u Calculate e

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

Total Submission(s): 35137 Accepted Submission(s): 15824

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

开始错了几次，后来在网上说第八个后面有个０，最后决定用ｐｒｉｎｔｆ控制格式

#include <iostream>

#include <iomanip>

#include <cstdio>

using namespace std;

double a;

double sum[10];

int main()

{

    a=1;

    sum[0]=1;

    sum[1]=2;

    for(int i=2; i<=9; i++)

    {

        a=a*(1.0/i);

        sum[i]=a+sum[i-1];

    }

    cout<<"n e"<<endl;

    cout<<"- -----------"<<endl;

    cout<<"0 1"<<endl;

    cout<<"1 2"<<endl;

    cout<<"2 2.5"<<endl;

    for(int i=3;i<=9;i++)

    {

        printf("%d %.9f\n",i,sum[i]);

    }

    return 0;

}

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