[POJ1004]Financial Management 试题描述 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 pr…

题目来源:http://poj.org/problem?id=1004 题目大意: Larry今年毕业并找到了工作.他开始赚很多的钱,然而他似乎总觉得不够.Larry决定好好掌控他的资产,解决他的财务问题.第一步就是找出他的钱到底发生了什么.Larry有一个银行账户,他希望知道他有多少钱.帮助Larry写一个程序,读入他过去12个月里每个月结束时的账户余额,计算出他每月平均账户余额.(实质就是求12个数的平均数.) 输入:由12行组成,每行一个正的浮点数,计到小数点后两位.表示每月结束时的账户余…

Financial Management Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 179458   Accepted: 67604 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 deci…

一.简介 (一)什么是jbpm JBPM,全称是Java Business Process Management(业务流程管理),它是覆盖了业务流程管理.工作流.服务协作等领域的一个开源的.灵活的.易扩展的可执行流程语言框架. (二)为什么用jbpm 业务分析师和开发人员使用的是同一种语言来交谈,大大降低了开发的风险,如果要开发一个项目,速度也更快了,因为开发人员不用再将用户需求转化成软件设计了. 其次,JBPM采用的不是一般的开发工具,而是自己的图形化开发工具,非常方便随时了解和掌握运行的进程…

Financial Management Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 126087   Accepted: 55836 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 deci…

Financial Management Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 164431   Accepted: 61085 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 deci…

Recently,i am learning some useful things about financial management by reading <Essentials of Corporate Finance> writed by Ross/Westerfield/Jordan.Now i will share with you. Chapter One : Introduction to Financial Management Traditionally,finance a…

一. 题目 Financial Management Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 173910   Accepted: 65186 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 ha…

Financial Management 时间限制:3000 ms  |  内存限制:65535 KB 难度:1   描述 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…

Financial Management POJ - 1004 解题思路:水题. #include <iostream> #include <cstdio> #include <iomanip> #include <cmath> using namespace std; int main() { double sum=0.0; double now=0.0; while(cin>>now){ sum+=now; ;i<;i++){ cin&…

Financial Management Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 acm.hust.edu.cn/vjudge/problem/visitOriginUrl.action?id=20184 Description Michael graduated of ITESO this year and finally has a job. He's making a lot of money, but somehow never seem…

Financial Management Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 182193   Accepted: 68783 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 deci…

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 3100 Root of the Problem http://poj.org/problem?id=3100 大意: 给定B和N,求一个数A使得A^N最接近B 太水了..... #include<cstdio> #include<cmath> int main()…

Financial Management Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 165062   Accepted: 61316 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 deci…

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 内存限制: 65536kB 描述 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 decid…

题目描述 Problem 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. Th…

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 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 s…

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 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 f…

Problem 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 fir…

描述 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 fig…

我承认这是一道水的不能再水的题,今天一下就做到了,还是无耻的帖上来吧 #include <stdio.h> int main() { double sum=0; for(int i=1;i<13;i++) { double tmp=0; scanf("%lf",&tmp); sum+=tmp; } printf("$%.2lf\n",sum/12); return 0; }…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1064 解题报告:用来凑个题数吧,看题的时间比过题的时间多的多,就是输入12个浮点数,然后输出平均数,只是前面加个美元符号就行了,另外保留两位小数,快去水吧. main() { ,x,tot = ; ;i <= ;++i) scanf("%lf",&x),tot += x; printf(); }…

原题链接 题目大意:给出12个月的收入,求一个平均值. 解法:没什么好说的,就是一个除法. 参考代码: #include<stdio.h> int main(){ int i; float sum=0,k; for(i=0;i<12;i++){ scanf("%f",&k); sum+=k; } sum=sum/12; printf("$%.2f",sum); return 0; }…

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 i; ; i < ; ++i) { scanf("%lf",&num); sum += num; } printf("$%.2f",sum/12.0); }…

http://acm.hdu.edu.cn/showproblem.php?pid=1064 思路:看懂英文就很简单,就是12个数相加求平均数就ok了. 扩展: C++ 标准输入输出流的控制符 #include<stdio.h> #include<iostream> #include<string> #include<iomanip> using namespace std; int main() { ]; double sum=0.0 ,average=0…

参考:https://www.cnblogs.com/BTMaster/p/3525008.html #include <iostream> #include <cstdio> #include <cstring> using namespace std; int main() { float a,sum; while (cin>>a)//要注意格式,能从文件读入多组数据! { sum=a; ;i<;i++) { cin>>a; sum+=…

这两题比较简单,就不做分析了,描述下题目,就上代码吧. [题目描述] 1003,其实就是求这个方程的最小n:1/2 + 1/3 + 1/4 + ... + 1/(n + 1) >= c: 1004,其实就是算个平均数,直接除12 [附:完整代码] 1003题: /* POJ-1003 Hangover */ #include <iostream> using namespace std; int main() { double c; while(cin>>c) { if (c…

poj1000-1009小结 poj1000-1009小结 poj1000 AB poj1001 Exponentiation poj1002 poj1003 poj1004 Financial Management poj1005 poj1006 poj1007 poj1008 poj1009 蒟蒻一枚,开始从poj做起.废了n久时间且参考有些参考大神博客才写完poj10题.但总算是写完了,小结一下. poj1000 A+B 经典题目,不解释,过. poj1001 Exponentiation…

Lesson: Overview of the JMX Technology (The Java™ Tutorials > Java Management Extensions (JMX)) https://docs.oracle.com/javase/tutorial/jmx/overview/index.html Lesson: Overview of the JMX Technology The Java Management Extensions (JMX) technology is…

POJ 排序的思想就是根据选取范围的题目的totalSubmittedNumber和totalAcceptedNumber计算一个avgAcceptRate. 每一道题都有一个value,value = acceptedNumber / avgAcceptRate + submittedNumber. 这里用到avgAcceptedRate的原因是考虑到通过的数量站的权重可能比提交的数量占更大的权重,所以给acceptedNumber乘上了一个因子. 当然计算value还有别的方法,比如POJ上…