hdu 1178数学問題Heritage from father

Heritage from father
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131070/65535 K (Java/Others) Total Submission(s): 5859    Accepted Submission(s): 2107
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 
Output
入力Nごとに1行出力し,科学的記数法を用いて金貨の総数を計算する(3桁の有効数字を保持する)
 
Sample Input

   
   
   
   

    
    
    
    1
3
0
   
   
   
   

 
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.
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
           i layer  i*（i+1）/2 coins
   
   
   
   
   
   
   
   

    
    
    
            sum=n*(n+1)*(n+2)/6
   
   
   
   
   
   
   
   

    
    
    
        i*（i+1）/2        （i*i）/2     i/2   
   
   
   
   
   
   
   
   

    
    
    
    i*i     n(n+1)(n+2)/6; i     n（n+1）/2;        sum；
   
   
   
   
   
   
   
   

    
    
    
    
    
    
    
    
#include <stdio.h>
#include <iostream>
using namespace std; int main() { double  n; while(scanf("%lf",&n)!=EOF) { double ans=1; if(n==0) break;
        ans=(n*(n+1)*(n+2))/6; int count=0; while(ans>=10) {
            ans/=10;
            count++; }
        printf("%.2lfE%d
",ans,count); } return 0; }