poj 1517 u Calculate e
| Time Limit: 1000MS | Memory Limit: 10000K | |||
| Total Submissions: 19465 | Accepted: 11362 | Special Judge | ||
Description
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.
Input
Output
Sample Input
no input
Sample Output
n e
- -----------
0 1
1 2
2 2.5
3 2.666666667
4 2.708333333
...
#include<stdio.h>
#include<string.h>
#include<cstdio>
#include<string>
#include<math.h>
#include<algorithm>
#define LL long long
#define PI atan(1.0)*4
#define DD double
#define MAX 2002000
#define mod 100
#define dian 1.000000011
#define INF 0x3f3f3f
using namespace std;
int main()
{
LL n,m,j,i,t,k;
printf("n e\n");
printf("- -----------\n");
DD sum=0;
printf("0 1\n");
for(i=1;i<=9;i++)
{
printf("%d ",i);
DD ans=1;
for(j=1;j<=i;j++)
ans*=j;
sum+=1/ans;
printf("%.10g\n",sum+1);
}
return 0;
}
poj 1517 u Calculate e的更多相关文章
- OpenJudge/Poj 1517 u Calculate e
1.链接地址: http://bailian.openjudge.cn/practice/1517 http://poj.org/problem?id=1517 2.题目: 总时间限制: 1000ms ...
- poj 1517 u Calculate e(精度控制+水题)
一.Description A simple mathematical formula for e is e=Σ0<=i<=n1/i! where n is allowed to go t ...
- 1517 u Calculate e
1. 最前面的格式要记得输入. 2. 计算的时候要从3开始重新计算, 否则会丢失精度. 3. 更快的方式就是打表. #include <iostream> using namespace ...
- 【转】POJ百道水题列表
以下是poj百道水题,新手可以考虑从这里刷起 搜索1002 Fire Net1004 Anagrams by Stack1005 Jugs1008 Gnome Tetravex1091 Knight ...
- 算法之路 level 01 problem set
2992.357000 1000 A+B Problem1214.840000 1002 487-32791070.603000 1004 Financial Management880.192000 ...
- POJ题目细究
acm之pku题目分类 对ACM有兴趣的同学们可以看看 DP: 1011 NTA 简单题 1013 Great Equipment 简单题 102 ...
- POJ 3254. Corn Fields 状态压缩DP (入门级)
Corn Fields Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 9806 Accepted: 5185 Descr ...
- POJ 2752 Seek the Name, Seek the Fame [kmp]
Seek the Name, Seek the Fame Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 17898 Ac ...
- POJ 2478 Farey Sequence
名字是法雷数列其实是欧拉phi函数 Farey Sequence Time Limit: 1000MS Memory Limit: 65536K Total Submi ...
随机推荐
- Codeforces 672
题目链接:http://codeforces.com/contest/672/problem A. Summer Camp(打表) 题意:123456789...一串字符串,问第n个是什么数字. 塞一 ...
- iOS学习笔记: 使用CAShapeLayer创建带有空心区域的遮罩层
CAShapeLayer是用来接受矢量Path,直接使用GPU来进行渲染的特殊图层.看下面效果: 对应代码: let markLayer = CAShapeLayer(); markLayer.fra ...
- 循环中不要放入openSession()
for(Shop s:list) { System.out.println(s.getName()); String sql="select shopId,sum(ele_bank+ele_ ...
- bzoj1053: [HAOI2007]反素数ant
51nod有一道类似的题...我至今仍然不会写暴搜!!! #include<cstdio> #include<cstring> #include<iostream> ...
- 实现窗口逐渐增大(moveTo(),resizeTo(),resizeBy()方法)
moveTo()方法格式:window.moveTo(x,y); 功能:将窗口移动到指定坐标(x,y)处; resizeTo()方法格式:window.resizeTo(x,y); 功能:将当前窗口改 ...
- Material Design 设计--阴影的重要性
<LinearLayout android:layout_width="match_parent" android:layout_height="wrap_cont ...
- Java [Leetcode 198]House Robber
题目描述: You are a professional robber planning to rob houses along a street. Each house has a certain ...
- 转载:浅析Java中的final关键字
谈到final关键字,想必很多人都不陌生,在使用匿名内部类的时候可能会经常用到final关键字.另外,Java中的String类就是一个final类,那么今天我们就来了解final这个关键字的用法.下 ...
- 【转】iOS学习之Autolayout(代码添加约束) -- 不错不错
原文网址:http://www.cnblogs.com/HypeCheng/articles/4192154.html DECEMBER 07, 2013 学习资料 文章 Beginning Auto ...
- 常用的PL/SQL开发原则
(1)广泛使用绑定变量,特别是批量绑定,因为这可以有效的避免sql的硬解析和PL/SQL引擎和SQL引擎的上下文切换!(2)广泛使用UROWID来处理DML语句(UROWID是ROWID扩展,ORAC ...