结题报告：POJ_1517

一、原题：

链接：http://poj.org/problem?id=1517

u Calculate e

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 12583		Accepted: 7674		Special Judge

Description
A simple mathematical formula for e is
e=Σ0<=i<=n1/i!
where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of n.

二、Source Code

Source Code

Problem: 1517		User: mervyn_diane
Memory: 208K		Time: 0MS
Language: C++		Result: Accepted

#include<iostream>
#include<iomanip>

using namespace std;

double formula( int );

int main()
{
cout << "n e" << endl;
cout << "- -----------" << endl;

double result = 0.0;
for( int i = 0; i < 10; ++i )
{
result += formula( i );

cout << setprecision(10) << i << " " << result << endl;
}

return 0;
}

double formula( int n )
{
if( n == 0 )
return 1;
else
{
int under = 1;
for( int i = 1; i <= n; ++i )
under *= i;

return 1.0 / under;
}
}

三、总结：

1、依次迭代，就好；

2、格式控制。