输入正整数n，当精确度分别取0.1，0.01，0.001.。。。。。时，计算并输出下列算式的值，s=1+1/2!+1/3!+1/4!.........
#include<stdio.h>
#include<math.h>
int x;
double y;
double cal(int n)
{
    double t=0,m,z;
    int i,j,s=1;
    {
     while((int)m%10!=0){
       i=1;
        for(j=1;j<=i;j++)
          s=j*s;
          t=1/s+t;
          z=1/s;
          s=1;
          i++;
          m=t*pow(10,n+1);      
                      }
                       t=t-z;           
                      return t;
}
void main()
{
     printf("请输入精确度x=\n");
     scanf("%d",&x);
     y=cal(x);
     printf("%.%df\n",n,y);
}

printf("%.%df\n",n,y);
printf("%d%f\n",n,y);

好像不至这一个错...

#include<stdio.h>
#include<math.h>
int x;
double y;
double cal(int n)
{
    double t=0,m,z;
    int i,j,s=1;
    {       //这个大括符多余的
     while((int)m%10!=0){
       i=1;
        for(j=1;j<=i;j++) // 由于前面的 i=1，for循环只会执行一次，请吧i赋值成自己需要的值，或者用scanf从键盘获取
          s=j*s;
          t=1/s+t;
          z=1/s;
          s=1;
          i++;
          m=t*pow(10,n+1);      
                      }
                       t=t-z;           
                      return t;
}
void main()
{
     printf("请输入精确度x=\n");
     scanf("%d",&x);
     y=cal(x);
     printf("%.%df\n",n,y);  //格式有误，应该是  printf("%d%f\n",n,y);
}

printf("%.%df\n",n,y);
那怎么让他小数保留n位

那哪位帮我把这题目解下

#include<stdio.h>
#include<math.h>
int x;
double y;
double cal(int n)
{
    double t=0,m,z;
    int i,j,s=1;
      {
        for(i=1;(int)m%10!=0;i++)
        for(j=1;j<=i;j++)
          s=j*s;
          z=1/s;
          t=z+t;
          s=1;
           m=t*pow(10,n+1);      
      }
                       t=t-z;           
                      return t;
}
void main()
{
     printf("请输入精确度x=\n");
     scanf("%d",&x);
     y=cal(x);
     printf("%f\n",y);
}
为什么运行结果总是0.000000，怎么样让他 printf("%f\n",y);的结果精确值是保留n位小数