题目：V - u Calculate e
A simple mathematical formula for e is
链接：https://cn.vjudge.net/contest...
where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of n.
Output
Output the approximations of e generated by the above formula for the values of n from 0 to 9. The beginning of your output should appear similar to that shown below.
Sample Output
n e

0 1
1 2
2 2.5
3 2.666666667
4 2.708333333

思路：这个比较简单就是循环相乘就可以了；

小技巧：但这个有一个特别要注意的点，就是因为要输出有限位，即多余的0不可以输出，所以对于这种题目的时候，因为C中没有特定的直接控制只能输出有限位，所以需要用if分类讨论，即有限循环的把特定的位数给标记出来；

代码：

#include<stdio.h>

int main()
{
    double sum=1;
    printf("n e\n");
    printf("- -----------\n");
    printf("0 1\n");
    double tum=1;
    for(int i=1;i<=9;i++)
    {
        sum*=1.0/i;
        tum+=sum;
        if(i==1)
            printf("%d %.0f\n",i,tum);
        else if(i==2)
            printf("%d %.1f\n",i,tum);
        else
            printf("%d %.9f\n",i,tum);
    }
    return 0;
}