实现自定义函数(超级详细)

#include<stdio.h>
#include<math.h>
double fact(int n)
{
	int i;
	double sum=1;
	for(i=1;i<=n;i++)
	{
		sum*=i;
	}
	return sum;
}
double mypow(double x,int n)
{
	
	double S;
	S=pow(-x,n);
	return S;
}
double fact(int n);
double mypow(double x,int n);
int main()
{
	double x;
	int n,i,t=-1;
	double f=0;
	scanf("%lf%d",&x,&n);
	for(i=1;i<=n;i++)
	{
		f+=t*mypow(x,i)/fact(i);
	}
	printf("%0.4lf",f);
	return 0;
}

#include <stdio.h>

double mypow(double a, long long int b){
    double q = a;
    if( b==1 ) return a;
    else return a * mypow(q,b-1);
}

double fac(long long int m){
    if( m==1 ) return 1;
    else return m * fac(m-1);
}

int main(){
    int t = -1;
    long long int i, n;
    double x, S = 0;
    scanf("%lf %lld", &x, &n);
    for( i=1; i<=n; i++ ){
        t *= -1;
        S += t * mypow(x,i) / fac(i);
    }
    printf("%.4lf", S);
}

double  fact(int n)
{
	int i = 1;
	double sn = 1;
	for (; i <= n; i++)
	{
		sn *= i;
	}
	return sn;
}

double mypow(double x, int n)
{
	int i = 1;
	double sx = 1;
	for (; i <= n; i++)
	{
		sx *= x;
	}
	return sx;
}

int main()
{
	double x = 0;
	int n = 0;
	int j = 0;
	int s = -1;
	double sum = 0;
	scanf("%lf %d", &x, &n);
	for (j = 1; j <= n; j++)
	{
		s *= -1;
		sum += s * (mypow(x, j) / fact(j));
	}
	printf("%.4lf\n", sum);
	return 0;
}

double fact(double n);
double mypow(double x, double n);
int main()
{
	double n, x;
	int i;
	
	scanf("%lf%lf", &n, &x);
	

		double sum = 0.0000;
		double t = 1;

	for (i = 1, t = 1; i <= n; i++)

	{

		sum += t * mypow(x, n) / fact(n);

		t = -t;

	}
	printf("%.4lf\n", sum);
}
double fact(double n)
{
	double c = 0;
	if (n == 1)
	{
		return 1;
	}
	else
	{
		c=1 * fact(n - 1)*n;
		return c;
	}
}
double mypow(double x, double n)
{
	int i = 0;
	double sum = 1;
	for (i = 0; i < n; i++)
	{
		sum *= x ;
	}
	return sum;
}
为什么我最后的结果是0.000

#include<stdio.h>
double fact(long long n)
{
    long long i;
    i=1;
    while(n>0)
    {
        i=i*n;
        n--;
    }
    return i;
}
double mypow(double x, long long n)
{
    long long i, j;
    i=j=1;
    while(i<=n)
    {
        j=j*x;
        i++;
    }
    return j;
}
int main()
{
    int t;
    t=1;
    long long a, b;
    b=1;
    double c, sum;
    sum=0;
    scanf("%lf %lld", &c, &a);
    while(b<=a)
    {
        sum=sum+t*mypow(c,b)/fact(b);
        t=-t;
        b++;
    }
    printf("%.4lf", sum);
    return 0;
}
这个有什么问题？

#include<stdio.h>
int main()
{
	double fact(long long n);
	double mypow(double x,long long n);
	double x,s=0.0,a,b;
	long long n,i,t=-1;
	scanf("%lf %d",&x,&n);
	for(i=1;i<=n;i++)
	{
		a=mypow(x,n);
		b=fact(n);
		if(i%2==0)
		{
			b=t*b;
		}
	
		s=s+a/b;
	}
	printf("%.4lf",s);
	return 0;
}
double mypow(double x,long long n)
{
	double z=1.0;
	long long j;
	for(j=1;j<=n;j++)
	{
		z=z*x;
	}
	return z;
}
double fact(long long n)
{
	double w=1.0;
	long long h;
	for(h=1;h<=n;h++)
	{
		w=w*h;
	}
	return w;
}

为什么提交上去不对呢？

#include <stdio.h>
#include <math.h>>
double fact(double n)
{
	int i;
	double j=1;
	for(i=1;i<=n;i++)
	{
		j=i*j;
	
	}
		return j;
}


double mypow(double x,int n)
{
	int i;
	double m,fac;
	for(i=1;i<=n;i++)
	{	fac=fact(n);
		m=(pow((-1),i-1)*pow(x,i))/fac+m;
	}
	return m;
}
int main ()
{ int n;
double x,w;
scanf("%lf %d",&x,&n);
w=mypow(x,n);
printf("%.4lf",w);
}

输出1.000问题在哪求解

#include <stdio.h>
#include <math.h>
int fact(int n)
{
	int i;
	double p=1;
	for(i=1;i<=n;i++)
	p=(double)p*i;
	return p;
}
int mypow(int x,int n)
{
	double y=1;
	int i;
	for(i=0;i<n;i++)
	y=(double)y*x;
	return y;
}
int main()
{
	double x,r=0,t=0,sum=0;
	int n,i,q=1;
	scanf("%lf%d",&x,&n);
	for(i=1;i<=n;i++)
	{
		q=1;
		if(i%2==0)
		{
			q=-1;
		}
		r=fact(i);
		t=mypow(x,i);
		sum=(double)sum+q*t/r;	
	}
	printf("%.4lf",sum);
}



这个哪里有问题呢？？大佬们拜托拜托！！