Heritage from father

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解决一道算法题目，关于计算特殊金字塔形状金币堆的总数。利用数学公式快速得出答案，并使用科学记数法表示结果。

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

原题：

Heritage from father

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131070/65535 K (Java/Others)
Total Submission(s): 5074 Accepted Submission(s): 1793

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.

Source

Gardon-DYGG Contest 1

解题思路：

求Sn=(1*2+2*3+....+n*(n+1))/2;

因为n*(n+1)=n^2+n，又1^2+2^2+3^2+...+n^2=n(n+1)(2n+1)/6 1+2+3+...+n=n(n+1)/2;

所以S=[n(n+1)(2n+1)/6+n(n+1)/2]/2=n(n+1)(n+2)/6;

科学计数法没办法直接按要求输出，得直接求出指数

code 1：

------------------------------------------------------------------------------------------------------------------------------------------------

#include <stdio.h>
int main ()
{
int n,c;
double sum;
while (scanf("%d",&n)!=EOF)
{
if (n==0)break;
sum=1.0/6*n*(n+2)*(n+1);
c=0;
while (sum>=10)
{
c++;
sum/=10;
}
printf ("%.02lfE%d\n",sum,c);
}
return 0;
}

------------------------------------------------------------------------------------------------------------------------------------------------------

code 2：

#include<stdio.h>
#include<math.h>
int main()
{
int m,i;
double n,sum;
while(scanf("%lf",&n)!=-1,n)
{
sum=(n*(n+1)*(2*n+1)/6+n*(n+1)/2)/2;
m=(int)log10(sum);
for(i=1;i<=m;i++)
sum*=0.1;
printf("%.2fE%d\n",sum,m);
}
return 0;
}

-------------------------------------------------------------------------------------------------------------------------------------------

code 3：

这是我自己写的代码，不知为何在hdu oj 上是wa，没通过。求路过的大神指点一下。。。

#include<stdio.h>
#include<math.h>
int main()
{
int m,n;
double a,x;
while(scanf("%d",&n)!=EOF)
{
if(n==0)
break;
x=log10(n*1.0/6)+log10((n+1)*1.0)+log10((n+2)*1.0);
m=(int)x;
a=pow(10.0,x-(int)x);
printf("%.02lfE%d\n",a,m);
}
return 0;
}