数学—杭电1178 Heritage from father

最新推荐文章于 2024-11-28 07:31:15 发布

原创最新推荐文章于 2024-11-28 07:31:15 发布 · 956 阅读

0 ·

CC 4.0 BY-SA版权

本文探讨了哈利波特故事中关于父亲遗留下来的魔法金币的计算问题，通过科学计数法揭示了不同层数金币总数的计算方法，深入解读了魔界银行Goblin的独特金币排列方式。

http://acm.hdu.edu.cn/showproblem.php?pid=1178

Heritage from father

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131070/65535 K (Java/Others)
Total Submission(s): 4847 Accepted Submission(s): 1727

Problem Description

Famous Harry Potter,who seemd to be a normal and poor boy,is actually a wizard.Everything changed when he had his birthday of ten years old.A huge man called 'Hagrid' found Harry and lead him to a new world full of magic power.
If you've read this story,you probably know that Harry's parents had left him a lot of gold coins.Hagrid lead Harry to Gringotts(the bank hold up by Goblins). And they stepped into the room which stored the fortune from his father.Harry was astonishing ,coz there were piles of gold coins.
The way of packing these coins by Goblins was really special.Only one coin was on the top,and three coins consisted an triangle were on the next lower layer.The third layer has six coins which were also consisted an triangle,and so on.On the ith layer there was an triangle have i coins each edge(totally i*(i+1)/2).The whole heap seemed just like a pyramid.Goblin still knew the total num of the layers,so it's up you to help Harry to figure out the sum of all the coins.

Input

The input will consist of some cases,each case takes a line with only one integer N(0<N<2^31).It ends with a single 0.

Output

对于每个输入的N,输出一行,采用科学记数法来计算金币的总数(保留三位有效数字)

Sample Input

Sample Output

  
   1.00E0
1.00E1


   
    
     Hint
    Hint
   
    
when N=1 ,There is 1 gold coins. 
when N=3 ,There is 1+3+6=10 gold coins.

#include<iostream>
#include<stdio.h>
#include<iomanip>
using namespace std;
int main()
{
    double n,sum;
    int digit;
    while(cin>>n&&n)
    {
        sum=( (n+2)*(n+2)*(n+2)-(3*n*n+10*n+8) )/6;
        /*1^2+2^2+3^2+...+n^2的前n项和为n*(n+1)*(2n+1)/6
        1+2+3+4+....+n的前n项和为n*(n-1)/2
        sum=( (n+2)*(n+2)*(n+2)-(3*n*n+10*n+8) )/6;*/
        digit=0;
        while( sum/10>=1 )//转换为科学计数法计数
        {
            digit++;
            sum/=10;
        }
        printf("%.2lfE%d\n",sum,digit);
    }
}