printf("Enter mean u: ");
scanf("%lf", &mu);
//scanning value into sigma
printf("Enter standard deviation: ");
scanf("%lf", &sigma);
//if sigma = 0 then exit
if(sigma == 0)
{
printf("error you entered: 0");
exit(0);
}
//storing number of x values in no_x
printf("Number of x values: ");
scanf("%d", &no_x);
//the for loop where i am calling function normal N times
for(int i = 1; i <= no_x; i++)
{
//printing i for the counter in prompted x values
printf("x value %d : ", i);
// scanning in x
scanf("%lf", &x);
x = normal(x,sigma,mu);
printf("f(x) = : %lf.12", x);
printf("\n");
}
return 0;
}
C:> \ A.EXE输入意味着U:3.489输入标准偏差S:1.203数量的x值:3×值1:3.4 F(X)= 0.330716549275 x值2:-3.4 F(X)= 0.000000025104 x值3:4 F(X)= 0.303015189801
但是,这是我收到
C:\ Csource> A.EXE输入意味着U:3.489输入标准偏差:1.203 x值的数目:3×值1:3.4 F(X)=:15086080.000000 x值2:-3.4 F(X)=:15086080.000000 X值3:4 F(X)=:1610612736.000000
插入这些行:
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
更改:
const double sigma, mu;
至:
double sigma, mu;
更改:
unsigned int x;
至:
double x;
更换normal
函数的定义:
double normal(double x, double sigma, double mu)
{
double func = 1.0/(sigma * sqrt(2.0*M_PI));
double t = (x-mu)/sigma;
return func * exp(-t*t/2);
}
#define _CRT_SECURE_NO_WARNINGS
#define _USE_MATH_DEFINES
#ifndef M_PI
#define M_PI (3.14159265358979323846)
#endif
#include<math.h>
#include<stdio.h>
#include <stdlib.h>
double normal(double x, double sigma, double mu)
{
double func = 1.0/(sigma * sqrt(2.0*M_PI));
double t = (x-mu)/sigma;
return func * exp((-0.5*t)* t);
}
我终于得到了与上述调整它字面上整天笑,C数学可以是相当棘手的工作后,这个代码,谢谢你上面还有帮助。