POJ 1004:Financial Management
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 165062 | Accepted: 61316 |
Description
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
Output
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
$1581.42
水题,求十二个数的平均值。
代码:
#include <iostream>
using namespace std;
int main()
{
float sum=0.0,a;int i;
for(i=1;i<=12;i++)
{
cin>>a;
sum+=a;
}
printf("$");
printf("%.2lf\n",sum/12);
return 0;
}
版权声明:本文为博主原创文章,未经博主允许不得转载。
POJ 1004:Financial Management的更多相关文章
- 九度OJ 1148:Financial Management(财务管理) (平均数)
与1141题相同. 时间限制:1 秒 内存限制:32 兆 特殊判题:否 提交:843 解决:502 题目描述: Larry graduated this year and finally has a ...
- 九度OJ 1141:Financial Management (财务管理) (平均数)
时间限制:1 秒 内存限制:32 兆 特殊判题:否 提交:939 解决:489 题目描述: Larry graduated this year and finally has a job. He's ...
- Financial Management POJ - 1004
Financial Management POJ - 1004 解题思路:水题. #include <iostream> #include <cstdio> #include ...
- 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....."--------------------------- ...
- poj 1004:Financial Management(水题,求平均数)
Financial Management Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 126087 Accepted: ...
- Financial Management 分类: POJ 2015-06-11 10:51 12人阅读 评论(0) 收藏
Financial Management Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 164431 Accepted: ...
- [POJ] #1004# Financial Management : 浮点数运算
一. 题目 Financial Management Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 173910 Acc ...
- [POJ] Financial Management
Financial Management Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 182193 Accepted: ...
- 练习英语ing——[POJ1004]Financial Management
[POJ1004]Financial Management 试题描述 Larry graduated this year and finally has a job. He's making a lo ...
随机推荐
- Day3-D-Protecting the Flowers POJ3262
Farmer John went to cut some wood and left N (2 ≤ N ≤ 100,000) cows eating the grass, as usual. When ...
- 最简单的mybatis增删改查样例
最简单的mybatis增删改查样例 Book.java package com.bookstore.app; import java.io.Serializable; public class Boo ...
- leetcode236 Lowest Common Ancestor of a Binary Tree
""" Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in ...
- zookeeper加Kafka集群配置
官方 https://zookeeper.apache.org/doc/r3.5.6/zookeeperStarted.html#sc_Prerequisites https://www.cnblog ...
- 吴裕雄--天生自然JAVA面向对象高级编程学习笔记:对象的多态性
class A{ // 定义类A public void fun1(){ // 定义fun1()方法 System.out.println("A --> public void fun ...
- Java的SPI机制
目录 1. 什么是SPI 2. 为什么要使用SPI 3. 关于策略模式和SPI的几点区别 4. 使用介绍或者说约定 4.1 首先介绍几个名词 4.2 约定 5. 具体的demo实现 5.1 创建服务提 ...
- java虚拟机开篇01
一直以来对java 基础设施都啥都不知道啊,感觉有时候挺费力,挺吃劲的. 一下是一些很好的参考资料: http://blog.csdn.net/bingduanlbd/article/details/ ...
- 1 初识JVM
- 在 .net 中释放嵌入的资源
private static void ExtractResourceToFile(string resourceName, string filename) { if (!Syste ...
- mysql合并结果集