一、Description

The king pays his loyal knight in gold coins. On the first day of his service, the knight receives one gold coin. On each of the next two days (the second and third days of service), the knight receives two gold coins. On each
of the next three days (the fourth, fifth, and sixth days of service), the knight receives three gold coins. On each of the next four days (the seventh, eighth, ninth, and tenth days of service), the knight receives four gold coins. This pattern of payments
will continue indefinitely: after receiving N gold coins on each of N consecutive days, the knight will receive N+1 gold coins on each of the next N+1 consecutive days, where N is any positive integer.



Your program will determine the total number of gold coins paid to the knight in any given number of days (starting from Day 1).

Input

The input contains at least one, but no more than 21 lines. Each line of the input file (except the last one) contains data for one test case of the problem, consisting of exactly one integer (in the range 1..10000), representing
the number of days. The end of the input is signaled by a line containing the number 0.

Output

There is exactly one line of output for each test case. This line contains the number of days from the corresponding line of input, followed by one blank space and the total number of gold coins paid to the knight in the given
number of days, starting with Day 1.

二、题解

       水题不多说!!值得祝贺,今天连克三道水题,水果30。

三、Java代码

import java.util.Scanner; 

public class Main {
public static void main(String[] args) {
Scanner cin = new Scanner(System.in);
int n,i,j,sum,total;
while((n=cin.nextInt())!=0){
i=0;
sum=0;
j=0;
total=0;
while(sum<=n){
i++;
sum+=i;
}
j=n-(sum-i);
for(int m=1;m<i;m++){
total+=m*m;
}
total=total+j*i;
System.out.println(n+" "+total);
}
}
}

版权声明:本文为博主原创文章,未经博主允许不得转载。

poj 2000 Gold Coins(水题)的更多相关文章

  1. OpenJudge/Poj 2000 Gold Coins

    1.链接地址: http://bailian.openjudge.cn/practice/2000 http://poj.org/problem?id=2000 2.题目: 总Time Limit: ...

  2. poj 2000 Gold Coins

    题目链接:http://poj.org/problem?id=2000 题目大意:求N天得到多少个金币,第一天得到1个,第二.三天得到2个,第四.五.六天得到3个....以此类推,得到第N天的金币数. ...

  3. POJ 1488 Tex Quotes --- 水题

    POJ 1488 题目大意:给定一篇文章,将它的左引号转成 ``(1的左边),右引号转成 ''(两个 ' ) 解题思路:水题,设置一个bool变量标记是左引号还是右引号即可 /* POJ 1488 T ...

  4. POJ 3641 Oulipo KMP 水题

    http://poj.org/problem?id=3461 直接KMP就好.水题 #include<cstdio> #include<cstring> const int M ...

  5. poj 1006:Biorhythms(水题,经典题,中国剩余定理)

    Biorhythms Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 110991   Accepted: 34541 Des ...

  6. poj 1002:487-3279(水题,提高题 / hash)

    487-3279 Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 236746   Accepted: 41288 Descr ...

  7. poj 1003:Hangover(水题,数学模拟)

    Hangover Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 99450   Accepted: 48213 Descri ...

  8. poj 2105 IP Address(水题)

    一.Description Suppose you are reading byte streams from any device, representing IP addresses. Your ...

  9. POJ 2365【YY水题】

    题目链接:POJ 2365 Rope Rope Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7488   Accepted ...

随机推荐

  1. linux shell脚本: 自动监控网站状态并发送提醒邮件

    1.创建监控脚本:$ vi /alidata/shell/webcheck.sh #!/bin/sh weblist="/alidata/shell/weblist.txt" my ...

  2. 浅谈命令查询职责分离(CQRS)模式---转载

    在常用的三层架构中,通常都是通过数据访问层来修改或者查询数据,一般修改和查询使用的是相同的实体.在一些业务逻辑简单的系统中可能没有什么问题,但是随着系统逻辑变得复杂,用户增多,这种设计就会出现一些性能 ...

  3. HIVE 总结

    http://blog.csdn.net/wisgood/article/details/17186181 常见错误 http://blog.csdn.net/sunnyyoona/article/d ...

  4. spring项目命名

    groupId 一般分为多个段,最简单的分两段,第一段为域,第二段为公司名称.域又分为org.com.cn等等许多, 举个apache公司的tomcat项目例子:这个项目的groupId是org.ap ...

  5. 20165101刘天野 2018-2019-2《网络对抗技术》Exp6 信息搜集与漏洞扫描

    目录 20165101刘天野 2018-2019-2<网络对抗技术>Exp6 信息搜集与漏洞扫描 1.实验内容 1.1 各种搜索技巧的应用 1.2 DNS IP注册信息的查询 1.3 基本 ...

  6. vo优化总结

    问题1:位姿估计用的ransac,只用了几个点,如果3d_2d点存在噪声,不行.优化:把这值当做初值,用非线性优化问题2:深度图有误差,深度过近或过远不行,有误差.而特征点往往在物体边缘处,深度测量值 ...

  7. eclipse从Git获取项目更新

    1.项目上右键 ——> Team ——> pull   如果报错:   解决方法:   依次打开:Window ——> Preferences ——> Team ——> ...

  8. java:Eclipse插件springsource-tool-suite的下载和安装

    1.打开下载页面http://spring.io/tools/sts/all 找到这个,后补全部版本链接http://spring.io/tools/sts/legacy 插件压缩包下载安装: 链接下 ...

  9. 分享知识-快乐自己:Liunx 根目录结构

  10. 再次理解WCF以及其通信(附加一個編程小經驗)

    一.概述 Windows Communication Foundation(WCF)是由微软发展的一组数据通信的应用程序开发接口,可以翻译为Windows通讯接口,它是.NET框架的一部分.由 .NE ...