• poj 2000 Gold Coins(水题)


    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。

  • 相关阅读:
    python使用urllib2抓取防爬取链接
    Python 标准库 urllib2 的使用细节
    源代码阅读利器:Source Navigator — LinuxTOY
    python程序使用setup打包安装 | the5fire的技术博客
    mctop: 监视 Memcache 流量 — LinuxTOY
    nload: 监视网络流量及带宽占用
    Learn Linux The Hard Way — LinuxTOY
    使用virtualenv创建虚拟python环境 | the5fire的技术博客
    autocompleteredis 基于redis的自动补全 开源中国
    atool home
  • 原文地址:https://www.cnblogs.com/AndyDai/p/4734181.html
Copyright © 2020-2023  润新知