n e
- -----------
0 1
1 2
2 2.5
3 2.666666667
4 2.708333333

#include<iostream>
using namespace std;
int main()
{
    printf("n e\n");
    printf("- -----------\n");
        printf("0 1\n");
        printf("1 2\n");
        printf("2 2.5\n");
        printf("3 ");
        printf("%.9lf\n",2+1.0/2+1.0/(3*2));
        printf("4 ");
        printf("%.9lf\n",2+1.0/2+1.0/(3*2)+1.0/(4*3*2));
        printf("5 ");
        printf("%.9lf\n",2+1.0/2+1.0/(3*2)+1.0/(4*3*2)+1.0/(5*4*3*2));
        printf("6 ");
        printf("%.9lf\n",2+1.0/2+1.0/(3*2)+1.0/(4*3*2)+1.0/(5*4*3*2)+1.0/(6*5*4*3*2));
        printf("7 ");
        printf("%.9lf\n",2+1.0/2+1.0/(3*2)+1.0/(4*3*2)+1.0/(5*4*3*2)+1.0/(6*5*4*3*2)+1.0/(7*6*5*4*3*2));
        printf("8 ");
        printf("%.9lf\n",2+1.0/2+1.0/(3*2)+1.0/(4*3*2)+1.0/(5*4*3*2)+1.0/(6*5*4*3*2)+1.0/(7*6*5*4*3*2)+1.0/(8*7*6*5*4*3*2));
        printf("9 ");
        printf("%.9lf\n",2+1.0/2+1.0/(3*2)+1.0/(4*3*2)+1.0/(5*4*3*2)+1.0/(6*5*4*3*2)+1.0/(7*6*5*4*3*2)+1.0/(8*7*6*5*4*3*2)+1.0/(9*8*7*6*5*4*3*2));
    return 0;
}

下面copy一下。。是比较简便的

<pre name="code" class="cpp"> 1: #include<stdio.h>
  2: void main()
  3: {
  4: 	double arr[10] = {1};
  5: 	int i = 1,j = 3;
  6: 	while(i < 10)
  7: 	{
  8: 		arr[i] = i * arr[i - 1];
  9: 		i++;
 10: 	}
 11: 	printf("n e\n");
 12: 	printf("- -----------\n");
 13: 	printf("0 1\n");
 14: 	printf("1 2\n");
 15: 	printf("2 2.5\n");
 16: 	double result = 2.5;
 17: 	while(j < 10)
 18: 	{
 19: 		result = result + 1/arr[j];
 20: 		printf("%d %11.9f\n",j,result);
 21: 		j++;
 22: 	}
 23: }