HDOJ题目地址：传送门

u Calculate e

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 41480    Accepted Submission(s): 18881

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

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<memory.h>
using namespace std;
int jiecheng(int m){
    if(m<=1)
        return 1;
    else
        return m*jiecheng(m-1);
}
int main(){
    printf("n e\n- -----------\n");
    for(int i=0;i<=9;i++){
        double result=0;
        for(int j=0;j<=i;j++){
            result+=(1.0/jiecheng(j));
        }
        if(i==0||i==1)
            printf("%d %.0lf\n",i,result);
        else if(i==2)
            printf("%d %.1lf\n",i,result);
        else
            printf("%d %.9lf\n",i,result);
    }
}