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=Σ_0<=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。