Description

Larry graduated this year and finally has a job. He's making a lot of money, but somehow never seems to have enough. Larry has decided that he needs to grab hold of his financial portfolio and solve his financing problems. The first step is to figure out what's been going on with his money. Larry has his bank account statements and wants to see how much money he has. Help Larry by writing a program to take his closing balance from each of the past twelve months and calculate his average account balance.

Input

The input will be twelve lines. Each line will contain the closing balance of his bank account for a particular month. Each number will be positive and displayed to the penny. No dollar sign will be included.

Output

The output will be a single number, the average (mean) of the closing balances for the twelve months. It will be rounded to the nearest penny, preceded immediately by a dollar sign, and followed by the end-of-line. There will be no other spaces or characters in the output.

Sample Input

100.00
489.12
12454.12
1234.10
823.05
109.20
5.27
1542.25
839.18
83.99
1295.01
1.75

Sample Output

$1581.42
求平均数;
#include<iostream>
using namespace std;
int main()
{
int a=0;
double b,s=0;
while(a<12)
{
cin>>b;
s+=b;
a++;
}
double ave=0;
ave=s/12;
cout<<'$'<<ave<<endl; return 0;
}

  

POJ-1004-Finanical Management的更多相关文章

  1. OpenJudge/Poj 1004 Financial Management

    1.链接地址: http://poj.org/problem?id=1004 http://bailian.openjudge.cn/practice/1004 2.题目: 总时间限制: 1000ms ...

  2. poj 1004:Financial Management(水题,求平均数)

    Financial Management Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 126087   Accepted: ...

  3. [POJ] #1004# Financial Management : 浮点数运算

    一. 题目 Financial Management Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 173910   Acc ...

  4. poj 1004 Financial Management

    求平均数,记得之前在杭电oj上做过一个求平均数的题目,结果因为题目是英文的,我就懵逼了 #include <stdio.h> int main() { ; double num; int ...

  5. [POJ 1004] Financial Management C++解题

    参考:https://www.cnblogs.com/BTMaster/p/3525008.html #include <iostream> #include <cstdio> ...

  6. Financial Management POJ - 1004

    Financial Management POJ - 1004 解题思路:水题. #include <iostream> #include <cstdio> #include ...

  7. POJ 3100 Root of the Problem || 1004 Financial Management 洪水!!!

    水两发去建模,晚饭吃跟没吃似的,吃完没感觉啊. ---------------------------分割线"水过....."--------------------------- ...

  8. POJ 1004:Financial Management

    Financial Management Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 165062   Accepted: ...

  9. POJ 1004 解题报告

    1.题目描述: http://poj.org/problem?id=1004 2.解题过程 这个题目咋一看很简单,虽然最终要解出来的确也不难,但是还是稍微有些小把戏在里面,其中最大的把戏就是float ...

  10. poj: 1004

    简单题 #include <iostream> #include <stdio.h> #include <string.h> #include <stack& ...

随机推荐

  1. 相位噪声 dBc/Hz

    相位噪声和抖动是对同一种现象的两种不同的定量方式.在理想情况下,一个频率固定的完美的脉冲信号(以1 MHz为例)的持续时间应该恰好是1微秒,每500ns有一个跳变沿.但不幸的是,这种信号并不存在.如图 ...

  2. php中的单引号、双引号和转义字符

    PHP单引号及双引号均可以修饰字符串类型的数据,如果修饰的字符串中含有变量(例$name):最大的区别是: 双引号会替换变量的值,而单引号会把它当做字符串输出. 例如: <?php        ...

  3. RHEL64 缺少ISO 9660图像 安装程序试图挂载映像#1,在硬盘上无法找到该映像

    用光盘安装Linux,很容易,按照提示一步一步就好.如果没有光驱,只好想办法用硬盘或者U盘安装了. 首先说说怎样用U盘启动Linux的安装程序:1.将ISO镜像文件拷贝到U盘中,并解压到U盘根目录.将 ...

  4. Xenu-web开发死链接检测工具应用

    Xenu 是一款深受业界好评,并被广泛使用的死链接检测工具. 时常检测网站并排除死链接,对网站的SEO 非常重要,因为大量死链接存在会降低用户和搜索引擎对网站的信任,web程序开发人员还可通过其找到死 ...

  5. gplots heatmap.2和ggplot2 geom_tile实现数据聚类和热图plot

    主要步骤 ggplot2 数据处理成矩阵形式,给行名列名 hclust聚类,改变矩阵行列顺序为聚类后的顺序 melt数据,处理成ggplot2能够直接处理的数据结构,并加上列名 ggplot_tile ...

  6. hihocoder Challenge 29 A.序列的值

    我现在就感觉我这人现在真的没有dp的意识 其实真写起来也不难,但是把就是练的少思维跟不上,dp从根本上就是一种状态的提炼和聚集. 按照题解的意思来,表示二进制第i位的值为j(0,1)的组合有多少,然后 ...

  7. JustMock .NET单元测试利器(一)

    1.什么是Mock? Mock一词是指模仿或者效仿,用于创建实例和静态模拟.安排和验证行为.在软件开发中提及"mock",通常理解为模拟对象.模拟对象的概念就是我们想要创建一个可以 ...

  8. css样式--表格

    1.示例源码 <!DOCTYPE html><html><head><meta charset="utf-8"> <title ...

  9. jpgraph 折线图--解决中文乱码的问题(标题和图例)

    在jpgraph根目录中: 如Jpg\jpgraph_ttf.inc.php 中开头添加 define('CHINESE_TTF_FONT','SIMYOU.TTF'); \Jpg\jpgraph_l ...

  10. Heavy Transportation POJ - 1797

    题意 给你n个点,1为起点,n为终点,要求所有1到n所有路径中每条路径上最小值的最最值. 思路 不想打最短路 跑一边最大生成树,再扫一遍1到n的路径,取最小值即可,类似Frogger POJ - 22 ...