HDU 1012累乘累加

/*
Problem Description
A simple mathematical formula for e is

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

这题也是简单题，思路就是先累乘再累加

---

代码（OJ用C++）C和C++一起使用比较方便

#include <iostream>
#include <stdio.h>
using namespace std;

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